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Daniele Dini 2014 Imperial College London

What is Stress Analysis? Stress Analysis is the short version of Stress-Strain Analysis, the engineering discipline related to the theory and the methods developed by physicists, engineers and mathematicians to determine the stresses and strains in materials and structures subjected to forces or loads. This is one of the fundamental subjects for civil, mechanical and aeronautical engineers and enables quantitative analysis of the stresses and deformations caused by external loads applied to a body. In this context, stresses are local measures of forces and are defined as forces per unit area, while strains are local measures of deformations and usually defined as linear deformations per unit length. Given an engineering component or a structure, the properties of the materials they are made of, the links between the parts of the structure, and the typical forces applied to them, the laws and methods taught in the Stress Analysis course will enable you to obtain a quantitative description of the stress over all those parts and joints, and the deformation caused by those stresses. There is a strong link between stresses and strains as they cannot exist independently from each other: stress causes strain and vice versa. The main difference between Stress Analysis and Mechanics (see ME1 Mechanics) is that in the Mechanics syllabus deformations are neglected and all parts and structures are considered as rigid bodies. This is an idealisation which enables to study the kinematics and dynamics of objects without having to consider deformations.

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Contents Chapter 1 Static equilibrium . . . . . . . . . . . . . . . . . 1 1.1 Dynamics and statics . . . 1.1.1 Dynamics of a particle

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1.2 Static equilibrium . . . . . . . . 1.2.1 Example: See-saw in equilibrium .

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1.3 Free body diagram . . . . . . . . . 1.3.1 Analysis of free body diagrams . . . . 1.3.2 Example: FBD analysis of bicycle . . . 1.3.3 Example: FBD analysis of block on wedge

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1.4 Notes.

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Chapter 2 Loaded frames . . . . . . . . . . . . . . . . .

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2.1 Plane frame supports.

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2.2 Example: plane frames .

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2.3 Pin-jointed frames. . . . . . 2.3.1 Analysis of a two-bar frame .

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2.4 Statical determinacy . . . . . . . . . . . 2.4.1 Evaluating statical determinacy . . . . . 2.4.2 Example: Statical determinacy of 6-bar frame .

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2.5 Example: Five-bar frame .

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2.6 Example: Generalised, statically determinate frame .

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2.7 Analysing frames by the method of sections

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2.8 Maximum allowable load on a frame . . . . . . . . . 2.8.1 Example: maximum allowable load on a 2-bar frame . . 2.8.2 Example: maximum allowable load on a complex frame .

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2.9 Notes.

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Chapter 3 Stress . . . . . . . . . . . . . . . . . . . .

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3.1 Stress and strength .

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3.2 Stresses in statically determinate systems . 3.2.1 Thin, pressurised cylinders: hoop stress 3.2.2 Thin, pressurised cylinders: axial stress 3.2.3 Thin, pressurised spherical shell: stress 3.2.4 Example: pressure vessel . . . . 3.2.5 Example: pressure vessel bolts . . . 3.2.6 Thin rotating ring . . . . . . .

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3.3 Notes.

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Chapter 4 Elastic strain . . . . . . . . . . . . . . . . . .

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4.1 Direct strain.

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4.2 Elastic stress-strain relationships .

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Contents

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4.3 Poisson’s ratio . . . . . . . 4.3.1 Typical Poisson’s ratio values

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4.4 Stress and strain in three dimensions . . . . . . . . . . 4.4.1 Hydrostatic stress and volumetric strain . . . . . . . 4.4.2 Relating elastic modulus, Poisson’s ratio and bulk modulus . 4.4.3 Example: volume change due to stress and temperature .

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4.5 Strains in cylindrical and spherical shells.

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4.6 Apparent modulus of elasticity.

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4.7 Volumetric strain within shells . . . . . . . . 4.7.1 Volumetric strain within a cylindrical shell . . 4.7.2 Volumetric strain within a spherical shell . . 4.7.3 Stress-strain relationships in polar coordinates .

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4.8 Elastic strain energy .

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4.9 Solving statically indeterminate stress-strain problems . 4.9.1 Example: statically determinate problem . . . 4.9.2 Example: statically indeterminate problem . . . 4.9.3 Example: hydraulic cylinder with tie rods . . . 4.9.4 Example: pressure vessel with hemispherical ends

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Chapter 5 Bending of beams . . . . . . . . . . . . . . . .

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5.1 Beam supports.

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5.2 Modelling of loaded beams.

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5.3 Shear force and bending moment.

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5.4 Example: Shear force and bending moment 5.4.1 Example: SF and BM diagrams (1) . . 5.4.2 Example: SF and BM diagrams (2) . . 5.4.3 Example: SF and BM diagrams (3) . .

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5.5 Moment applied to a beam at a point

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5.6 Relating shear force and bending moment to applied loads 5.6.1 Example: Problem 5.1 . . . . . . . . . .

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5.7 Notes.

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Chapter 6 Bending of beams: stress and strain

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6.1 Bending stress and strain . . . . . . . . . . . . 6.1.1 Stress in beams of symmetrical cross-section . . . . 6.1.1.1 Example: bending stress in a rectangular beam . 6.1.2 Stress in beams of arbitrary cross-section . . . . . 6.1.2.1 Example: stress in beams of arbitrary cross-section

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.72 .74 .75 .76 .77

6.2 Flexural rigidity of a beam . . . . . . . . . . . . 6.2.1 Second moment of area . . . . . . . . . . . 6.2.1.1 Composite areas . . . . . . . . . . . 6.2.1.2 The parallel axis theorem . . . . . . . . 6.2.1.3 Example: Second moment of a non-symmetric area

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Contents

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6.3 Example: Problem 5.3

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6.4 Example: stress in an I-beam

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6.5 Bars under combined axial and bending loads . . 6.5.1 Example: combined axial and bending loads .

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6.6 Notes.

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Chapter 7 Bending of beams: deflections 7.1 Bending deflections . . . . . . 7.1.1 Example: Bending deflections (1) 7.1.2 Example: Bending deflections (2) 7.1.3 Example: Bending deflections (3) 7.1.4 Example: Bending deflections (4)

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7.2 Macaulay’s method . . . . . . . . . 7.2.1 Example: Macaulay’s method (1) . . . 7.2.2 Macaulay’s method for distributed load . 7.2.3 Distributed loads over a restricted length

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7.3 Beam bending solutions by superposition . . . . . . . . . 7.3.1 Example: end deflection of a cantilever . . . . . . . . 7.3.2 Example: end loaded cantilever with uniformly distributed load 7.3.3 Example: end-supported cantilever . . . . . . . . .

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7.4 Statically indeterminate beams

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Chapter 8 Torsion of bars . . . . . . . . . . . . . . . . . 106 8.1 What is torsion?

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8.2 Analysis of torsional stress and strain . 8.2.1 Torsion of a thin-walled cylinder 8.2.2 Solid and hollow circular shafts 8.2.3 Example: Torsion bar design (1) 8.2.4 Example: Torsion bar design (2)

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8.3 Shafts of varying diameter . . . . . . 8.3.1 Example: torsion of a stepped shaft .

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8.4 Bars of non-circular cross-section . . . . 8.4.1 Example: torsion of a non-circular bar

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8.5 Analysis: torsion of thin-walled, non-circular tubes

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Contents

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Chapter 1 Static equilibrium

1 Static equilibrium

Stress analysis begins by dealing with structures and components which are in static equilibrium — neither moving nor accelerating rapidly enough for inertial forces to significantly load them. On successfully completing this section, you will be able to: •

Define statics and dynamics

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Recall Newton’s laws of motion

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Define forces and moments

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Define static equilibrium

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Sketch free body diagrams for very simple systems

1.1 Dynamics and statics Classical Mechanics is historically divided in three branches: 1.

Statics — the science which studies equilibrium of bodies and its relation to forces;

2.

Kinetics — the science which studies motion of bodies and its relation to forces;

Kinematics — the science which deals with the implications of observed motions of bodies without regard for circumstances causing them. These three topics have been linked to dynamics in different ways. Earlier approaches to this subjects combined statics and kinetics under the name, Dynamics, which became the branch dealing with determination of the motion of bodies resulting from the action of specified forces. A more recent approach — more common in engineering books on mechanics and still in widespread use among mechanicians — separated Statics, and combined kinetics and kinematics under Dynamics.

3.

In the ME1 Solid Mechanics syllabus we follow the latter approach; however, while ME1 Mechanics will deal with both Statics and Dynamics but will not treat deformation of bodies under the action of external forces, ME1 Stress Analysis will only deal with Statics and will concentrate on the study of the effect that forces have in terms of deformations, stresses and strains. Links will also be made between stresses and material characteristics to assess the potential failure of structures.

1.1.1 Dynamics of a particle Newton’s laws of motion were originally formulated for a particle, but they apply in a general sense to all bodies under certain conditions.

Chapter 1 Static equilibrium

1

First law

Second law The acceleration of a particle as produced by a net force is directly proportional to the magnitude of the net force in the same direction as the net force, and inversely proportional to the mass of the particle: ¨. F = mx Third law For every action, there is an equal and opposite reaction.

1.2 Static equilibrium In a static problem, there is no translational acceleration and no rotational motion. For a body of mass m to which several forces Fi and moments Mi are applied: ¨=0 ∑i Fi = ∑ mx

and ∑jMj = 0

i.e. both the net force and the net moment in every direction is zero. Such a system is in static equilibrium.

1.2.1 Example: See-saw in equilibrium The frictionless see-saw illustrated below is balanced with no linear or angular acceleration — i.e. in static equilibrium — in the position shown. Find F and N, neglecting the see-saw’s own mass. 1200 N

1m

2m

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Draw a free body diagram.

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Apply the condition that the net vertical force is zero: ∑i Fi = 0 = − F + N − 1200

1.2 Static equilibrium

2

1 Static equilibrium

A particle at rest tends to stay at rest and a particle in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced (net) force.

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Apply the condition that the net moment is zero in the clockwise direction:

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∑i Mi = 0 = − 1 × F + 2 × 1200

Solve for F and N: F = 2400 N, N = 3600 N.

1.3 Free body diagram To analyse any complex engineering system, it is first broken down into the interacting components it consists of, while noting the points at which each is connected to others. The whole system can then be predicted from a knowledge of how each component behaves in isolation. To analyse mechanisms and structures, they are disassembled into separate free bodies. All of the points at which each free body contacted those surrounding it are then replaced by equivalent forces and moments, and the free body is analysed using static equilibrium equations.

1.3.1 Analysis of free body diagrams If a mechanism or structure is in equilibrium, it can first be represented as a free body diagram by removing it from outside influences and replacing them by the forces they exert. 1.

Construct a free-body diagram (FBD) of the system, removing any external components which interact with it — including constraints such as supports or enclosing boundaries.

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Mark onto the diagram all known external forces and moments which act on the free body.

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Mark onto the diagram, as variables, any unknown reaction forces and moments (e.g. from the ground, a wall) which act on it. Remember Newton’s Third Law: When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.

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Apply the static equilibrium condition of forces and moments equations to form: •

For a 2D system, form no more than three equations

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For a 3D system, form no more than six equations.

Solve, as far as possible: •

If there are as many equilibrium equations as unknowns, solve completely

1.3 Free body diagram

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Static equilibrium

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If there are more unknowns than equations, separate one section of the free body as a new free body and return to Step 2.

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1.3.2 Example: FBD analysis of bicycle

Static equilibrium

Analyse the forces acting on the frame of a bicycle of mass MB loaded by a rider of mass MR.

Before attempting to analyse stresses using a static equilibrium free body diagram, we first need to be sure that the bike is in static equilibrium — i.e. it is stationary or being ridden at constant speed, not accelerating or hitting a bump. 1.

Construct a free-body diagram (FBD) of the bicycle:

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Mark the known forces acting on the bike frame: W 1 = M1g, W B = MBg

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Add to the diagram variable reaction forces R1, R2 from the ground

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Apply the static equilibrium condition of forces and moments equations to form: 1.3.2 Free body diagram

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A horizontal force equilibrium equation ∑i FH = 0

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A vertical force equilibrium equation ∑i FV = 0

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A moment equilibrium equation about some chosen point.

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In this case there are more unknowns than equations and we continue by separating one section of the system as a new free body and returning to Step 2. Using Newton’s 3rd Law to separate subcomponents

1.3.3 Example: FBD analysis of block on wedge A block of mass m1 is stationary on a wedge of angle θ and mass m2 (or slides on it at constant speed). The wedge slides at constant or zero speed along a horizontal surface. Determine the forces acting along, and normal to each surface.

It is not always necessary to begin with a free body diagram of the whole system. Because both masses move at constant or zero velocity, there are no accelerations and the system is in static equilibrium. 1.

Begin by drawing each component — block and wedge — as a separate free body:

1.3.3 Free body diagram

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Static equilibrium

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a.

1 Static equilibrium

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Mark the known forces m1g, m2g acting on each component.

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Add to the diagram the variable, equal and opposite pairs of action reaction forces (normal R1 and R2, tangential F1 and F2) between the block and wedge surfaces. Note It doesn’t matter which direction is chosen for any action force arrow: if the action and reaction arrows were both reversed, the resulting value of each force would have the opposite sign.

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Apply the static equilibrium condition of forces and moments equations to form: a.

A horizontal force equilibrium equation for the block: −F2cosθ + R2sinθ = 0

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A vertical force equilibrium equation for the block: −F2sinθ − R2cosθ − m2g = 0

Hence R2 = − m2gcosθ and F2 = − m2gsinθ. c.

A horizontal force equilibrium equation for the wedge: F2cosθ − R2sinθ + F1 = 0

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A vertical force equilibrium equation for the wedge: F2sinθ + R2cosθ − m1g = 0

Hence R1 = m1 + m2 g and F1 = 0. 5.

In this case we have fewer unknowns than possible equations: no moment balance was required.

As a check on the result, a free-body diagram for the entire system could be drawn, as is usually done first:

1.3 Free body diagram

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1 Static equilibrium

θ

Hence for vertical force equilibrium: R1 = m1 + m2 g and for horizontal force equilibrium: F1 = 0 as before.

1.4 Notes

1.4 Notes

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1

Static equilibrium

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1.4 Notes

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Static equilibrium

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1.4 Notes

Chapter 2 Loaded frames Pin jointed frames consist of straight bars loaded in tension or compression, assembled with joints which are (or might become) free to rotate. Examples include bridges, cranes and roof support structures for warehouses etc..

2 Loaded frames

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On successfully completing this module, you will be able to: •

Assess the static equilibrium of plane frames

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Recognise different types of frames

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List possible supports and reaction forces

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Build free body diagrams of simple pin-jointed frames

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Discuss static determinacy of pin-jointed frames

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Determine forces in individual bars of pin-jointed frames

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Calculate stresses in individual bars and assess the likelihood of failure under the prescribed load

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Assess potential failure of the bars by buckling

2.1 Plane frame supports The forces which act on bars throughout a plane frame may depend on the exact nature of each of its supports. A built-in or encastré support can support a vertical force without movement, and a moment without rotation. Chapter 2 Loaded frames

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V

2 V

H

H

Loaded frames

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A pin joint can support a horizontal or vertical force without movement but cannot resist rotation.

V

V

V H

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M or sliding joint can support only a vertical load without moveA roller joint ment.

