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Pure Bending

4-3

Pure Bending: Prismatic members subjected to equal and opposite couples acting in the same longitudinal plane

Other Loading Types

4-4

• Principle of Superposition: The normal stress due to pure bending may be combined with the normal stress due to axial loading and shear stress due to shear loading to find the complete state of stress.

• Transverse Loading: Concentrated or distributed transverse load produces internal forces equivalent to a shear force and a couple

• Eccentric Loading: Axial loading which does not pass through section centroid produces internal forces equivalent to an axial force and a couple

Mz

My

Fx

³ yV x dA

³ V x dA 0 ³ zV x dA 0 M 4-5

• These requirements may be applied to the sums of the components and moments of the statically indeterminate elementary internal forces.

• The moment is the same about any axis perpendicular to the plane of the couple and zero about any axis contained in the plane.

• The sum of the components of the forces in any direction is zero.

• From statics, a couple M consists of two equal and opposite forces.

• Internal forces in any cross section are equivalent to a couple. The moment of the couple is the section bending moment.

Symmetric Member in Pure Bending

4-6

• stresses and strains are negative (compressive) above the neutral plane and positive (tension) below it

• a neutral surface must exist that is parallel to the upper and lower surfaces and for which the length does not change

• length of top decreases and length of bottom increases

• cross-sectional plane passes through arc center and remains planar

• bends uniformly to form a circular arc

• member remains symmetric

Beam with a plane of symmetry in pure bending:

Bending Deformations

Strain Due to Bending

Hx

Hm

Hx

Lc

UT

yT

or

y Hm c

U

c

L

G

ȡ

c

U

y

Hm

yT (strain varies linearly)

U y T G L Lc U y T UT

After deformation, the length of the neutral surface remains L. At other sections,

Consider a beam segment of length L.

MECHANICS OF MATERIALS

4-7

y EH m c

y V m (stress varies linearly) c

EH x

³ V x dA

V m ³ y dA c

0

First moment with respect to neutral plane is zero. Therefore, the neutral surface must pass through the section centroid.

0

Fx

y ³ c V m dA

• For static equilibrium,

Vx

• For a linearly elastic material,

Stress Due to Bending

M S

2 ³ y dA

Mc I

c

Vm

Vx

My I

y Vm c

c

V mI

§ y · V y dA y ³ ³ ¨ c V m ¸ dA x © ¹

Substituting V x

Vm

M

M

• For static equilibrium,

MECHANICS OF MATERIALS

4-8

h2

1 bh3 12

1 bh3 6

1 6

Ah

4-9

• Structural steel beams are designed to have a large section modulus.

Between two beams with the same cross sectional area, the beam with the greater depth will be more effective in resisting bending.

S

I c

• Consider a rectangular beam cross section,

A beam section with a larger section modulus will have a lower maximum stress

Vm

Mc M I S I section moment of inertia I S section modulus c

• The maximum normal stress due to bending,

Beam Section Properties

MECHANICS OF MATERIALS

Properties of American Standard Shapes

MECHANICS OF MATERIALS

4 - 10

Vm Ec

Hm c M EI

1 Mc Ec I

QH x

Qy U

Hz

QH x

Qy U

1 Uc

Q U

anticlastic curvature

4 - 11

• Expansion above the neutral surface and contraction below it cause an in-plane curvature,

Hy

• Although cross sectional planes remain planar when subjected to bending moments, in-plane deformations are nonzero,

U

1

• Deformation due to bending moment M is quantified by the curvature of the neutral surface

Deformations in a Transverse Cross Section

MECHANICS OF MATERIALS

A cast-iron machine part is acted upon by a 3 kN-m couple. Knowing E = 165 GPa and neglecting the effects of fillets, determine (a) the maximum tensile and compressive stresses, (b) the radius of curvature.