V

V H

M

V H

2.2 Example: plane frames Find the reaction force at B in the plane frame shown: B 10 kN

20 kNm

6m

3m A

C 10 m

Free body diagram: B 6m

20

10 HA

P in-joint

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A VA

VC

HC

2.2 Example: plane frames

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Static equilibrium. Resolve horizontally: HA + 10 − HC = 0 Resolve vertically: V A + V C = 0

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Take moments clockwise about A: 10 × 3 + 20 − 10V C = 0

Loaded frames

and V C = 5 kN and V A = − 5 kN +5VA = 0

MB=0 by deﬁnition!

10 HA VA

For member AB, take moments clockwise about B: 10 × 3 − 6HA + 5V A = 0 Hence HA = 55/6 kN and HC = 5/6 kN and V C = 5 kN and V A = − 5 kN MB=0 by deﬁnition!

HC

20 VC

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For member BC, take moments clcokwise about B: 20 + 6HC − 5V C = 0 Hence HC = 5/6 kN and HA = 55/6 kN and V C = 5 kN and V A = − 5 kN. For forces at B, use free body diagram: HB

B VB

20 kNm

6m

C HC

5m VC

Resolve horizontally: HB = HC Resolve vertically: V B + V C = 0 Hence HC = HB = 5/6 kN 2.2 Example: plane frames

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and V B = −V C = − 5 kN. To verify previous calculations: For forces at B, use free body diagram:

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VB HB

10 kN HA

Loaded frames

B

6m

A VA

5m

Resolve horizontally: 10 + HA = HB Resolve vertically: V A = V B = − 5 kN. Take moments clockwise about A: 10 × 3 + 5V B − 6HB = 0 Hence HB = 30 − 25 = 5/6 kN.

2.3 Pin-jointed frames A pin-jointed frame consists of straight bars, connected at joints which cannot be relied on to prevent rotation. The joints are therefore assumed to be free to rotate. As well as assuming that joints are free to rotate, we make two other assumptions: •

External forces are applied only at the joints

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The weight of the bars themselves is assumed to be negligible. B

A

The reaction moment at a pin is therefore zero: hence M = Qr = 0 and Q = 0. Another way to consider the bar is to recognise that the bar is a two-forces body and, therefore it can only be in equilibrium if the forces are colinear:

2.3 Pin-jointed frames

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F2 2 Loaded frames

F1

RELATED LINKS Canadian Institute of Steel Construction educational website on steel framed structures

2.3.1 Analysis of a two-bar frame In the frame shown: 0.9m HA

1.6m

VA

HC

A 1.5m

VC C 1.2m

2m B 10kN

What are the tensions in bars AB and BC and the horizontal and vertical reactions at A and C? 1.

Draw an FBD of the joint. TAB

TBC B 10kN

2.

Resolve horizontally. −T ABsin α + T BCsin β = 0

Using the information that sin α = 35 , cos α = 45 , sin β = 45 , cos β = 35 : 2.3.1 Pin-jointed frames

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4 T 5 BC

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= 5 T AB

hence T BC = 34 T AB. 3.

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Resolve horizontally.

4 3 T +4 5 AB

Loaded frames

−T ABcos α + T BCcos β − 10 = 0 3

× 5 T AB = 10

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4 + 4 T AB = 50.

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Hence solve for tensions in bars: T AB = 8 kN and T BC = 6 kN.

5.

Resolve at A to find reactions there. Horizontally: HA+T ABsin α = 0 hence HA = − 4.8 kN. Vertically: V A+T ABcos α = 0 hence V A = 8 × 45 = 6.4 kN.

6.

Resolve at C to find reactions there. Horizontally: HC+T BCsin β = 0 hence HC = 6 ×

4 5

= 4.8 kN.

Vertically: V C−T BCcos β = 0 hence V C = 6 × 35 = 3.6 kN. Important Notice that all tensions and reactions have been found by static equilibrium alone: this system is statically determinate.

Not all systems are statically determinate. RELATED LINKS Animations and additional material on pin-jointed frames

2.4 Statical determinacy A structure which is statically determinate is one in which all forces can be calculated from a knowledge of external, applied forces using equilibrium alone. Not all structures are statically determinate. Forces in the frame shown below, in which A, D and C are in the same horizontal plane, cannot be found by equilibrium alone. Even if the vertical force at B is zero, forces can be generated in the three bars if one is slightly shorter or longer than the diagram suggests.

2.4 Statical determinacy

15

VA HA

VD A

HD

D

VC HC

C

2 Loaded frames

B

Attempting to solve by equilibrium: •

Each joint provides two equations, so there are 2 × 4 = 8 equilibrium equations

•

There are 6 unknown reaction forces and 3 unknown bar tensions, hence 6 + 3 = 9 unknowns

Hence the problem cannot be solved by equilibrium: it is statically indeterminate. To determine forces for a statically indeterminate structure, we need to consider the compatibility of displacements at B: i.e. the need for the three B ends to remain in the same position as the three bars stretch under load.

•

2.4.1 Evaluating statical determinacy A simple function of the number of bars, joints and reactions in a frame can help to assess whether it is statically determinate or not. Consider the problem of determining whether the frame shown below is statically determinate or not. D

A

B

C

1.

Count the number of bars, b.

2.

Count the number of reactions at supports, n.

3.

Count the number of bar-end joints (i.e. pins), j.

4.

Since there are b + n unknown forces and reactions and each joint gives two equilibrium equations (2 j in total), calculate b + n − 2 j to determine statical determinacy.

There are three possible distinct results: •

b + n − 2 j = 0 is a necessary but not sufficient condition for statical de-

terminacy (the placing of bars is important).

2.4.1 Statical determinacy

16

•

b + n − 2 j > 0 indicates that the structure is statically indeterminate.

•

b + n − 2 j < 0 indicates that the structure is a mechanism.

In the system shown b = 3, n = 4 and j = 4. Hence b + n − 2 j = 3 + 4 − 2 × 4 = − 1 and the structure is a mechanism.

2 Loaded frames

2.4.2 Example: Statical determinacy of 6-bar frame This example demonstrates that the statical determinacy test for frames must be used with care. Determine whether the structure shown is statically determinate:

1.

Count the number of bars, b = 6.

2.

Count the number of reactions at supports, n = 8.

3.

Count the number of bar-end joints, j = 7.

4.

b + n − 2 j = 0, which is a necessary condition for statical determinacy.

This is an example of where the rule fails: the structure is not statically determinate. It consists of two regions: 1.

The upper three bars form a statically indeterminate structure

2.

The lower three bars form a mechanism, the point at which it meets the upper three bars being effectively a fixed support. The b + n − 2 j = 0 values for these two regions cancel.

2.5 Example: Five-bar frame Determine the force in each member of the truss. Indicate whether the members are in tension or compression.

2.4.2 Statical determinacy

17

10kN

B

2 Loaded frames

2m

A

1m

D

1m

C

60kN

1.

2.

Determine whether the structure is statically determinate. a.

b=5

b.

n=3

c.

j=4

d.

b + n − 2 j = 8 implies that the structure is statically determinate, as neither a statically indeterminate sub-structure nor a mechanism sub-structure can be identified.

Resolve horizontally for the entire structure. HC = 10 kN

3.

Resolve vertically for the entire structure. V A + V C = 60 kN

4.

Take moments around C 10 × 2 + 2V A − 60 × 1 = 0 hence V A = 20 kN and V C = 60 − V A = 40 kN.

5.

Resolve forces at A

TAB A

45° 20

TAD

Vertically: T ABsin 45° + 20 = 0 hence T AB = −

40 √2

= − 28.3 kN

Horizontally: T AD+T ABcos 45° = 0 hence T AD = 20 kN 2.5 Example: Five-bar frame

18

6.

Resolve forces at C

TCB TCD

C

2

10

Loaded frames

40

Vertically: T CB = − 40 kN Horizontally: hence T CD = − 20 kN 7.

Resolve forces at D

TDB 20

α=63.44° 10 60

D Vertically: T DBsin α − 60 = 0 kN hence T DB =

60 sin α

= 67.1 kN.

2.6 Example: Generalised, statically determinate frame Find all reaction forces and the tension in every bar for the frame shown. Every bar is of length L. B

A HA RA

C

G

D

F W

E

RE

•

Note that all angles are 60°

•

Initially, assume that all bars are in tension, exerting outward forces on joints

1.

Assume static equilibrium and resolve forces and moments for the entire frame:

2.6 Example: Generalised, statically determinate frame

19

Moments clockwise about E: RA × 3L − W × 2L = 0, hence RA = 23 W Horizontal forces: HA = 0 Vertical forces: RA + RE − W = 0, hence RE = 13 W . Begin to move through the structure joint by joint, resolving vertically and horizontally at each:

Loaded frames

2.

2

In this example, start at A TAB 60°

TAG

RA

Resolve vertically: T ABsin 60° + RA = 0 Resolve horizontally: T ABcos 60° + T AG = 0 Hence T AB = − and T AG = 3.

2 R 3 A

4 Wcos 3 3

= −

2 3

60° =

2

× 3W = −

4 W 3 3

2 W 3 3

If possible, choose the sequence of joints so that each one has only two remaining unknowns. In this example A, B, G, C, F, D, E. a.

Sketch an FBD at B: 60o TAB

60o

TBC

TBG

Resolve vertically: T ABcos 30° + T BGcos 30° = 0 Resolve horizontally: T BGcos 60° + T BC − T ABcos 60° = 0 Hence T BG = − T AB =

4 W 3 3

and T BC = 12 T AB − 12 T BG = − b.

4 W 3 3

Sketch an FBD at G:

2.6 Example: Generalised, statically determinate frame

20

TBG TAG

TCG

60o!

60o

TGF

2

W

Resolve horizontally: T CGcos 60° + T GF − T BGcos 60° − T AG = 0 Hence T CG =

2 W 3

− T BG =

2 W 3 3

and T GF = 12 T BG − 12 T CG + T AG =

1 W 3

And then onwards to joins C, F, D and E. Defining k =

W 3 3

the bar tensions and reactions are: -4k

B -4k

-2k

C

-2k

2k

4k G!

A

D 2k

-2k E

F

2k

k

3k W

2W/3

W/3

2.7 Analysing frames by the method of sections The method of sections is particularly useful when the forces in only a few specified bars are required. An appropriately chosen portion of the frame is cut out and analysed using a Free Body Diagram. Find the forces in bars CD and CF for the frame shown. Every bar is of length L. B

A HA RA

C

G

D

F W

E

RE

2.7 Analysing frames by the method of sections

21

Loaded frames

Resolve vertically: T BGsin 60° + T CGsin 60° = W

1.

Assume static equilibrium and resolve forces and moments for the entire frame. Moments clockwise about E: RA × 3L − W × 2L = 0, hence RA = 23 W

2

Horizontal forces: HA = 0

2.

Loaded frames

Vertical forces: RA + RE − W = 0, hence RE = 13 W Cut out from the frame a free body on which the bar forces which you need to evaluate act as external forces: In this example, cut bars CD, CF and (to release the free body) GF. C

B

TCD TCF

RA=2W/3

3.

TGF

G

A

(√3/2)L

F

W

Apply static equilibrium to the extracted free body. Moments clockwise about F: RA × 2L + T CD × T CD =

2 3

W − 2RA = −

3 L − WL 2

= 0, hence

2 W 3 3

Vertical forces: RA − W − T CFsin 60° = 0, hence RA − W

2

T CF = sin 60° = − W 3 3

2.8 Maximum allowable load on a frame Possible failure modes in a pin-jointed frame are the shear failure of a pin, tensile or compressive yield of a bar or buckling of a bar. As the externally applied load is increased, the mode which occurs first in any member will determine the maximum allowable load. For each failure mode, the maximum load on the frame depends on the following factors: Shear failure of a pin Proportional to its cross-section and the shear strength of the material; inversely proportional to the total load on it

2.8 Maximum allowable load on a frame

22

Tensile yield of a bar Proportional to its cross-section and the tensile yield strength of the material; inversely proportional to the tensile load on it

2

Compressive yield of a bar

Loaded frames

Proportional to its cross-section and the compressive yield strength of the material; inversely proportional to the compressive load on it Buckling of a bar Proportional to the second moment of area of its cross-section and the Young’s modulus of its material; inversely proportional to the square of its length. Note The second moment of area is a function of the beam cross-sectional shape, characterising its resistance to bending.

2.8.1 Example: maximum allowable load on a 2-bar frame Both bars in the frame shown are made of the same material and have the same cross-sectional dimensions. The load W may be applied either upwards, as shown, or downwards. 0.9m HA

1.6m VC

VA

HC

A 1.5m

2m

C 1.2m

B W

Determine which member will fail first by: 1.

Tensile yield

2.

Compressive yield

3.

Buckling.

1.

Use static equilibrium analysis to determine the tension (here compression, i.e. negative tension) in each bar. T AB = − 0.8W T BC = − 0.6W

2.8.1 Maximum allowable load on a frame

23

2.

Calculate −TL2 for each beam. 2 −T ABL2 AB = − 0.8W × 1.5 = 1.8W 2 −T BCL2 BC = − 0.6W × 2.0 = 2.4W

2

1.

While W acts downwards, bar AB is under the greatest tensile load and will be the first to fail by tensile yield.

2.

While W acts upwards, bar AB is under the greatest compressive load and will be the first to fail by compressive yield.

3.

While W acts upwards, bar BC has the greatest value of −TL2 and will be the first to fail by buckling.

2.8.2 Example: maximum allowable load on a complex frame In this example the method of sections is used, and symmetry is exploited, to determine the bar tensions. In the symmetrical, pin-jointed structure shown below, all bars have the same cross-section. Angle ABH is 60°, angle BCH is 45° and all vertical bars are of length L. W B

C

D

A

E H

HA VA

G

F

L

VE

Determine which bar carries the greatest tensile load and which is most likely to buckle when 1.

W acts downwards

2.

W acts upwards.

1.

Assume static equilibrium of the entire structure to determine reaction forces. Take moments clockwise about A: W L + 3L − 2V E L + 3L = 0, hence V E = 12 W . Vertical equilibrium: V A + V E − W = 0, hence V A = 12 W . Horizontal equilibrium: HA = 0.

2.8.2 Maximum allowable load on a frame

24

Loaded frames

Using these results, as the magnitude of W increases:

2.

Make successive cuts across the structure to separate larger and larger sections around the left-hand end, solving for reactions at each newly added pin. a.

Cut AB, AH

2 B

A

Loaded frames

TAB

60o H

TAH

W/2

Take moments ↻ B: 12 W × 3L − T AH × L = 0 hence T AH =

1 2

3W .

Resolve vertically: 12 W + T ABcos 60° = 0 hence T AB = − W. b.

Cut BC, BH, AH B

TBC TBH

A

H

TAH

W/2

Resolve vertically: T BH = 12 W Resolve horizontally: T AH + T BC = 0 hence T BC = − c.

1 2

3W .

Cut BC, HC, HG B

TBC THC

A H THG W/2

Take moments ↻ C: 12 W 3L + L − T HG × L = 0 hence 1

T HG = 2 1 + 3 W

Resolve vertically: 12 W + T HCcos 45° = 0 hence T HC = − d.

1 W. 2

Cut AB, AH and invoke symmetry

2.8 Maximum allowable load on a frame

25

TCG THG G THG

2

Resolve vertically: T CG = 0.

2.

W acts downwards: Bar

Tension/W

Length/L

Tension × (length)2 / WL2

AB

–1

2

–4.0

AH

√3/2 = 0.87

√3

2.61

BH

1/2 = 0.5

1

0.5

BC

–√3/2 = –0.87

1

–0.87

HC

–1/√2 = 0.71

√2

–1.41

HG

(1 = √3)/2 = 1.37

1

1.37

•

Maximum tension T = 1.37W in HG (and FG)

•

Maximum value of − TL 2 = –4.0 in AB (and DE)

2

WL

W acts upwards: all bar tension signs are reversed Bar

Tension/W

Length/L

Tension × (length)2 / WL2

AB

1

2

4.0

AH

–√3/2 = 0.87

√3

–2.61

BH

–1/2 = 0.5

1

–0.5

BC

√3/2 = –0.87

1

0.87

HC

1/√2 = 0.71

√2

1.41

HG

–(1 = √3)/2 = 1.37

1

–1.37

•

Maximum tension W in AB (and DE)

•

Maximum value of − TL 2 = –2.61 in AH (and EF)

2

WL

2.9 Notes

2.9 Notes

26

Loaded frames

1.

2

Loaded frames

27

2.9 Notes

2

Loaded frames

28

2.9 Notes

Chapter 3 Stress Stress is the mechanical force per unit area of the material through which the force is transmitted. On successfully completing this module, you will be able to:

3

Define stress and strain

•

Discuss material strength and links to failure

•

Define statically determinate and statically indeterminate systems

•

Analyse statically determinate systems, e.g. pressure vessels and blocks

•

Build free body diagrams and sketch stresses acting over surfaces in simple systems

Stress

•

3.1 Stress and strength Under increasing stress any solid will first extend (strain) elastically and then, in most cases, will begin to yield (deform permanently). Ultimately the material will tear or fracture. The strength of a material is the maximum stress it can withstand before it fails, whether by yield, tearing or fracture. The applied load at which a component fails depends on the distribution of stress which transmits the load through it. The higher the load, the greater is the stress at every point. The objective of stress analysis is to determine the load at which the component will fail because its material has at some point reached a critical stress. A normal stress or direct stress acts on a surface when a force is applied normal to it:

Mean normal stress σn =

Fn An

in units of Nm–2 (MN–2 etc.).

Another name for a stress or pressure of one Nm–2 is a pascal, symbol Pa (kPa, MPa etc.). Stress is defined as positive when the normal load is tensile:

and tensile when the load is compressive:

Chapter 3 Stress

29

A shear stress acts on a surface area when a force is applied parallel to it: As

3

Fs

Fs As

Stress

The mean shear stress σs =

has the same units as direct stress.

Strength σmax for some common materials: Material

Strength (MPa)

Iron, steel

300

Aluminium

200

Cu-Be alloy

1000

Nylon

60

RELATED LINKS Missouri University of Science and Technology MecMovies: see Ch 1, Stress

3.2 Stresses in statically determinate systems In a statically determinate system under applied load, the load can be found at every point by considering equilibrium alone. If the component geometry at each point is known, stresses there can be determined too.

3.2.1 Thin, pressurised cylinders: hoop stress Many industrially important pressurised structures. e.g. pipelines, can be represented as thin-walled cylinders. This analysis determines the circumferential (‘hoop’) stresses acting within the wall. The cylinder shown below, whose thickness t is much smaller than its mean diameter 2Rm, is subjected to internal pressure P compared to the pressure outside it. We need to determine the hoop stress or circumferential stress which acts on any r, z plane (i.e. a plane of constant θ) within the wall.

3.2 Stresses in statically determinate systems

30

3 Stress

We will assume the wall thickness t to be small enough compared to the mean radius Rm for the assumption that Rm ≈ Ri, where Ri is the inner radius, is valid. 1.

The cylinder is in static equilibrium. Remove a semi-circular element of the wall, Δz in length, and draw the free-body diagram showing the forces acting on it. The lower half of the circular section is removed, and replaced by stresses

σθ acting on the upper half which acted across the plane joining them.

2.

To resolve forces F vertically upwards, first determine the upward force due to pressure acting on the internal surface. Radial force on an element of wall which subtends an angle δα: PRiδθΔz Vertical force on this element: PRiδθΔzsin θ. Hence total vertical force due to pressure FP = ∫π0 PRiΔzsin θ ⅆθ = PRiΔz − cos θ π0 = 2PRiΔz.

Tip This result can be derived more simply by noting that the total upward force FP due to pressure is simply the product of the pressure and the area, projected in the vertical direction, on which it acts.

3.2 Stresses in statically determinate systems

31

3.

Next, determine the downward force due to circumferential stress acting on the cut section of wall. Fσ = 2σθtΔz.

4.

Apply vertical equilibrium to show that the total vertical force is zero, and solve for σθ. Since FP = Fσ,2PRiΔz = 2σθtΔz PRi t

Stress

hence σθ =

.

3.2.2 Thin, pressurised cylinders: axial stress The forces due to pressure acting on the closed ends of a pressurised cylinder tend to push them off. The axial stress in the cylinder wall depends on whether or not the ends are attached to the wall, which then has to carry these forces. We will first analyse the case of a cylinder with both ends closed (i.e. both ends attached to the cylinder wall). The entire cylinder is in equilibrium. Assume that the wall thickness t is small enough compared to the mean radius Rm to assume that Rm ≈ Ri, where Ri is the inner radius. 1.

The pressurised shell is in static equilibrium. Cut across its diameter at some z = const., remove the right-hand part and draw a free-body diagram of the rest showing forces equivalent to it.

2.

To resolve forces F in the horizontall direction, determine the leftward force due to pressure. FP = πR2 iP

3.

Determine the rightward force due to normal stress σz acting on the cutthrough wall. Fσ = 2πRmtσz.

4.

Apply horizontal equilibrium to show that the total axial force is zero, and solve for σθ. Since FP = Fσ, πR2i P = 2πRmtσz hence σz =

2 P Ri 2t Rm

3

PRi

≈ 2t .

3.2.2 Stresses in statically determinate systems

32

In an open-ended cylinder, the axial pressure force FP is not transferred to the cylinder wall: one or two internal pistons transfer the force to some external structure. The axial stress in the cylinder wall is therefore zero.

3 Stress

Note A typical example of an open-ended cylinder is a pneumatic or hydraulic actuator, from which the pressure force is applied to an external load.

3.2.3 Thin, pressurised spherical shell: stress Thin-walled shells are used for pressure vessels wherever possible. Analysing stresses in the wall will help to show us why. We will assume the wall thickness t to be small enough compared to the mean diameter Rm for the assumption that Rm ≈ Ri, where Ri is the inner radius, to be valid. 1.

The pressurised shell is in static equilibrium. Cut across a horizontal diameter, remove the lower half and draw a free-body diagram of the upper half showing forces equivalent to the part removed.

t

r

A

y

r x

Pressure, P

2.

z

To resolve forces F in the vertical direction, determine the upward force due to pressure. FP = πR2 iP

3.

Determine the downward force due to normal stress σθ acting on the cutthrough wall circumference of width t. Fσ = 2πRmtσθ.

4.

Apply horizontal equilibrium to show that the total vertical force is zero, and hence solve for σθ. Since FP = Fσ,πR2i P = 2πRmtσθ hence σθ =

2 P Ri 2t Rm

PRi

≈ 2t . 3.2.3 Stresses in statically determinate systems

33

Since every diameter in a spherical shell is equivalent, the circumferential stress anywhere in a spherical shell is PRi /2t. The circumferential stress in a cylindrical shell of the same radius and thickness is twice this value. RELATED LINKS Missouri University of Science and Technology MecMovies: see Ch 14, Pressure vessels

3

The 300 mm diameter, 6 mm thick pressure vessel shown is to be made up of two hemispherical shells bolted together at flanges. The maximum allowable tensile stress in the material to be used is 150 MPa, what is the maximum allowable working pressure?

1.

Rearrange the equation for stress in a spherical shell as a function of pressure to give pressure as a function of stress. 2t

σθ = PRi /2t , hence Pmax = R σθmax i

Hence Pmax =

2 × 0.006 0.15

× 150 × 106 Pa = 12 MPa.

3.2.5 Example: pressure vessel bolts A 300 mm diameter, 6 mm thick pressure vessel shown is to be made up of two hemispherical shells bolted together at flanges. Twenty bolts of 6 mm diameter are used to hold the flanges together. What is

3.2.4 Stresses in statically determinate systems

34

Stress

3.2.4 Example: pressure vessel

3 Stress

3.2.6 Thin rotating ring If a cylindrical ring rotates about its axis at high speed every element experiences an outward centrifugal force, just as it would do from internal pressure. Like internal pressure, this develops a circumferential stress, which can be calculated by analogy. This analysis derives an equation for the circumferential stress in a ring of mean diameter Rm, thickness t ≪ Rm and material density ρ, at a rotational speed ω.

1.

Draw the free body diagram for a circumferential element of included angle δα, and determine the radial centrifugal force on it.

3.2.6 Stresses in statically determinate systems

35

Fr = mRmω2 = ρRmδαtΔz Rmω2

2.

In order to use the results of analysis for stress in an internally pressurised thin cylinder, convert this to an ‘equivalent internal pressure’, Pω. For the internal-pressure case, Fr = Pω RiδαΔz

3

so that

Pω =

3.

2 ρR2 mω t

Ri

Stress

ρRmδαtΔz Rmω2 = Pω RiδαΔz and hence

.

Use the result for an internally-pressurised cylinder to derive the circumferential stress due to rotation. σθ =

PωRi t

2 = ρR2 mω .

3.3 Notes

3.3 Notes

36

3 Stress

3.3 Notes

37

3 Stress

3.3 Notes

38

Chapter 4 Elastic strain When a solid is put under stress, its shape changes slightly as internal bonds stretch. The resulting small displacement of particles within a body — without movement of the body as a whole — are defined as strain; if it disappears when the stress is removed, the strain is elastic. On successfully completing this section, you will be able to: Recall Hooke’s law for uniaxial tension and its generalised form for threedimensional systems

•

Determine stress and strains in thin-walled pressure vessels and solid blocks

•

Define volumetric strain

•

Determine relationships between elastic constants

•

Define and calculate strain energy for individual bodies loaded in tension and shear

•

Identify and analyse statically indeterminate systems

4.1 Direct strain Direct strain is the extension of a line along a specified direction within the material. A bar of initial length L and cross-sectional area A is subject to an axial force F which causes an increase in length of ΔL.

The direct stress at any cross-section is σ =

F . A

The direct strain along the axis (in this case the longitudinal, or axial strain) is ΔL e= L . Sign convention for strain: an increase in length L is positive, giving positive strain. Positive strain is therefore associated with positive (tensile) stress. RELATED LINKS Missouri University of Science and Technology MecMovies: see Ch 1, Stress

Chapter 4 Elastic strain

39

4 Elastic strain

•

4.2 Elastic stress-strain relationships The relationship between stress and strain is a property of the material to which either stress or strain is applied. To measure this relationship, apply an increasing tensile force along the axis of a bar and measure the increase in length of the bar along the same axis. Plot as stress versus strain to obtain a uniaxial stress-strain plot:

4 Elastic strain

•

If e returns to zero as soon as the load is reduced again to zero (e.g. as for a rubber), the behaviour of the material is described as elastic.

If the relationship between σ and e is linear up to some point, the behaviour of the material there is described as linearly elastic. Many materials are linearly elastic at small strains. For these linearly elastic materials Hooke’s Law holds: σ e = E, •

where the constant E is the elastic modulus or Young’s modulus. RELATED LINKS Animation: linear elastic stress and strain Missouri University of Science and Technology MecMovies: see Ch 3, Mechanical Properties

4.3 Poisson’s ratio If a bar is loaded uniaxially in tension, its axial extension is accompanied by contraction of its cross-section. The strains in the two directions of its cross-sectional plane are negative, and proportional to the axial strain. To demonstrate Poisson’s ratio, load a bar of uniform cross-section in tension σ x. If the material is linearly elastic, it will extend axially by a strain ex = σ x /E. Important For most materials, every cross-sectional plane along the axis of a uniaxially extended bar will contract by an equal strain in every direction. These in-plane strains will be a constant proportion −ν of the axial strain.

For example:

4.2 Elastic stress-strain relationships

40

•

If the cross-section is rectangular, of width Ly and height Lz, these dimensions will change by a strain −νex: they will contract to Ly 1 − νex and Lz 1 − νex

•

If the cross-section is circular, of radius R, the radius will change by a strain −νex: it will contract to R 1 − νex .

The ratio ν is known as Poisson’s ratio, and is a constant of the material.

4

If Poisson’s ratio is 0.5, when the bar is extended its volume will not change (axial extension is balanced by lateral contraction)

•

If Poisson’s ratio is positive and less than 0.5, when the bar is extended in one direction its volume will increase

•

If Poisson’s ratio is negative, when the bar is extended in one direction its volume will increase. Such auxetic materials are very rare.

4.3.1 Typical Poisson’s ratio values Material

Poisson’s ratio

Rubber

0.5

Lead

0.48 to <0.5

Aluminium

0.44

Polystyrene

0.35

Brass

0.33

Ice

0.33

Polystyrene foam

0.30

Steel

0.29

Beryllium

0.08

4.4 Stress and strain in three dimensions A stress in one direction creates strains in both of the other dimensions. If the material is linearly elastic and stresses are applied in these directions too, the strains simply add. To analyse the interaction of stresses and strains in all three dimensions, consider a rectangular block with uniform stresses σ x, σ y, σz applied to opposite faces along all three directions.

4.3.1 Poisson’s ratio

41

Elastic strain

•

σz

4 Elastic strain

1.

Tabulate the strains due to each stress in turn acting alone. Direction x

2.

y

z

Strain

ex

ey

ez

Strain due to σ x

σ x /E

−νσ x /E

−νσ x /E

Strain due to σ y

−νσ y /E

σ y /E

−νσ y /E

Strain due to σz

−νσz /E

−νσz /E

σz /E

Add the strain contributions from each stress to give the total strain in each direction. σx

σy

σz

1

σy

σz

σx

1

σz

σx

σy

1

ex = E −ν E −ν E = E σ x − ν σ y + σz ey = E −ν E −ν E = E σ y − ν σz + σ x ez = E −ν E −ν E = E σz − ν σ x + σ y

3.

If the block experiences a uniform temperature increase ΔT , there are equal strains αΔT in all three directions where α is the coefficient of linear thermal expansion. Add these to these thermal strains to those due to stress. 1

+ αΔT

1

+ αΔT

1

+ αΔT

ex = E σ x − ν σ y + σz ey = E σ y − ν σz + σ x ez = E σ z − ν σ x + σ y

The resulting equations are referred to as the Generalised Hooke’s Law.

4.4 Stress and strain in three dimensions

42

4.4.1 Hydrostatic stress and volumetric strain A material is under hydrostatic stress when an equal normal stress acts on it in every direction. In this case we can define the hydrostatic stress σH as σ H = σ = σ y = σ z. x

ΔV

ev = V .

The bulk modulus, K is the ratio of volumetric strain to volumetric stress: σH

K= e V

and is a material property.

4.4.2 Relating elastic modulus, Poisson’s ratio and bulk modulus The generalised Hooke’s law equations can be used to derive the relationship between these three material constants. Stresses σ x, σ y and σz are applied to an elemental block of size Δx, Δy and Δz in the x, y and z directions. 1.