Sample Problem 4.2

¦ yA ¦A I xc

2 ¦ I Ad

Mc I

U

1

M EI

• Calculate the curvature

Vm

4 - 12

• Apply the elastic flexural formula to find the maximum tensile and compressive stresses.

Y

• Based on the cross section geometry, calculate the location of the section centroid and moment of inertia.

SOLUTION:

MECHANICS OF MATERIALS

Sample Problem 4.2

I

I xc

Y

yA, mm3 90 u103 24 u103 3 ¦ yA 114 u10

868 u 103 mm 868 u10-9 m 4 4 - 13

¦ 121 bh3 A d 2 121 90 u 203 1800 u122 121 30 u 403 1200 u182

38 mm

y , mm 50 20

114 u103 3000

2 ¦ I Ad

¦ yA ¦A

Area, mm 2 1 20 u 90 1800 2 40 u 30 1200 ¦ A 3000

Based on the cross section geometry, calculate the location of the section centroid and moment of inertia.

SOLUTION:

MECHANICS OF MATERIALS

Sample Problem 4.2

Mc I M c A 3 kN m u 0.022 m I 868 u109 mm 4 M cB 3 kN m u 0.038 m I 868 u109 mm 4

U

1

3 kN m

165 GPa 868 u10-9 m 4

M EI

• Calculate the curvature

VB

VA

Vm

U U

1

131.3 MPa

VB

47.7 m

4 - 14

20.95 u103 m -1

76.0 MPa

VA

• Apply the elastic flexural formula to find the maximum tensile and compressive stresses.

MECHANICS OF MATERIALS

My I

V1 V x

Vx

V2

nV x

U

y

E1H x

U

Ey 1

V2

E2H x

U

E y 2

U

E1 y

dF2

U

nE1 y dA

V 1dA

dA dF2

V 2 dA

U

E2 y

U

Ey 1 n dA

n

dA

4 - 15

E2 E1

• Define a transformed section such that

dF1

• Elemental forces on the section are

Neutral axis does not pass through section centroid of composite section.

V1

• Piecewise linear normal stress variation.

Hx

• Normal strain varies linearly.

• Consider a composite beam formed from two materials with E1 and E2.

Bending of Members Made of Several Materials

MECHANICS OF MATERIALS

Bar is made from bonded pieces of steel (Es = 29x106 psi) and brass (Eb = 15x106 psi). Determine the maximum stress in the steel and brass when a moment of 40 kip*in is applied.

Example 4.03

4 - 16

• Determine the maximum stress in the steel portion of the bar by multiplying the maximum stress for the transformed section by the ratio of the moduli of elasticity.

• Calculate the maximum stress in the transformed section. This is the correct maximum stress for the brass pieces of the bar.

• Evaluate the cross sectional properties of the transformed section

• Transform the bar to an equivalent cross section made entirely of brass

SOLUTION:

MECHANICS OF MATERIALS

Example 4.03

1.933

15 u 10 psi 0.4 in 1.933 u 0.75 in 0.4 in

6

29 u 106 psi

2.25 in

5.063 in 4

1 b h3 12 T

1 2.25 in. 3 in 3 12

Mc I

V b max V s max

11.85 ksi

1.933 u11.85 ksi

5.063 in

4

40 kip in 1.5 in

V b max V m V s max nV m

Vm

• Calculate the maximum stresses

I

4 - 17

22.9 ksi

11.85 ksi

• Evaluate the transformed cross sectional properties

bT

n

Es Eb

• Transform the bar to an equivalent cross section made entirely of brass.

SOLUTION:

MECHANICS OF MATERIALS

n As x n As d

0 0

Vx

Vx Vc

My I

Vs

nV x

• The normal stress in the concrete and steel

1 b x2 2

2

bx x n As d x

• To determine the location of the neutral axis,

4 - 18

• In the transformed section, the cross sectional area of the steel, As, is replaced by the equivalent area nAs where n = Es/Ec.

• The steel rods carry the entire tensile load below the neutral surface. The upper part of the concrete beam carries the compressive load.