Calculate the volumetric strain of the block in terms of strains ex, ey and ez : Δx increases by Δx × ex to a new length of Δx′ = Δx 1 + ex Δy increases by Δy × ey to a new length of Δy′ = Δy 1 + ey Δz increases by Δz × ez to a new length of Δz′ = Δz 1 + ez

Hence ΔV= Δx′Δy′Δz′ − ΔxΔyΔz = Δx 1 + ex Δy 1 + ey Δz 1 + ez = ΔxΔyΔz 1 + ex + ey + ez + exey + eyez + ezey − 1

Since strains are small, products of strains can be neglected so that ΔV = ΔxΔyΔz ex + ey + ez

and hence eV =

ΔV V

= e x + e y + ez .

4.4.1 Stress and strain in three dimensions

43

4 Elastic strain

The strain will be equal in every direction too, and will cause a change in volume. The volumetric strain, ev is defined as the change in volume V divided by the original volume:

2.

Use generalised Hooke’s law expressions to express strains in terms of applied stresses. 1

+ E σ y − ν σz + σ x

1

ν

eV= E σ x − ν σ y + σz

1

1

+ E σz − ν σ x + σ y

= E σ x + σ y + σz − E 2σ x + 2σ y + 2σz = σ x + σ y + σz

3.

1 − 2ν E

Set all three stresses equal to the hydrostatic stress σH to derive the bulk eV

Hence eV = 3σH

1 − 2ν E

Elastic strain

modulus K =

and

E

K = 3 1 − 2ν

or E = 3K 1 − 2ν .

4.4.3 Example: volume change due to stress and temperature Find the increase in volume of a 200 mm long and 100 mm wide rectangular steel plate, of 50 mm thickness, whose temperature increases by 20°C while a compressive stress of 120 MPa is applied to both faces and a tensile stress of 60 MPa is applied to the smallest edges. Take the tensile modulus E to be 200 GPa, Poisson’s ratio ν to be 0.3 and α = 10 × 10−6 K–1. 120 MPa

200 mm

100 mm 60 MPa

60 MPa

50 mm 120 MPa

What stress must be applied to the previously unstressed edges if there is to be no change in the 100 mm dimension due to the combined effect of all stresses and the increase in temperature? 1.

4

σH

Define coordinate axes. In-plane x (length) and y (width); thickness z

2.

Apply generalised Hooke’s law equations to calculate linear strains

3.

Add linear strains to find the volumetric strain, and multiply by the initial volume to find the change in volume

4.

Repeat steps 2 accounting for an additional, unknown stress σ y

5.

Set ey to zero to determine σ y.

4.4.3 Stress and strain in three dimensions

44

4.5 Strains in cylindrical and spherical shells Consider a thin-walled cylinder of radius R and thickness t ≪ R:

4 Elastic strain

If the cylinder expands under internal pressure, the axial strain is simply the increase in length divided by the original length: ΔL

ez = L .

Similarly the circumferential strain is the increase in circumference divided by the original circumference: eθ =

2π R + u − 2πR 2πR

u

=R

where u is the increase in radius caused by extension of the circumference.

Remember u/R is the circumferential strain, not the radial strain! There is no tensile stress to extend

any material in the radial direction.

For a spherical shell, the circumferential strain for any circumference is also u/R. The real radial strain in any shell is the increase in radial length, divided by the original radial length, of an element of material in the cylindrical shell. The average radial strain is the increase in wall thickness divided by the original wall thickness.

4.6 Apparent modulus of elasticity The extension of a uniaxially loaded bar under axial force will indicate the true modulus of elasticity only if the bar is free to contract in the lateral, unloaded directions. If it is not, a higher elastic modulus will be measured. If a stress is applied in the x direction and the lateral surfaces are free, there will be Poisson contraction in the y and z directions. If they are prevented

4.5 Strains in cylindrical and spherical shells

45

from contracting, there must be tensile stresses σ y and σz of the right magnitude to cancel out the contraction, and these stresses will cause a contraction which reduces the strain in the x direction.

σz

4 Elastic strain

From the generalised Hooke’s law equations with no temperature change: 1

ex = E σ x − ν σ y + σz

+ αΔT

The apparent modulus of elasticity is σx

E

Eapparent = e = 1 − ν σ y /σ x + σz /σ x x

and σ y and σz can be calculated by assuming that ey and ez are zero.

4.7 Volumetric strain within shells When the wall of a cylindrical or spherical shell stretches, the shell increase in size and the internal volume increases. Analysing this effect allows us to solve problems concerning, for example, the additional volume of liquid needed to pressurise a filled shell.

4.7.1 Volumetric strain within a cylindrical shell Derivation of the volumetric strain for the internal volume of a cylindrical shell when its length and circumference increase. A cylindrical shell of radius R and length L. 1.

The volume of the unstrained cylinder is V = πR2L

2.

Now let there be small changes in R and L, and determine the resulting change in volume. Using partial differentiation (δV =

∂V ∂V δR + ∂L δL): ∂R

δV = π ⋅ 2RLδR + πR2δL.

4.7 Volumetric strain within shells

46

3.

Divide by the original volume to express this change in volume as a volumetric strain. δV V

=

π ⋅ 2RLδR + πR2δL πR2L

δR

δL

= 2 R + L = 2eθ + ez

Hence ev = 2eθ + ez.

4.7.2 Volumetric strain within a spherical shell

4 Elastic strain

Derivation of the volumetric strain for the internal volume of a spherical shell when its circumference increases. A spherical shell has radius R. 1.

The volume of the unstrained sphere is V = 43 πR3L

2.

Let there be a small changes in R, and determine the resulting change in volume. Since δV =

ⅆV δR: ⅆR

4

δV = 3 π ⋅ 3R2δR.

3.

Divide by the original volume to express this change in volume as a volumetric strain. δV V

δR

= 3 R = 3eθ

Hence ev = 3eθ.

4.7.3 Stress-strain relationships in polar coordinates For any point in a cylindrical solid, the generalised Hooke’s law equations become: 1

+ αΔT

1

+ αΔT

1

+ αΔT

eθ = E σθ − ν σr + σz er = E σr − ν σz + σθ ez = E σ z − ν σ θ + σ r

For any point in a spherical solid, the generalised Hooke’s law equations become: 1

+ αΔT

1

+ αΔT

1

+ αΔT

eθ = E σ θ − ν σ r + σ ϕ er = E σ r − ν σ ϕ + σ θ eϕ = E σϕ − ν σθ + σr

4.7.2 Volumetric strain within shells

47

4.8 Elastic strain energy If a material extends elastically when stress is applied to it, potential energy has been stored in it. When the stress is reduced to zero, the energy is immediately recovered. An axial force is applied to a bar of linearly elastic material, L long and A in cross-sectional area, the point at which the force is applied will move due to direct strain.

4

Load

Elastic strain

Extension

The external work done during a small increment of extension δx is Fδx and the total area under the curve for an extension of ΔL is 21 FΔL. The stored strain energy per unit volume is

1 W ⅆL 2

AL

.

Since strain e = δL/L, stress σ = F/ A and tensile modulus E = σ/e, the elastic strain energy per unit volume stored by applying a direct stress is 21 σ2 /E. If a shear force is applied to a bar of linearly elastic material, L long and A in cross-sectional area, the force will be displaced due to shear strain.

Following the same argument, the elastic strain energy per unit volume stored by applying a direct stress is 12 τγ = 12 τ2 /G.

4.8 Elastic strain energy

48

4.9 Solving statically indeterminate stress-strain problems A statically indeterminate problem cannot be solved by stress or load equilibrium alone. However, if we know the stress/strain relationship for the material (e.g. linearly elastic) it can be solved by considering strain compatibility. A general method of solution is shown in the flow chart below:

4

System FBD, reactions, tensions, etc Apply static equilibrium

Elastic strain

More unknowns than equations

Statically indeterminate

Statically determinate

Equations in terms of strains

Geometry of deformation Stress-strain relationships Equations in terms of stresses

Solve all equations

1.

Define the system, identifying all possible external loads and reactions acting on it.

2.

Draw a free body diagram and apply static equilibrium conditions to form up to three equilibrium equations.

3.

Apply the method of sections to successive subcomponents (e.g. bars) of the system until there are as many equations as unknowns.

4.

If there are enough equations, the system is statically determinate: solve them all.

5.

If there are more unknowns than equations, analyse the geometry of deformation to derive compatibility equations in terms of strain, deflection or extension.

6.

Use stress-strain (or load-deflection) relatonships to convert these compatibility equations into terms of stress (or load), until there are as many equations as unknowns.

7.

Solve all equations.

4.9.1 Example: statically determinate problem

4.9 Solving statically indeterminate stress-strain problems

49

Bar 1, of length L1 and cross-sectional area A1, is made of a material with modulus E1. It is connected in series with bar 2, of length L2 and cross-sectional area A2, made of a material with modulus E2. One end of the combined bar is fixed and a force F is applied at the other. What are the axial displacements where the bars meet, and at the load point? 1.

Take the left-hand end as being fixed.

4

Draw a free body diagram and apply static equilibrium conditions to form up to three equilibrium equations.

3.

Apply the method of sections to successive subcomponents (e.g. bars) of the system until there are as many equations as unknowns. For horizontal equilibrium of each of bars 1 and 2, F1 = F2 = F.

4.

If there are enough equations, the system is statically determinate: solve them all. This problem is now already solved. Displacements can be calculated, but they did not have to be calculated in order to solve for forces and stresses.

For each bar, the extension d is: σ1 e1 σ2 e2

F1

d1

F1L1

F2

d2

F2L2

= E1, so that A = E1 × L and d1 = A E 1 1 1 1 = E2, so that A = E2 × L and d2 = A E . 2 2 2 2

From geometry, the displacement of the load point is FL1

FL2

d = d1 + d2 = A E + A E . 1 1 2 2

If A1 = A2 = A: d = d1 + d2 =

F L1 A E1

L2

+E . 2

4.9.2 Example: statically indeterminate problem A cylinder, made from a material with modulus E1, has a length L and crosssectional area A1. It is is mounted between rigid end-caps. A round bar of 4.9.2 Solving statically indeterminate stress-strain problems

50

Elastic strain

2.

equal length, made from a material with modulus E2 and of cross-sectional area A2, is mounted coaxially between the same end-caps. One end-cap is fixed and a force F is applied at the other: what is the resulting displacement at the load point? 1.

Take the left-hand end as being fixed.

2.

Draw a free body diagram and apply static equilibrium conditions to form up to three equilibrium equations.

4 Elastic strain

3.

Apply the method of sections to successive subcomponents (e.g. bars) of the system until there are as many equations as unknowns. Here, for either end-cap F = F1 + F2

4.

If there are enough equations, the system is statically determinate: solve them all. Here, no: one equation, two unknowns.

5.

If there are more unknowns than equations, analyse the geometry of deformation to derive compatibility equations in terms of strain, deflection or extension. From the geometry of this problem, for the cylinder and bar to remain attached to the rigid end-caps their extensions δ1 and δ2 must be equal so that δ1 = δ2 = δ.

6.

Use stress-strain (or load-deflection) relationships to convert these compatibility equations into terms of stress (or load), until there are as many equations as unknowns. Hence F1 = E1 A1 × δL and F2 = E2 A2 × δL and F = F1 + F2 = E1 A1 ×

7.

δ L

δ

δ

+ E2 A2 × L = L E1 A1 + E2 A2 .

Solve all equations.

Thus the load-point displacement is given by δ = F1 = F2 =

E1 A 1 E1 A1 + E2 A2 E2 A 2 E1 A1 + E2 A2

F E1 A1 + E2 A2

L, from which

F and F.

4.9.3 Example: hydraulic cylinder with tie rods 4.9.3 Solving statically indeterminate stress-strain problems

51

A hydraulic cylinder of 80 mm inside diameter and 4 mm wall thickness is welded to rigid end plates as shown below. The end plates are tied together by four rods of 8 mm diameter symmetrically arranged around the cylinder. 4 tie rods

4

Hydraulic cylinder

Elastic strain

Calculate the stresses in the rods and the cylinder at the cylinder design pressure of 20 MPa. The cylinder and tie rods are made from steel with a Poisson’s ratio of 0.3. 1.

Define the system, identifying all possible external loads and reactions acting on it. There are no significant external loads on the system. Unless the piston (not shown) is accelerating rapidly, the system is in static equilibrium. Inside radius of cylinder R = 12 dcyl =

1 2

× 80 = 40 mm

Thickness of cylinder t = 4 mm Diameter of each rod drod = 8 mm. 2.

Draw a free body diagram and apply static equilibrium conditions to form up to three equilibrium equations. Free body diagram of either end cap: end plate tie rod

Hydraulic cylinder

With stresses in MPa: Axial load in each rod Frod =

πd2 rod 4

σrod = 50.3σrod N

Axial load in cylinder wall 1

Fcyl = 2π R + 2 t tσz, cyl = 2π 40 + 2 ×4σz, cyl = 1055.6σz, cyl N

4.9 Solving statically indeterminate stress-strain problems

52

Axial force due to pressure = PπR2 = 100.5 × 103 N. Hence resolving horizontally: 4Frod + Fcyl = PπR2 Substituting: 4 × 50.3σrod + 1056σz, cyl = 100.5 × 103 N hence σrod + 5.25σz, cyl = 500 MPa. 3.

Apply the method of sections to successive subcomponents (e.g. bars) of the system until there are as many equations as unknowns.

4.

If there are enough equations, the system is statically determinate: solve them all. In this example there are two unknowns (σrod, σz, cyl) and one equation: not enough.

5.

If there are more unknowns than equations, analyse the geometry of deformation to derive compatibility equations in terms of strain, deflection or extension. For the cylinder of thickness t under internal pressure 20 MPa, geometry of deformation gives erod = ez, cyl.

6.

Use stress-strain (or load-deflection) relatonships to convert these compatibility equations into terms of stress (or load), until there are as many equations as unknowns. Stress-strain relationships give σrod

erod = E

1

ez, cyl = E σz, cyl − ν σθ, cyl + er, cyl PR

eθ, cyl = t =

20 × 40 4

= 200 MPa

σr, cyl ≈ − P ≈ − 20 MPa, which is small compared to σθ, cyl and can be

neglected. Hence 1

ez, cyl = E σz, cyl − 200ν MPa.

7.

Solve all equations. From deformation we have erod = ez, cyl from which σrod E

= σz, cyl − 0.3 × 200 MPa

σrod − σz, cyl = − 60 MPa …………… (Eq. 2) 4.9 Solving statically indeterminate stress-strain problems

53

Elastic strain

In this example there are effectively no subcomponents.

4

From equilibrium we have σrod + 5.25σz, cyl = 500 …………… (Eq. 1)

Solving Eqs. (1) and (2) simultaneously: 6.25σz, cyl = 560 MPa and finally σz, cyl = 89.6 MPa

4

σrod = 29.6 MPa.

The pressure vessel on an air-blast circuit breaker has a thin-walled cylindrical section of mean radius R and thickness t. The vessel has hemispherical ends and the radial expansion of the cylinder and hemispheres at their junction is to be the same when it is subjected to pressure P:

1.

Calculate the required wall thickness of the hemispheres, assuming ν = 1/3.

2.

If the length of the cylindrical section is 4R, what is the change in volume of the vessel in terms of R, t and tensile modulus E?

[0.4t, 1.

32πPR4 ] 3tE

Assume static equilibrium… RELATED LINKS

What's an air-blast circuit breaker? Air-blast circuit breaker in action Typical compressed air tank designs

4.10 Notes

4.9.4 Solving statically indeterminate stress-strain problems

54

Elastic strain

4.9.4 Example: pressure vessel with hemispherical ends

4

Elastic strain

55

4.10 Notes

4

Elastic strain

56

4.10 Notes

Chapter 5 Bending of beams A beam is a structural element which is slender relative to its length, but supports loads perpendicular to its longitudinal axis by resisting bending. In structural engineering, at least, beams are usually mounted horizontally and are usually loaded by bearing the weight of other elements. A beam is generally supported at points other than those at which it is loaded. Its function is: 1.

To limit deflection when the load is applied or moved (e.g. as a vehicle crosses a bridge)

2.

To prevent any risk of the load becoming unsupported because excessive force breaks the beam (e.g. an aircraft fuselage becoming unsupported when the wing root breaks).

5

On successfully completing this module, you will be able to: •

Define a beam

•

Recognise different types of beams

•

Recognise and list possible supports and reaction forces

•

Build free body diagrams of simple beams

•

List the possible loads and recognise their engineering symbols

•

Determine shear forces and bending moments in simple beams

•

Sketch diagrams for shear forces and bending moments

•

Derive mathematical expressions to link applied Loads, shear forces and bending moments.

5.1 Beam supports Because a beam is not usually designed to be loaded along its axis, only two types of support are important: built-in, and roller. A built-in or encastré support can support a vertical force without deflection and a moments without rotation.

Chapter 5 Bending of beams

57

Bending of beams

The objective of our analysis is to predict the deflections and strength of a beam given its length, cross-section and support conditions. The first step is to summarise the applied loading as a distribution of bending moment and of shear stress acting at every point along the beam length.

A cantilevered beam can bend freely except at its support (the wall), where its angle to the horizontal and its vertical position are fixed.

A simply supported beam can bend freely at and between its supports, but it can deflect only between supports. Note We did not consider whether or not each support could support loads along the axis, because these are beams (intended to resist bending) rather than bars (intended to resist axial force). A structural element can be both a beam and a bar — but we consider these two functions separately.

Because it is not important to a beam whether a support can resist axial load or not, a simple support is drawn simply as an abstract, triangular ‘point support’.

5.2 Modelling of loaded beams To model a loaded beam using a free-body diagram, each support is replaced by the normal (vertical) reaction and moment reaction it applies, loads applied over a short distance are represented as point forces and pressure-like distributed loads as functions of position. The first stage for analysing a loaded beam is to reduce it to a standardised diagram. Note the sign convention always used for beam loading: load acting downwards is positive. Example 1: A horizontal beam simply supported at its ends, and loaded by a single mass placed on it:

5.2 Modelling of loaded beams

58

Bending of beams

A simple support is equivalent to a pin joint or roller support. It can support a vertical force without deflection but cannot resist rotation.

5

Example 2: A horizontal built-in at one end (a cantilever), loaded by two masses:

5

w / unit length

5.3 Shear force and bending moment A loaded beam can be considered as a sequence of connected, infinitesimally narrow slices. In transmitting vertical forces horizontally along the beam, each slice is loaded in shear. For example, consider the cantilever below:

The beam can be considered as a horizontal row of connected slices:

Each slice is subjected to a downward shear force of F on its left-hand face and, in reaction, to an upward shear force of F on its right-hand face. Now section the beam at a point B, distance x from the left-hand end:

5.3 Shear force and bending moment

59

Bending of beams

Example 3: A horizontal beam built-in at both ends ends, and loaded by a continuous distribution of weight:

C A

B

and draw regions to the left and right of B as free body diagrams: C A

5

B

Bending of beams

For the free body AB: Resolve vertically: F x = P Resolve horizontally: H x = 0 Moments about B: M x = Px. We see that each slice of a beam in static equilibrium must be loaded through its surfaces by its neighbours in two ways: 1.

A shear force created by the equal and opposite vertical loads from the beam sections to its left and right

2.

A bending moment created by the equal and opposite moments acting on the beam sections to its left and right Remember

A consistent sign convention is used to represent distributed load w, shear force F and bending moment M on a beam element. All are positive as shown below:

w F

F M

M

5.4 Example: Shear force and bending moment A simply supported beam 15 m long is loaded by 12 force units and 10 force units at distances 3 and 6 units, respectively, from the left hand end. Calculate the shear force at each point along the beam. 10

12

B

A

3

3

6

5.4 Example: Shear force and bending moment

60

Remember Sign convention:

w F

F M

1.

M

Treat the entire beam as a free body to calculate the reaction forces.

5

Resolve vertically: RA + R = 12 + 10 B

Bending of beams

Take moments about A: 12 × 3 + 10 × 6 = RB × 12 hence RB = 8 and RA = 12 + 10 − 8 = 14. 2.

Calculate shear force F within each interval of beam length: a.

Within 0 < x < 3: A

=14

hence F + 14 = 0, F = − 14. b.

Within 3 < x < 6: 12 A

=14

hence F + 14 − 12 = 0, F = − 2. c.

Within 6 < x < 12: 12

10

A

=14 F + 14 − 12 − 10 = 0, F = 8.

3.

Draw shear force diagram:

5.4 Example: Shear force and bending moment

61

+8 A

–2

B

–14

4.

Calculate bending moment M within each interval of beam length: a.

Within 0 < x < 3:

5

A

Bending of beams

=14

hence M + 14x = 0, M = − 14x. b.

Within 3 < x < 6: 12 A

=14

hence M + 14x − 12 x − 3 = 0, M = − 36 − 2x. c.

Within 6 < x < 12: 12

10

A

=14 M + 14x − 12 x − 3 − 10 x − 6 = 0, M = − 96 + 8x.

5.

Draw a bending moment diagram: BM A

B –42

–48

5.4.1 Example: SF and BM diagrams (1) A simply-supported uniform beam of length 6 units is subjected to a point load of 6 units at a point 1 m from the left-hand end.

5.4.1 Example: Shear force and bending moment

62

6

5

1

Draw shear force and bending moment diagrams. 1.

Treat the entire beam as a free body to calculate the reaction forces.

5

C

B

A 1

Bending of beams

6

5

Resolve vertically: RA + RB = 6 Take moments about C: 6RA − 5 × 6 = 0 Hence RA = 5 and RB = 1. 2.

Calculate shear force F within each interval of beam length. Within 0 < x < 1: F + 5 = 0 hence F = − 5. Within 1 < x < 6: F + 5 − 6 = 0 hence F = 1.

3.

Draw shear force diagram. SF 1

–5

4.

Calculate bending moment M within each interval of beam length. Within 0 < x < 1: M + 5x = 0 hence M = − 5x. Within 1 < x < 6: M + 5 − 6 x − 1 = 0 hence M = x − 6.

5.

Draw a bending moment diagram.

5.4 Example: Shear force and bending moment

63

BM 1

–5

5.4.2 Example: SF and BM diagrams (2) A simply-supported uniform beam of length 10 units is subjected to point loads of 12 units at 3 units at distances of 1 and 4 units from the left-hand end.

Bending of beams

12

3

1

2

3

Draw shear force and bending moment diagrams. 1.

Treat the entire beam as a free body to calculate the reaction forces. 12

3 B

A 1

D

C 3

2

Resolve vertically: RA + RB = 15 Take moments about C: 6RA − 5 × 12 − 2 × 3 = 0 Hence RA = 11 and RB = 4. 2.

Calculate shear force F within each interval of beam length. Within 0 < x < 1: F + 11 = 0 hence F = − 11. Within 1 < x < 4: F + 11 − 12 = 0 hence F = 1. Within 4 < x < 6: F + 11 − 12 − 3 = 0 hence F = 4

3.

5

Draw shear force diagram.

5.4.2 Example: Shear force and bending moment

64

SF 14

–25

4.

Calculate bending moment M within each interval of beam length. Within 0 < x < 1: M + 11x = 0 hence M = − 11x.

5

Within 1 < x < 6: M + 11x − 12 x − 1 = 0 hence M = x − 12.

Bending of beams

Within 4 < x < 6: M + 11x − 12 x − 1 − 3 x − 4 = 0 hence M = 4x − 24

5.

Draw a bending moment diagram. BM –4 –11

5.4.3 Example: SF and BM diagrams (3) A simply-supported uniform beam of length 10 units is subjected to point loads of 12 units at 3 units at distances of 1 and 4 units from the left-hand end, and a load of 24 units uniformly distributed between them. 3

24 total

12

1

2

3

Draw shear force and bending moment diagrams. 1.

Treat the entire beam as a free body to calculate the reaction forces. 12 B

A 1

3 8 / unit length

3

D

C 2

5.4.3 Example: Shear force and bending moment

65

Resolve vertically: RA + RD = 12 + 3 + 24 Take moments about C: 6RA − 5 × 12 −

7 2

× 24 − 2 × 3 = 0

Hence RA = 25 and RD = 14. 2.

Calculate shear force F within each interval of beam length. Within 0 < x < 1: F + 25 = 0 hence F = − 25. Within 1 < x < 4: F + 25 − 12 − 8 x − 1 = 0 hence F = 8x − 21.

5

Within 4 < x < 6: F + 25 − 12 − 24 − 3 = 0 hence F = 14 Draw shear force diagram.

Bending of beams

3.

SF 14

–25

4.

Calculate bending moment M within each interval of beam length. Within 0 < x < 1: M + 25x = 0 hence M = − 25x. Within 1 < x < 6: M + 25x − 12 x − 1 − 8 x − 1

1 2

x − 1 = 0 hence

M = 4x2 − 21x − 8.

Within 4 < x < 6: M + 25x − 12 x − 1 − 3 × 8 x − hence M = 14x − 84 5.

5 2

−3 x−4 =0

Draw a bending moment diagram. BM 0

–25 –40

5.5 Moment applied to a beam at a point Shear force and bending moment diagrams reduce the beam to a straight line, to which only normal loads can be applied. A moment can only be represented as two equal and opposite, normal, point loads applied close together.

5.5 Moment applied to a beam at a point

66

The simply-supported beam shown below is drawn as having a real height. A moment could be applied to it as shown if — for example — it was bolted to another, vertical beam at this point. A

B

Resolving vertically: RA + RB = 0 Taking moments clockwise about A: M0 − R L = 0 B

Hence RB =

M0 L

and RA = −

M0 L

.

The value of e affects the shear force diagram only within a region of its own size: SF

A

B

The bending moment diagram is: BM

A

B

5.5 Moment applied to a beam at a point

67

5 Bending of beams

When loaded beams are reduced to free-body diagrams in order to model them using shear force and bending moment diagrams, their depth is neglected. The effect of a bending moment can be seen by representing the moment as a force couple F0 separated by an arbitrarily small small distance e, where F0 = eM0:

5.6 Relating shear force and bending moment to applied loads Analysis of a uniform beam under general distributed and/or point loading leads to analytical expressions for the distributions of shear force and bending moment along it. Consider the FBD of a small section AB of a uniform beam, where w is the distributed load per unit length:

5

1.

Bending of beams

A

B

Resolve vertically for equilibrium of section AB: ⅆF

F + ⅆx δx − F = wδx

hence w = 2.

ⅆF . ⅆx

Take moments B: ⅆM

1

M + ⅆx δx − M − Fδx − wδx × 2 δx = 0

so that

ⅆM ⅆx

and as δx

1

− F − 2 wδx = 0 0,

ⅆM

F = ⅆx .

We can immediately see that at local point of maximum or minimum bending moment, the shear force is zero.

5.6.1 Example: Problem 5.1 A beam 6 m long, simply supported at each end, carries a distributed load over the whole span. The loading distribution is represented by the equation w x = ax2 + bx + c

where w is the intensity of load at a distance x along the beam and a, b and c are constants. The intensity of load is zero at each end and has its maximum value of 4 kNm–1 at mid-span. Apply the differential equations of equilibrium relating w, F and M to obtain the maximum values and distributions of shear force and bending moment. [8 kN, 15 kNm] 5.6 Relating shear force and bending moment to applied loads

68

1.

Treat the entire beam as a free body to calculate the reaction forces.

w x = 0 at x = 0 and x = 6: provides 2 equations

5

w x = 4 at x = 3: provides 1 equation.

2.

Apply w =

ⅆF ⅆx

Hence F(x) = 3.

Apply F = beam. M(x) =

ⅆM ⅆx

to determine the shear force distribution along the beam.

∫w(x)dx = ∫ − 49 x2 + 83 x dx = − 274 x3 + 86 x2 + C. to determine the bending moment distribution along the

∫F(x)dx = ∫ − 274 x3 + 43 x2 + C dx = − 271 x4 + 49 x3 + Cx

+D

4.

Find sufficient boundary conditions to complete the solution. Both ends are simply supported so that M = 0 at x = 0 and x = 6.

5.7 Notes

5.7 Notes

69

Bending of beams

Hence w x = − 49 x2 + 83 x.

5

Bending of beams

70

5.7 Notes

5

Bending of beams

71

5.7 Notes

Chapter 6 Bending of beams: stress and strain Once the length, support conditions and applied loads for a beam have been expressed as shear force and bending moment distributions, stresses and strains can be calculated at every point along it. To do so, the cross-sectional shape is expressed as a single, geometrical property: the second moment of area. On successfully completing this module, you will be able to: •

Define bending stresses in beams

•

List the assumptions made to derive the mathematical expressions for stresses and strains in slender beams

•

Define neutral axis and second moment of area

•

Recall mathematical expressions to locate the neutral axis and determine the second moment of area for regular and arbitrary cross-sections

•

Apply the parallel axis theorem to find the second moment of area for a simple composite shape or non-symmetric cross section

•

Determine stresses in beams characterised by regular and arbitrary crosssections

•

Assess the potential failure of beams by relating the maximum stresses to the material strength

•

Find stresses in beams subjected to combined axial load and bending moment.

6

Deflections of a loaded beam are caused mainly by the distribution of bending moment along it. At every point, a straight beam bends to a radius of curvature which, by stress analysis, can be determined from the local bending moment. Consider an elemental length AB along a beam between two planes A and B normal to the axis. Under the applied bending moment M, the beam bends to a sector of a circle whose radius of curvature is to be determined. The beam can be of any cross-section.

Chapter 6 Bending of beams: stress and strain

72

Bending of beams: stress and strain

6.1 Bending stress and strain

Cross-section:

A1

B1

A

B

A0

B0

A2

B2

Side view of beam: A'1 B'1

B'

A' A'0

B'0

A'2

B'2

6

•

There are no axial loads or constraints on the beam.

•

There are no out-of-plane loads or constraints on the beam.

•

When the beam is bent, every plane cross-section remains plane and normal to the axis.

•

The curvature R−1 remains small.

1.

Determine the strain distribution as a function of distance y across the section: The axial strain in AB is eAB =

A′B′ − AB AB

This strain increases with radius, but the average strain must be zero if the beam bears no axial load. Important There must be a neutral axis between the top and bottom of the beam at which the strain is zero.

This axis passes through points A0 and B0, so that AB = A0B0 = A′0B′0 = Rδθ.

Hence for the element shown at a height y from the neutral axis, A′B′ = R + y δθ

and eAB =

2.

R + y δθ − Rδθ Rδθ

y

= R.

Determine the stress distribution as a function of distance y across the section:

6.1 Bending stress and strain

73

Bending of beams: stress and strain

Assumptions:

Since, as usual, we assume there to be no axial or out-of-plane loads on the beam, axial stress is simply given by: E

σAB = EeAB = R y. RELATED LINKS Animation: assumptions for bending analysis

6.1.1 Stress in beams of symmetrical cross-section The axial stress in a bent beam varies linearly from bottom to top. If its cross-section is symmetrical about a horizontal line midway between bottom and top, then that line is the neutral axis and stresses can be calculated directly.