• Concrete beams subjected to bending moments are reinforced by steel rods.

Reinforced Concrete Beams

MECHANICS OF MATERIALS

• Calculate the maximum stresses in the concrete and steel.

• Evaluate geometric properties of transformed section.

4 - 19

• Transform to a section made entirely of concrete.

SOLUTION:

A concrete floor slab is reinforced with 5/8-in-diameter steel rods. The modulus of elasticity is 29x106psi for steel and 3.6x106psi for concrete. With an applied bending moment of 40 kip*in for 1-ft width of the slab, determine the maximum stress in the concrete and steel.

Sample Problem 4.4

MECHANICS OF MATERIALS

4.95 in 2

8.06

2 8.06 u 2 ªS4 85 in º »¼ «¬

3.6 u10 psi

6

29 u106 psi

1 12 in 1.45 in 3 3

4.95 in 2 2.55 in 2

x 1.450 in

Vs

Vc

n

Mc2 I

Mc1 I

44.4 in 40 kip in u 2.55 in 8.06 44.4 in 4

4

40 kip in u 1.45 in

• Calculate the maximum stresses.

I

§ x· 12 x¨ ¸ 4.954 x 0 © 2¹

Vs

4 - 20

18.52 ksi

1.306 ksi

44.4 in 4

Vc

• Evaluate the geometric properties of the transformed section.

nAs

n

Es Ec

• Transform to a section made entirely of concrete.

SOLUTION:

Sample Problem 4.4

MECHANICS OF MATERIALS

• in the vicinity of abrupt changes in cross section

• in the vicinity of points where the loads are applied

Stress concentrations may occur:

Stress Concentrations

MECHANICS OF MATERIALS

Vm

K

Mc I

4 - 21

y Hm c

strain varies linearly across the section

My I

³ V x dA 0

M

³ yV x dA

4 - 22

• For a member with vertical and horizontal planes of symmetry and a material with the same tensile and compressive stress-strain relationship, the neutral axis is located at the section centroid and the stressstrain relationship may be used to map the strain distribution from the stress distribution.

Fx

• For a material with a nonlinear stress-strain curve, the neutral axis location is found by satisfying

and

Vx

• If the member is made of a linearly elastic material, the neutral axis passes through the section centroid

Hx

• For any member subjected to pure bending

Plastic Deformations

MECHANICS OF MATERIALS

MU c I

4 - 23

• RB may be used to determine MU of any member made of the same material and with the same cross sectional shape but different dimensions.

RB

• The modulus of rupture in bending, RB, is found from an experimentally determined value of MU and a fictitious linear stress distribution.

• When the maximum stress is equal to the ultimate strength of the material, failure occurs and the corresponding moment MU is referred to as the ultimate bending moment.

Plastic Deformations

MECHANICS OF MATERIALS

MY

Vm

I VY c

Mc I maximum elastic moment

©

yY

elastic core half - thickness

k

Mp

MY

Mp

plastic moment

4 - 24

shape factor (depends only on cross section shape)

3M 2 Y

• In the limit as the moment is increased further, the elastic core thickness goes to zero, corresponding to a fully plastic deformation.

M

· ¸ c ¸¹

2 § 3 M ¨1 1 yY 2 Y¨ 3 2

• If the moment is increased beyond the maximum elastic moment, plastic zones develop around an elastic core.

VY

Vm

V x d VY

• Rectangular beam made of an elastoplastic material

Members Made of an Elastoplastic Material

MECHANICS OF MATERIALS

A2V Y

R2

Mp

12 AV Y d

• The plastic moment for the member,

4 - 25

The neutral axis divides the section into equal areas.

A1V Y

R1

• Resultants R1 and R2 of the elementary compressive and tensile forces form a couple.

• The neutral axis cannot be assumed to pass through the section centroid.

• Fully plastic deformation of a beam with only a vertical plane of symmetry.