Bending of beams: stress and strain

Consider a beam of rectangular cross-section of depth d and thickness b: Tension

B

A

Compression

Side-view

1.

Stress

Cross-section

Apply the condition that the beam carries no axial force: The axial force is given by +d/2

P=

∫

σ y b ⅆy

−d/2

and we know that the stress is given by σ y = αy where α = E/R. Hence αb

2.

1 2 +d/2 y = 0. 2 −d/2

Calculate the bending moment from the stress distribution: The bending moment at any cross-section is obtained by integrating the product of •

The force σ y bδy on every element AB and

•

The moment arm, y, of that force.

Hence

6.1.1 Bending stress and strain

6

74

+d/2

∫

σbydy = αb

−d/2

1 3 +d/2 αb d3 d3 y − − . = 3 3 8 8 −d/2

This must be equal to the local bending moment M, and since from Step 1 α = E/R: M=

E bd3 . R 12

We have shown that the distribution of stress σ and the local radius of bending curvature R at any point along a rectangular beam where the moment is M are given by: 12M

σ

6

E

= y = R. bd 3

40 mm

What is the maximum tensile stress in a 2 m long horizontal cantilever of negligible weight, subjected to a vertical end load of 9 kN if it has a rectangular cross-section 150 mm deep and 40 mm thick? 9 kN

150 mm

2m

1.

Calculate bending moment distribution. At any cross-section x, M = 9x.

2.

Identify point of maximum tensile stress. The maximum bending moment is therefore at the beam root, x = 2 m, where Mmax = 9 × 2 kNm.

Since

σ y

=

12M bd3

,

the maximum stress will be at ymax = 0.075 m. 3.

Hence calculate maximum stress. σmax =

where

Mmaxymax bd3 /12

bd3 0.04 × 0.153 = = 1.125 × 10−5 m4. Hence: 12 12

6.1.1.1 Stress in beams of symmetrical cross-section

75

Bending of beams: stress and strain

6.1.1.1 Example: bending stress in a rectangular beam

σmax =

Mmaxymax bd3 /12

=

9 × 103 × 2 × 0.075 1.125 × 10−5

= 120 kNm.

6.1.2 Stress in beams of arbitrary cross-section For an arbitrary cross-section: 1.

The beam width is not constant, but a function b y of vertical distance from the neutral axis

2.

The cross-section is no longer symmetrical about any horizontal axis

3.

The neutral axis cannot therefore be identified at the outset.

tension

B

compression

Side view

Cross-section

Stress

Given conditions (1) and (2), our task is to find the location of the neutral axis. 1.

Calculate the force transmitted through the cross-section for an arbitrary choice of neutral axis. The force transmitted through the elemental strip shown is σb y δy, where σ = αy where σ = E/R. Hence the total axial force is +y1

+y1

+y1

−y2

−y2

−y2

∫ σb y ⅆy = ∫ αyb y ⅆy =α ∫ b y y ⅆy.

2.

Determine the neutral axis position by setting the transmitted force to zero. Hence the neutral axis position is given by solving the equation +y1

∫ b y y ⅆy = 0,

−y2

which gives a relationship between y1 and y2. 3.

Calculate the bending moment for the same choice of neutral axis. The moment due to the force calculated in Step 1 is σb y δy × y. Hence the total bending moment is:

6.1.2 Bending stress and strain

76

Bending of beams: stress and strain

A

6

+y1

+y1

−y2

−y2

∫ σb y ydy = α ∫ b y y2dy = α I

where I is the second moment of area of the beam cross-section about the chosen neutral axis.

6.1.2.1 Example: stress in beams of arbitrary cross-section The procedure for analysing a beam of arbitrary cross-section should work equally well, of course, for symmetrical cross-sections in which the neutral axis position is already known. Show that for a rectangular cross section with depth d and constant thickness b the neutral axis lies at the centre of the beam, and determine the second 1.

Bending of beams: stress and strain

moment of area.

Apply Eq. () to the case b y = b: +y1

+y1

−y2

−y2

∫ bydy = b ∫

y2 ydy =b 2

+y1 −y2

y2 y2 1 2 − =b =0 2 2

hence y1 = y2 = 12 d, defining the neutral axis. 2.

Apply Eq. () to the case b y = b and y1 = y2 = 12 d: M=

σ y

+y1

∫

−y2

σ by dy = b y 2

σ

σ y

∫

y2dy =

−d/2

= yb

Hence

d/2

M

d/2 3

3

−

−d/2 3

3

σ y3 d/2 b y 3 −d/2

σ bd

x

3

= y 12

bd3

= I where I = 12 as derived previously.

6.2 Flexural rigidity of a beam The flexural rigidity of a beam (also known as bending stiffness) is a measure of the resistance of a beam to bending — one of its most important properties. It is defined as the product of the elastic modulus of the beam material and the second moment of area of its cross section. When a moment M is applied to a beam whose second moment of area around the neutral axis is I and whose material has a modulus E, it will bend to a radius R where M I

E

= R.

6.1.2.1 Stress in beams of arbitrary cross-section

6

77

The greater the radius of curvature, per unit bending moment, of an initally straight beam the smaller displacements over a given length will be. Thus, EI characterises the bending stiffness or flexural rigidity of the beam.

6.2.1 Second moment of area The second moment of area of a beam, alongside the elastic modulus of its material, determines the resistance of the beam to bending. It depends on the size and shape of the beam cross section. The second moment of area is defined as y2

Ixx =

∫by2 ⅆy,

6

y1

The centroid of an area is identical to the centre of mass of a flat plate of the same shape and of uniform thickness and density. Its position can be determined from the shape of the area alone. In the table below, xx and yy are the orthogonal neutral axes which pass through the centroid of the section. Table 6.2-1 Second moments of area: simple shapes Cross section

Second moments of area

y

x

x

bd3

Ixx = 12

bd3

Iyy = 12

y y

x

x

πa4

Ixx = Iyy = 4

y

6.2.1 Flexural rigidity of a beam

78

Bending of beams: stress and strain

where the line y = 0 passes through the centroid and y1 and y2 are the maximum and minimum extent of the area.

Cross section

Second moments of area

y

Ixx =

x

x

Iyy =

b1 + b2 h3 36 2 b1 + b2 b2 1 + b1b2 + b2 h

36

y

6.2.1.1 Composite areas

6

Two common beam cross-sections which can be treated as composite areas are a round tube, and a hollow rectangular section (a box section). Round tube y

x

x y

For the outer area: Iext = For the inner area: Iint =

πa4 4 πb4 4

For the composite: Itube =

π 4

a4 − b4

6.2.1.1 Second moment of area

79

Bending of beams: stress and strain

Where the shape of a ‘composite’ area can be expressed as the sum or difference of component areas whose second moments of area are known for the same axes, that of the composite area can be derived by adding or subtracting those of the components.

Box section y

x

x

y

ec3 12

For the composite: Ibox =

1 12

bd3 − ec3 .

6.2.1.2 The parallel axis theorem Where the shape of a ‘composite’ area can be expressed as the sum or difference of component areas whose second moments of area are known for axes which are are parallel but not the same, the axis of one can be shifted using the parallel axis theorem. The parallel axis theorem is used to derive the new second moment Iaa of an area after its axis has been shifted from ‘xx’ by a distance a to another, parallel axis ‘aa’:

Area A x a

x a

For the original axis xx passing through the centroid of this area the moment of area was y2

Ixx =

∫b y y2 ⅆy y1

which can be written as Ixx =

∫ y2 ⅆA. A

6.2.1.2 Second moment of area

80

Bending of beams: stress and strain

For the outer area: Iext = For the inner area: Iint =

6

bd3 12

The new second moment of area is given by Iaa =

∫ y + a 2 ⅆA = ∫ y2 ⅆA + 2a∫ y ⅆA + a2∫ ⅆA. A

A

A

A

But because the original axis passed through the centroid, 2a

∫ y ⅆA = 0 A

by definition; hence Iaa = Ixx + a2

∫dA = Ixx + a2A. A

6

The second moment of area for an I-beam (RSJ) cross section can be analysed using the parallel axis theorem by two methods subtraction from the outer rectangle, or addition of the component rectangles. I-beams are designed for high flexural rigidity per unit weight, and are very widely used in construction. The cross-section shown has a width of b, a total height of d + 2t and a uniform thickness of t. Derive an expression for its second moment of area referred to the neutral axis shown.

web N

A flange

1.

Subtraction method: the I-beam cross-section is treated as two rectangular side blocks removed from the outer cross-section block. The second moments of areas of the side blocks are therefore subtracted from that of the outer block.

6.2.1.2.1 The parallel axis theorem

81

Bending of beams: stress and strain

6.2.1.2.1 Example: Parallel axis theorem (1)

1 b − t d3 b d + 2t 3 2 I = −2 12 12 Hence NA 1 = b d + 2t 3 − b − t d3 12

2.

Addition method: the I-beam cross-section is constructed from rectangular flange blocks added to the web block. The second moments of area of the flange blocks are therefore added to that of the web block after the parallel axis theorem has been used to move them away from the NA by a distance ± d + 12 t .

6 Bending of beams: stress and strain

Hence INA =

td3 bt 3 d+t 2 . +2 + bt 12 12 2

Both methods lead to the same result.

6.2.1.3 Example: Second moment of a non-symmetric area A T-beam provides one common example of a cross-section for which the neutral axis position must first be found. Find a general expression for the second moment of area for the cross-section shown below.

1.

Find location of neutral axis. Let the location be a distance y from the base. Then A1+ A2 y = A1y1 + A2y2 td + td y = td ×

d t + td × d + 2 2

6.2.1.3 Second moment of area

82

d

t

2y = 2 + d + 2 x

hence 3d t + . 4 4

y=

2.

Determine second moment of area around the axis now located. Using the parallel axis theorem I=I

a1a′1

x + A1 y − y1 2 +I

a2a′2

x + A2 y − y 2 2

hence

6

6.3 Example: Problem 5.3 A beam ABCD is simply supported at A and C and carries concentrated loads of 40 kN at B and 50 kN at D, together with a load of 60 kN uniformly distributed, w, from B to D. AB = BC = 4 m, CD = 2m. Sketch the bending moment diagram and quote maximum values. [+120 kNm, –60 kNm] ln order to reduce the greatest bending moment, a jack is placed under the beam at D. What upward jacking force, P, will reduce the greatest bending moment in the beam to its smallest possible value? [20 kN]

1.

6.3 Example: Problem 5.3

83

Bending of beams: stress and strain

td3 3d t d 2 dt 3 t 3d t 2 + dt + − + + dt d + − + . I= 12 4 4 2 12 2 4 4

6

Bending of beams: stress and strain

84

6.3 Example: Problem 5.3

6.4 Example: stress in an I-beam

I 120 kN m–1

5m

528.8 mm

If not, determine the length and thickness of flange reinforcement plates needed to make it so. Flange plates are fixed to the outer surfaces of the upper and lower flanges of an I-beam, and of equal width. 1.

Since

σ y

M

= I and all variables are specified except the bending moment

M, determine the maximum allowable bending moment:

Mallow =

2.

ymax 160 × 106 × 404 × 10−6 σmax = = 245 kNm. I 528.8/2 × 10−3

Treat entire beam as FBD to determine reactions: By symmetry, RA = RB. Resolve vertically: 2RA = 5 × 120 hence RA = 300 kN.

3.

Determine bending moment distribution: Take moments around a point x from the left-hand end: 1

M + 300x − 120x × 2 x = 0 hence M = − 300x − 60x2 kNm.

4.

Compare maximum bending moment to maximum allowable: Absolute maximum bending moment is at x = 2.5 m, where Mmax = − 300 × 2.5 + 60 × 2.52 = − 375 kNm.

6.4 Example: stress in an I-beam

85

6 Bending of beams: stress and strain

A beam is to bear a uniformly distributed load of 120 kNm–1 over the 5 m span between its simply supported ends. If the maximum allowable stress for the material is 160 MPa determine whether an I-beam 528.8 mm in height, 200 mm wide and having a second moment of area of 404×10–6 m4 will be strong enough.

5.

If the beam will fail anywhere, determine the length over which the maximum allowable moment must be increased. The region within which the bending moment exceeds 245 kNm is given by M = − 300x − 60x2 = − 245

hence 60x2 − 300x + 245 = 0 x=

6.

300 ± 3002 − 4 × 60 × 245 so that x = 2.5 ± 1.47 m. 120

6

The necessary plate length will be 2 × 1.47 = 2.94 m. Let its thickness be

t.

Using the parallel axis theorem: Mallow = I = I0 + 2

σmaxI where ymax

0.2t 3 D+t 2 + 0.2t 12 2

1

ymax = 2 D + t , and D is the original beam depth (528.8 mm).

Try different values of t… t = 0 mm: Mallow = 245 kNm t = 10 mm: Mallow = 405 kNm

t = 8 mm: Mallow = 375 kNm, OK.

6.5 Bars under combined axial and bending loads For a bar made of linear elastic material the stress, strain and extension due to axial load strains and the bending stress and strain due to bending moments can simply be added together.

6.5 Bars under combined axial and bending loads

86

Bending of beams: stress and strain

Determine the necessary length and thickness of flange reinforcement plates.

1

2

+

Bending

=

3

Axial tension

6 The neutral axis for bending of the cross section is determined only by the cross-sectional shape: it does not move if additional axial load moves the line of zero stress.

There are two practical cases in which combined bending and tension can arise by a single force: 1.

A bar, of finite thickness, end-loaded at a point which does not lie on the centroid:

Axial stresses along the bar are the sum of those due to

2.

•

An axial force θ, plus

•

Bending by a clockwise moment M = Fe applied at the end.

A bar which is end-loaded at an angle θ to its axis:

Axial stresses along the bar are the sum of those due to •

An axial force Fcos θ, plus

•

Bending by a normal end load Fsin θ.

RELATED LINKS Animation: bending moment, stress and strain

6.5.1 Example: combined axial and bending loads

6.5.1 Bars under combined axial and bending loads

87

Bending of beams: stress and strain

Note

Determine the smallest value of end load W which will induce compressive stress anywhere in a round cantilever bar of diameter 60 mm subjected to axial tensile force of 10 kN? 1m 10 KN 60 mm

1.

Determine axial tensile stress due to 10 kN end load: σa =

π 0.03 2

= 3.537 MPa.

6

Determine maximum compressive stress due to bending load W :

Bending of beams: stress and strain

2.

10 × 103

= σ M = y I

Bending moment M at distance x from support M=W 1−x

hence Mmax = W × 1 Nm at support. I=

π 0.06 4 = 6.362 × 10−7 m4. 64

Hence σmax = ±

Mmax W ymax = ± × 0.03 Pa I 6.362 × 10−7

and for this to become equal to σa, 3.537 × 106 × 6.362 × 10−7 W × 0.03 = = 75 N. 0.03 6.362 × 10−7

This result illustrates that for a typical bar geometry, stresses (and strains) due to bending are much higher than those developed by stretching. If you want to break a bar you don’t stretch it: you bend it against a support.

6.6 Notes

6.6 Notes

88

6

Bending of beams: stress and strain

89

6.6 Notes

6

Bending of beams: stress and strain

90

6.6 Notes

Chapter 7 Bending of beams: deflections Beam analysis allows the radius of curvature to be calculated at every point along a beam of known material, geometry and loading conditions. The final step is to integrate this curvature distribution to yield deflections at every point; in statically indeterminate cases, however, these deflections may themselves influence loading. On successfully completing this module, you will be able to: •

Define bending deflections and recall the moment-curvature relationship

•

Apply the moment-curvature relationship to find slopes and deflections at any point along the beam

•

Solve standard beam problems and recall expressions for significant deflection values values for simple beams

7

•

Use Macaulay’s method to solve for deflections in statically determinate beams

•

Apply superposition principles to find deflections in simple beams

•

Use Macaulay’s Method and superposition principles to solve for deflections in statically indeterminate beams.

Bending of beams: deflections

7.1 Bending deflections When a beam is loaded, it deflects — partly due to shear strains and partly due to axial strains set up by the bending moment. Unless the beam is very short and deep, bending deflections greatly exceed shear deflections, which can therefore be ignored. Given a description of… •

A beam geometry (i.e. length and cross-section)

•

The modulus E of its material

•

The positions and types (simple, built-in) of its supports and

The position and distribution of applied loads, • axial stresses at any point x along the beam and at any point y across it can be determined using the bending equation: M x E σ x, y . = = y I R x

The third term in this equation gives the radius of beam curvature due to loading, and can be used to determine the profile of deflection profilev x . The curvature at a point for any function is given by 3/2 ⅆ2v ⅆv 2 1 = / 1 + R ⅆx ⅆx2 Chapter 7 Bending of beams: deflections

91

but deflections and slopes are small, and so give

ⅆv ⅆx

≪ 1 and can be ignored to

1 ⅆ2v = R ⅆx2

and hence the moment-curvature relationship for a beam EI

ⅆ2v = M. ⅆx2 RELATED LINKS

Missouri University of Science and Technology MecMovies: see Ch 10, Beam Deflections

7

7.1.1 Example: Bending deflections (1)

Bending of beams: deflections

End-loaded cantilever

!

1.

A

X

B

Analyse free-body diagram of the entire system. Resolve vertically: V A = P Moments about A: MA = PL.

2.

Consider FBD for beam cut at an arbitrary point X: A

X

F and M are shear force and bending moment, respectively, at X.

Take moments about X (to avoid F appearing) M = M A − V Ax = PL − Px = P L − x .

3.

Apply the moment-curvature relationship: EI

4.

ⅆ2v = M = PL − Px. ⅆx2

Integrate twice to obtain the displacement profile: EI

ⅆv x2 = PLx − P +A ⅆx 2

7.1.1 Bending deflections

92

EIv x = PL

x2 x3 −P + Ax + B 2 6

where A and B are arbitrary constants of integration to be determined by boundary conditions. 5.

Apply boundary conditions to determine constants. a.

At x = 0, deflection v = 0 and therefore B = 0.

b.

At x = 0, slope

ⅆv ⅆx

= 0 and therefore A = 0.

Thus the deflection of the cantilever under this loading is given by v x =

1 x2 x3 . PL −P EI 2 6

7

The end deflection is 2

3

=

PL 3EI

Bending of beams: deflections

1 L L PL −P EI 2 6

3

Note the strong dependence of end deflection on L. The end slope is ⅆv 1 L2 PL2 . = PL2 − P = ⅆx EI 2 2EI

7.1.2 Example: Bending deflections (2) Uniformly-loaded cantilever A

1.

X

B

Analyse free-body diagram of the entire system. Resolve vertically: V A = wL Moments about A: MA = wL × 12 L.

2.

Consider FBD for beam cut at an arbitrary point X: A

/unit length X

F and M are shear force and bending moment, respectively, at X.

Take moments about X (to avoid F appearing) 7.1.2 Bending deflections

93

1 1 1 1 M = MA − V Ax + 2 wx2 = 2 wL2 − wLx + 2 wx2 = 2 w L − x 2.

3.

Apply the moment-curvature relationship: EI

4.

ⅆ2v 1 = M = w L − x 2. 2 2 ⅆx

Integrate twice to obtain the displacement profile: EI

ⅆv 1 = − w L−x 2+A ⅆx 6

EIv x =

1 w L − x 4 + Ax + B 24

where A and B are arbitrary constants of integration to be determined by boundary conditions. 5.

Apply boundary conditions to determine constants. At x = 0, deflection v = 0 and therefore B = −

b.

At x = 0, slope

ⅆv ⅆx

Bending of beams: deflections

a.

1 wL4. 24

1

= 0 and therefore A = 6 wL3.

Thus the deflection of the cantilever under this loading is given by v x =

1 L−x 4 L3 L4 w −w + wx . EI 6 24 24

The end deflection is 1 L3 L4 wL4 wL −w = EI 6 24 8EI

Note the strong dependence of end deflection on L. The end slope is ⅆv wL3 . = ⅆx 6EI

7.1.3 Example: Bending deflections (3) Simply supported beam with uniformly distributed load. A

1.

B

Analyse free-body diagram of the entire system. Resolve vertically (and accounting for symmetry): V A = V B = 12 wL

2.

Consider FBD for beam cut at an arbitrary point X:

7.1.3 Bending deflections

7

94

A

X

F and M are shear force and bending moment, respectively, at X.

Take moments about X (to avoid F appearing) 1

1

1

1

M = − V Ax + 2 wx2 = − 2 wLx + 2 wx2 = 2 wx x − L .

3.

Apply the moment-curvature relationship: EI

4.

ⅆ2v w 2 =M= x − Lx . 2 2 ⅆx

7

Integrate twice to obtain the displacement profile: ⅆv w x3 x2 = +A −L ⅆx 2 3 2

EIv x =

w x4 x3 + Ax + B −L 2 12 6

where A and B are arbitrary constants of integration to be determined by boundary conditions. 5.

Apply boundary conditions to determine constants. a.

At x = 0, deflection v = 0 and therefore B = 0.

b.

At x = L, deflection v = 0 and therefore A =

1 wL3. 24

Thus the deflection of the cantilever under this loading is given by v x =

wx x3 x2 L3 −L + . 6EI 4 2 4

The deflection at the mid-point (x = 12 L) is wL L3 L2 L3 5wL4 . −L + = 12EI 32 8 4 384EI

7.1.4 Example: Bending deflections (4) Simply supported beam with point load:

B

1.

Analyse free-body diagram of the entire system.

7.1.4 Bending deflections

95

Bending of beams: deflections

EI

Moments about A: V B =

aP L

Resolve vertically: V A + V B = P hence V A = 2.

bP . L

Consider FBD for beam cut at an arbitrary point X:

A X F and M are shear force and bending moment, respectively, at X.

a.

0 < x < a:

7

b

M1 = − L Px

Bending of beams: deflections

b.

a < x < L: b

M2 = P x − a − L Px.

3.

Apply the moment-curvature relationship: a.

0 < x < a: EI

b.

ⅆx

2

= −

bP x L

a < x < L: EI

4.

ⅆ 2v 1

ⅆ 2v 2 ⅆx

2

=P x−a −

bP x L

Integrate twice to obtain the displacement profile: a.

0 < x < a: EI

ⅆv1 bP 2 = − x + A1 ⅆx 2L

EIv1 x = −

b.

bP 3 x + A1x + B1 6L

a < x < L: EI

ⅆv2 P bP 2 = x−a 2− x + A2 ⅆx 2 2L

EIv2 x =

P bP 3 x−a 3− x + A 2 x + B2 6 6L

where A1, A2, B1, B2 are arbitrary constants of integration to be determined by boundary conditions. 5.

Apply boundary conditions to determine constants. 7.1 Bending deflections

96

a.

At x = 0, deflection v1 = 0 and therefore B1 = 0.

b.

At x = L, deflection v2 = 0 and therefore 0=

P bP 3 P bP 2 L−a 3− L + A2L + B2 = b3 − L + A 2L + B 2 6 6L 6 6

hence

bP 2 b − L2 + A2L + B2 = 0. 6

c.

From continuity conditions at x = a (i.e. there is no break or ‘kink’ in the beam) at x = a, deflection v1 = v2, and

d.

At x = a, slope

−

ⅆv1 ⅆx

ⅆv2

= ⅆx .

bP 3 P bP 3 a + A1a + 0 = a−a 3− a + A2a + B2 6L 6 6L

7

and −

Bending of beams: deflections

hence A1a = A2a + B2 bP 2 P bP 2 a + A1 = a−a 2− a + A2 2L 2 2L

hence A1 = A2, B2 = 0 bP 2 b − L2 + A2L = 0 6 bP 2 L − b2 A1 = A2 = 6L

Thus the deflection of the cantilever under this loading is given by bP 2 L x − b2x − x3 6L P bP 2 x−a 3+ L x − b2x − x3 . EIv2 x = 6 6L EIv1 x =

7.2 Macaulay’s method Macaulay’s method is a mathematical technique used to integrate the beam moment-curvature relationship through a point at which the load changes discontinuously. It avoids the need to match deflection and slope values at such points. This method uses angle brackets to enclose functions which need to be ‘switched off’ when their value is negative. For example: f x = x−a n=

0

x≤a

x−a n x>a

f(x)

a

x

7.2 Macaulay’s method

97

We can integrate these ‘step functions’ w.r.t. x in the same way:

∫ x−a

n

ⅆx =

∫0 ⅆx ∫ x − a n ⅆx

const. =

x≤a

.

x>a

x≤a

n+1

x−a n+1

+ const.

x>a

7.2.1 Example: Macaulay’s method (1) Point load applied to a simply supported beam: A

Determine the bending moment equation in Macaulay form for a point x which lies beyond the right-most load.

A X b

M = − LP x + P x − a

2.

Apply the moment-curvature relationship: EI

3.

ⅆ2v bP =P x−a − x. 2 L ⅆx

Integrate twice to obtain the displacement profile: EI

ⅆv P bP 2 = x−a 2− x +A ⅆx 2 2L

EIv x =

4.

P bP 3 x−a 3− x + Ax + B 6 6L

Apply boundary conditions to determine constants: a.

At x = 0, deflection v = 0 and therefore B = 0.

b.

At x = L, deflection v = 0 and therefore A= −

P bP 2 P 3 bP 2 L−a 3+ L = − b + L . 6L 6 6L 6

Substitution into the deflection equation reduces this to the result found by direct integration: 7.2.1 Macaulay’s method

98

Bending of beams: deflections

1.

7

B

EIv1 x =

bP 2 L x − b2x − x3 for 0 < x < a 6L

EIv2 x =

P bP 2 x−a 3+ L x − b2x − x3 for a < x < L. 6 6L

7.2.2 Macaulay’s method for distributed load Point load applied to a simply supported beam with distributed load: A

1.

B

Analyse free body diagram for the whole system:

7

Taking moments clockwise about A: a

A

Resolve vertically: RA + RB = w L − a

X

hence R A = w L − a 2.

1 2

a

− 2L .

Determine the bending moment equation in Macaulay form for a point x which lies beyond the right-most load.

A X 1 a 1 M = − w L − a 2 − 2L x + 2 wP x − a 2.

3.

Apply the moment-curvature relationship

4.

Integrate twice to obtain the displacement profile

5.

Apply boundary conditions to determine constants

…as before.

7.2.3 Distributed loads over a restricted length Macaulay's method treats loads at different points along a beam as events which occur as integration proceeds rightwards from the left-hand. A load distributed over only part of the beam is treated by following its appearance at its left-hand limit by the appearance of an equal and opposite distribution at the right-hand limit.

7.2.2 Macaulay’s method

99

Bending of beams: deflections

1

RB = w L − a 2 + 2L

In the case illustrated below, the load is w per unit length between x = a and x = b.

B

A

1.

Analyse free-body diagram of the entire system. Hence RA, RB.

2.

Represent the removal of the distributed load w per unit length from x = b to x = L as the addition of a distributed load −w per unit length.

=

+ B

A

B

A

Consider FBD for beam cut at an arbitrary point X:

A

Hence 1 1 M = − RAx + 2 w x − a 2 − 2 w x − b 2

4.

Apply the moment-curvature relationship: EI

5.

ⅆ2v w w = M = − R Ax + x−a 2− x − b 2. 2 2 2 ⅆx

Integrate twice to obtain the displacement profile: EI

ⅆv x2 w w = − RA + x−a 3− x−b 3+A ⅆx 2 6 6

7.2 Macaulay’s method

100

Bending of beams: deflections

B

A

3.

7

x3 w w EIv x = − R A x−a 4− x − b 4 + Ax + B + 24 24 6

where A and B are arbitrary constants of integration to be determined by boundary conditions. 6.

Apply boundary conditions to determine constants. a.

At x = 0, deflection v = 0

b. At x = L, deflection v = 0 etc..

7.3 Beam bending solutions by superposition

7.3.1 Example: end deflection of a cantilever Determine the end deflection of the cantilever shown.

1.

Determine (e.g. from Handbook of Data and Formulae) end deflection and slope for a cantilever of length L1. End deflection δ1 = End slope θ =

2.

wL3 1 6EI

wL4 1 8EI

.

Determine δ2 by geometry. The beam in L1 < x < L2 is not loaded and therefore has no curvature at any point.

7.3.2 Example: end loaded cantilever with uniformly distributed load Determine the end deflection of the cantilever shown.

7.3 Beam bending solutions by superposition

101

7 Bending of beams: deflections

The bending profile of a loaded beam is determined by solving a differential equation whose terms are proportional to applied loads. As a result, solutions can be superposed.

1.

Determine the end deflection from distributed load alone. wL4

δw = 8EI .

2.

Determine the end deflection from point load alone. PL3

δP = 3EI .

3.

Determine total deflection as sum of those to individual loads. 4

wL

7

3

PL

Bending of beams: deflections

δw + δP = 8EI + 3EI .

7.3.3 Example: end-supported cantilever The free end of the cantilever shown is supported at the same height as its root. Determine the maximum deflection of the beam.

1.

Treat the entire system as a free body to determine reaction forces. Resolve vertically: wL = V A + V B Take moments about A: 12 wL2 − V BL − MA = 0. Attention 2 equations and 3 unknowns! The reactions cannot be found by static equilibrium alone.

This problem is statically indeterminate and can only be solved by considering deflections. 2.

Replace the support at B by a force sufficient to cancel out deflection due to w using known solutions. The solution to the problem shown below is known:.

7.3.3 Beam bending solutions by superposition

102

End deflection δw + δP =

wL4 8EI

PL3

+ 3EI .

Hence for zero end deflection, P = − 38 wL and V B = − P = 38 wL. Equilibrium gave two equations to determine the three unknowns and compatibility has given a third. Hence V A and MA can now be determined.

7.4 Statically indeterminate beams If a loaded beam has more constraints than are needed to support the applied loads, reaction forces cannot be determined by static equilibrium alone. We need to know the flexural rigidity of the beam and to calculate deflections in order to know all the forces needed to restrain them.

Resolve vertically: P = RA + RB. Take moments about A: aP − LRB − MA + MB = 0. To impose the additional condition that the beam slope at both ends is zero, the displacement profile v x must be determined from the bending moment distribution M = − RAx + M A + P x − a . RELATED LINKS Missouri University of Science and Technology MecMovies: see Ch 11, Indeterminate beams

7.5 Notes

7.4 Statically indeterminate beams

103

7 Bending of beams: deflections

The beam shown below needs only simple supports to support the load. Because it is built-in at both ends there are also reaction moments, and these make the problem statically indeterminate.

7

Bending of beams: deflections

104

7.5 Notes

7

Bending of beams: deflections

105

7.5 Notes

Chapter 8 Torsion of bars Torsion is the twisting of an object by an applied moment. It is important in engineering both as an unwanted result of the need to transmit a torque along a shaft, and as the source of useful compliance in a coil spring. On successfully completing this module, you will be able to: Define torsion, shear stress and shear strain

•

Determine stresses and deformations in thin-walled circular and non-circular shafts under torsion

•

Determine stresses and deformations in solid circular and hollow shafts under torsion

•

Define torsional stiffness

•

Analyse circular shafts of varying cross-section subjected to torsional load

•

Recall formulae for the study of stress and deformation in non-circular shafts under torsion.

8.1 What is torsion? Examples of engineering systems for which torsional analysis is needed include gearbox shafts, torsion bar suspensions and coil springs. 1.

A torque is a moment applied about the axis of a shaft, usually to transmit power as in a gearbox. The moment is applied to a gearbox transmission shaft as a force acting on the radius of a gear fixed to the shaft. A reaction force acts on the radius of a second gear fixed to the same shaft. Clutch case

Outer transmission shaft

Inner transmission shaft

FROM ENGINE

TO DIFFERENTIAL

Clutch 1

Clutch 2

Chapter 8 Torsion of bars

8 Torsion of bars

•

106

1

2

3

Equilibrium of the system implies: T 1 − T 2 + T 3 = 0 The torque on the shaft 1-2 is: T 1 (equilibrium of shaft 1) The torque on the shaft 2-3 is: T 3 = T 2 − T 1 (equilibrium of shaft 1). Note

8

In this case torsion of each shaft is unwanted, and must be minimised.

Torsion of bars

2.

Torsion bar suspension for vehicles.

3.

Torsion of the wire or bar which a torsion spring is made of allows the spring to change in length when a force is applied to it.

Note In both cases 2 and 3, torsion is required — and must be predicted by engineering analysis. RELATED LINKS Video on torsion bar suspension Online textbook on torsion bar suspension

8.1 What is torsion?

107

8.2 Analysis of torsional stress and strain Torque generates shear stresses, and torsional ‘twist’ generates shear strains. For a material of given elastic modulus (i.e. stress/strain ratio), the relationship between torque and twist depends on the cross section of the bar. Analysis to obtain the relationship between torque, twist and shear stress for a torsion bar is very much like that to obtain the relationship between bending moment, radius of curvature and direct stress for a beam. Shear strain γ (measured in radians) and for small γ , τ

γ = G,

where G is the shear modulus. It was shown in that E = 2G 1 + ν .

8.2.1 Torsion of a thin-walled cylinder

Consider a cylinder or tube wall whose thickness t is small compared to its mean radius Rm

P

Q

R

S

Assume that: •

Assume the angle of twist is small, so that tan ϕ =

•

The wall is so thin that shear strain γ can be considered independent of radius.

•

There is axial symmetry, so γ does not vary around the circumference.

•

Each cross-section rotates by an angle α without changing shape

1.

Calculate the shear strain for the element PQRS, which was rectangular before torque was applied: P

Q

P'

R

S

R'

Rθ . L

Q'

S'

Shear strain γ = tan ϕ ≈ ϕ 8.2 Analysis of torsional stress and strain

108

8 Torsion of bars

If the wall of a long cylinder or tube is sufficiently thin and its cross section is uniform, stress and strain under torsion can be analysed very simply.

But ϕ =

Roθ L

so that shear strain γ ≈ 2.

Roθ L

.

Calculate the moment (element of torque T ) needed to shear this element of the wall: The shear stress on the element is τ = Gγ , so that the force on the element is τtRm ⅆα. Hence ⅆT = τtRm ⅆα Rm.

3.

Integrate circumferentially to obtain the total torque: T=

∫0

2π

2 τR2 mt ⅆα = τ2πRmt

so that τ =

2πR2 mt

.

8

Use the assumption that t ≪ Rm so that Rm ≈ Ro ≈ Ri ≈ R. Hence

τ R

=

T

Gθ

Torsion of bars

4.

T

Gϕ

= L = R . 2πR t 3

8.2.2 Solid and hollow circular shafts To analyse and design solid and hollow circular shafts subjected to torsion, we need to know the angle of twist for a given torque and the distribution of shear stress and strain through the material. We can analyse the torsion of thick-walled and solid circular bars using the result for a cylinder whose wall thickness is infinitesimal, so that shear stress and strain can be assumed uniform through it.

8.2.2 Analysis of torsional stress and strain

109

1.

Recall the shear stress distribution and torque/twist relationship for an infinitely thin-walled tube of radius r and thickness t:

8

θ

τ = Gr L , and

2.

Integrate from the axis r = 0 to the outside radius Ro to obtain the total torque for a solid shaft consisting of coaxial, thin-walled tubes with thickness ⅆr under equal twist: θ T=G L

Ro

∫2πr3dr = G θL J 0

where Ro

J=

∫

2πr 3dr =

0

3.

π 4 R0 πD4 . [r ]0 = 2 32

For a hollow shaft, change the lower limit of the integral for J from zero to the internal radius Ri: R0

J=

Ro

∫2πr3dr = 2π [r4]R 0

i

=

π D4 − D4 i . 32 o

Result: again we have τ T Gθ = = r J L

where in this case J =

πD4 . Compare this with the bending equation 32

M E σ = = . y I R

8.2 Analysis of torsional stress and strain

110

Torsion of bars

θ

T = 2πr 3t × G L .

Note that in 8.2.1 Torsion of a thin-walled cylinder on page 108, we effectively approximated J by 2πR3mt. The table below compares this approximation to the exact result obtained by integration from an infinitesimal thickness. t/R0

Japprox / J

0.1

0.997

0.2

0.987

0.3

0.97

8.2.3 Example: Torsion bar design (1) An aluminium alloy tube of constant cross-section is to satisfy: •

Length L = 0.5 m

•

Maximum thickness t = 50 MPa

•

Maximum torque T = 1.0 kNm

•

Torsional stiffness at least 30 kNm rad–1.

8 Torsion of bars

If the outer diameter must be D0 = 50 mm, find the largest allowable inner diameter, Di. We know that

τ T Gθ = = , and maximum shear stress at outer diameter r J L

τmax T max = . D0 /2 J

1.

Calculate the J value required to satisfy torque and maximum shear stress limits. J=

2.

T max D0 /2 1000 × 0.025 = = 0.5 × 10−6 m4 6 τmax 50 × 10

Hence determine the necessary inner diameter Di. J=

π −6 , D4o − D4i = 5.09 × 10−6, D4 − D4 i = 0.5 × 10 32 o

4 −6 = 1.16 × 10−6 so that Di = 32.8 mm. D4 i = 0.05 − 5.09 × 10

3.

Check whether this Di gives us the required torsional stiffness (K T) of at least 30 kNm rad–1: τ = r TL of Di J = = Gθ

Re-arrange

T Gθ = to give the torsional stiffness condition in terms J L T L L . = KT G θ G

Hence for Al alloy of shear modulus G = 26.5 GPa, J = 30000 ×

0.5 = 5.66 × 10−7 m4. 9 26.5 × 10 8.2.3 Analysis of torsional stress and strain

111

Hence determine the J =

π −7 D4 − D4 , i = 5.66 × 10 32 0

4 −6 and with D4 0 − Di = 5.76 × 10 4 −6 D4 = 4.9 × 10−7 hence with Do = 50 mm i = 0.05 − 5.76 × 10

Di = 26.5 mm.

Therefore, for maximum T ≥ 1 kNm: Di ≤ 32.5 mm; but for torsional stiffness to exceed 30 kNm: Di ≤ 26.5 mm and we must choose the thicker tube on the second condition.

8.2.4 Example: Torsion bar design (2) An aluminium alloy tube of constant cross-section is to satisfy: 1.

Length L = 0.5 m

2.

Maximum thickness t = 50 MPa

3.

Maximum torque T = 1.0 kNm

4.

Torsional stiffness at least 30 kNm rad–1.

8

We know that

τ T Gθ = = , and maximum shear stress at outer diameter r J L

τmax T max = . D0 /2 J

1.

Apply the torsional stiffness (K T) required (at least 30 kNm rad–1): re-arτ = r TL J= = Gθ

range

2.

T Gθ = to give the torsional stiffness condition J L T L L . = KT G θ G

Hence for Al alloy of shear modulus G = 26.5 GPa, J = 30000 ×

3.

Hence J =

0.5 = 5.66 × 10−7 m4. 9 26.5 × 10

π −7 , D40 − D4i = 5.76 × 10−6 and D4 − D4 i = 5.66 × 10 32 0

with D40 = 6.25 × 10−6 − 5.76 × 10−6 = 1.2 × 10−5 m4 hence D0 = 58.8 mm is required. 4.

Check whether this is strong enough to give the required maximum

T = 1.0 kNm. Maximum shear stress is at outer diameter, hence τmax T max = must satisfy maximum shear stress condition τmax = 50 D0 /2 J

MPa.

8.2.4 Analysis of torsional stress and strain

112

Torsion of bars

If the internal diameter Di = 50 mm, find the smallest allowable outer diameter, Do.

50 × 106 1000 = gives 4 D0 /2 π D0 − D4 i /32 32 × 1000 × D0 4 D4 = 1.0186 × 10−4D0, hence hence 0 − Di = 6 100π × 10

5.

−6 D4 + 1.0186 × 10−4D0. 0 = 6.25 × 10

6.

To solve this for D0, rewrite as D0 = 6.25 × 10−6 + 1.0186 × 10−4D0

1/4

and solve by iteration.

Hence D0 = 59.2 mm. To exceed a torsional stiffness of 30 kNm requires D0 ≥ 58.8 mm, but to ensure a maximum torque of T = 1 kNm requires D0 = 59.2 mm. The thicker tube must be chosen, hence D0 = 59.2 mm.

8

Torsion of a shaft which consists of several sections with different constant diameters can be analysed by treating each section separately, matching torque and rotation at each junction. Consider the case illustrated, which consists of just two sections. B

A

C

We know that for each section, T J

Gθ

= L

T2 θ2

G2 J2

= L = K T2 = torsional stiffness of section 2. 2

Hence TL1

T

TL2

T

θ1 = G J = K and 1 1 T1 θ2 = G J = K 1 2 T2

1.

For section 1, between A and B, the rotation θ1 at B relative to that at A (considered fixed) is

2.

T1 J1

G1θ1

= L . 1

The torsional stiffness of section 1 is defined as

T1 θ1

G1 J1

= L = K T1 1

8.3 Shafts of varying diameter

113

Torsion of bars

8.3 Shafts of varying diameter

3.

For section 2, between B and C, the rotation θ2 at C relative to that at B is T2 J2

4.

G2θ2

= L . 2

The torsional stiffness of section 2 is defined as

T2

G2 J2

= L = K T2 θ2 2

Hence the total rotation of C relative to A is θ = θ1 + θ2 = But

T θ

T K T1

T

+K . T2

≡ K T, the torsional stiffness of the entire shaft.

Hence

1 KT

1

1

=K +K . T1 T2

8.3.1 Example: torsion of a stepped shaft

D

2D L 2

1

T GJ G πD4 = = L L 32 θ

1.

The original torsional stiffness was

2.

Therefore the required stiffness is

3.

The stiffness of section 1 after the collar is fitted will be K T1 =

4.

Stiffness of section 2 after the collar is fitted will be K T2 =

Hence, since

2GπD4 . 32L G π 2D 4 a 32

G π 4 D L − a 32

1 1 1 , = + KT KT KT 1 2

32L 32a 32 L − a = + 4 4 2GπD Gπ 2D GπD4 a L = + L−a 2 16 8L = a + 16 L − a 15a = 8L

and the required collar length isa =

8 L. 15 8.3.1 Shafts of varying diameter

114

8 Torsion of bars

A torsion bar consists of a solid shaft of length L and diameter D. It is required to double the torsional stiffness by shrinking onto one end of this shaft a collar of the same material and of diameter 2D. How long should the collar be? a

8.4 Bars of non-circular cross-section When a bar of non-axisymmetric cross-section is twisted, its cross-sections do not remain flat: they warp. This makes the analysis of torsion more complicated, but its results are no more difficult to use. The additional shear stresses set up by warpage make the section stiffer.

8 Shape of cross-section

τmax

Square

4.81

Eqilateral triangle

20

Ellipse

T a3

Torsion of bars

Table 8.4-1 Analytical results for torsion of square, triangular and elliptical cross-sections θ T a3

7.10

46

2 T π a3

TL a4G

TL a4G

a2 + b2 TL πa3b3 G

8.4.1 Example: torsion of a non-circular bar The 6061-T6 aluminum shaft shown has cross-sectional area in the shape of equilateral triangle.

8.4 Bars of non-circular cross-section

115

Determine the greatest torque T that can be applied to the end of this shaft if the maximum allowable shear stress is τallow = 56 MPa and the maximum allowable twist is θallow = 0.02 rad. Take the shear modulus of 6061-T6 to be GAl = 26 GPa. How much torque can be applied to a shaft of circular cross-section made from same amount of material?

8

By inspection, the torque at any internal cross-section along the shaft axis is also T .

2.

Use τallow =

20T a3

from Table 8.4-1 Analytical results for torsion of square,

triangular and elliptical cross-sections on page 115 to apply the stress limit. This condition requires T = 179.2 Nm. 3.

Use θallow = 46

TL a4GAl

from Table 8.4-1 Analytical results for torsion of

square, triangular and elliptical cross-sections on page 115 to apply the twist limit. This condition requires T = 24.12 Nm 4.

Comparison these results to find the lowest torque limit: Torque is thereore limited by the angle of twist, to T = 24.12 Nm.

5.

For a circular cross-section, first calculate the radius of the equivalent cross-section: For Acircle = Atriangle, R = 14.850 mm.

6.

Use results from 8.2.2 Solid and hollow circular shafts on page 109 to implement the stress limit: From τallow =

7.

TR , this requires T J

= 288.06 Nm.

Similarly, impose the twist limit: Hence from θallow =

TL :T JGAl

= 33.10 Nm.

8.4 Bars of non-circular cross-section

116

Torsion of bars

1.

Hence in this case the torque is limited by angle of twist. Comparing both results, we can see that a shaft of circular cross-section can support 37% more torque than a triangular one. RELATED LINKS Missouri University of Science and Technology MecMovies: see Ch 6, Torsion

8.5 Analysis: torsion of thin-walled, non-circular tubes Thin-walled tubes of noncircular shape are used to construct lightweight frameworks such as those in aircraft. Thin-walled tubes of noncircular shape are used to construct lightweight frameworks such as those used in aircraft. This section will analyse such shafts with a closed cross-section. As walls are thin, we assume stress is uniformly distributed across the thickness of the tube.

8 Torsion of bars

Average shear stress: τavg =

T , where 2tAm

τavg is the average shear stress acting over thickness of tube (changes with

location if t changes)

T is the resultant internal torque at cross-section t is the thickness of the tube where τavg is to be determined

Am is the mean area enclosed within the boundary of the centreline of the

tube thickness

8.5 Analysis: torsion of thin-walled, non-circular tubes

117

The angle of twist can be determined using energy methods. θ=

TL 4A2 mG

ⅆs

∮ t

For constant thickness,this is the perimeter of Am.

8 Torsion of bars

8.6 Notes

8.6 Notes

118

8

Torsion of bars

119

8.6 Notes

8

Torsion of bars

120

8.6 Notes

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