Plastic Deformations of Members With a Single Plane of Symmetry

MECHANICS OF MATERIALS

Residual Stresses

• The final value of stress at a point will not, in general, be zero. 4 - 26

• Residual stresses are obtained by applying the principle of superposition to combine the stresses due to loading with a moment M (elastoplastic deformation) and unloading with a moment -M (elastic deformation).

• Since the linear relation between normal stress and strain applies at all points during the unloading phase, it may be handled by assuming the member to be fully elastic.

• Plastic zones develop in a member made of an elastoplastic material if the bending moment is large enough.

MECHANICS OF MATERIALS

4 - 27

After the loading has been reduced back to zero, determine (c) the distribution of residual stresses, (d) radius of curvature.

Determine (a) the thickness of the elastic core, (b) the radius of curvature of the neutral surface.

A member of uniform rectangular cross section is subjected to a bending moment M = 36.8 kN-m. The member is made of an elastoplastic material with a yield strength of 240 MPa and a modulus of elasticity of 200 GPa.

Example 4.05, 4.06

MECHANICS OF MATERIALS

MY

I c

2 3

I V Y 120 u10 6 m3 240 MPa c 28.8 kN m

2

50 u103 m 60 u103 m

120 u10 6 m3

2 bc 2 3

• Maximum elastic moment:

Example 4.05, 4.06

©

yY 60 mm

U

HY

HY

0.666

200 u109 Pa E

HY

yY

1.2 u103

40 u103 m

1.2 u 103 yY

240 u 106 Pa

VY

U

©

· ¸ c ¸¹

U

2 yY

2 § 3 28.8 kN m ¨1 1 yY 2 ¨ 3 2

• Radius of curvature:

yY c

36.8 kN m

M

· ¸ c ¸¹

2 § 3 M ¨1 1 yY 2 Y¨ 3 2

• Thickness of elastic core:

MECHANICS OF MATERIALS

4 - 28

33.3 m

80 mm

40 mm

240 MPa

yY

VY

• M = 36.8 kN-m c Vm

Mc 36.8 kN m I 120 u 106 m3 306.7 MPa 2V Y

• M = -36.8 kN-m

Example 4.05, 4.06

MECHANICS OF MATERIALS

U

Hx

200 u 109 Pa E

Hx

yY

U

4 - 29

225 m

177.5 u10 6

40 u 103 m

177.5 u 10 6

35.5 u 106 Pa

Vx

At the edge of the elastic core,

• M=0

F M

P Pd

• Eccentric loading

P My A I

V x centric V x bending

4 - 30

• Validity requires stresses below proportional limit, deformations have negligible effect on geometry, and stresses not evaluated near points of load application.

Vx

• Stress due to eccentric loading found by superposing the uniform stress due to a centric load and linear stress distribution due a pure bending moment

Eccentric Axial Loading in a Plane of Symmetry

MECHANICS OF MATERIALS

• Evaluate the maximum tensile and compressive stresses at the inner and outer edges, respectively, of the superposed stress distribution.

• Superpose the uniform stress due to the centric load and the linear stress due to the bending moment.

• Find the equivalent centric load and bending moment

SOLUTION:

Beer • Johnston • DeWolf

4 - 31

An open-link chain is obtained by bending low-carbon steel rods into the shape shown. For 160 lb load, determine • Find the neutral axis by determining the location where the normal stress (a) maximum tensile and compressive is zero. stresses, (b) distance between section centroid and neutral axis

Example 4.07

MECHANICS OF MATERIALS

P 160 lb M Pd 160 lb 0.6 in 104 lb in

• Equivalent centric load and bending moment

Example 4.07

MECHANICS OF MATERIALS

815 psi

Vm

I

1 S 0.25 4 4 3 4

8475 psi

4 - 32

3.068 u10 in Mc 104 lb in 0.25 in I .068 u103 in 4

1 Sc 4 4

• Normal stress due to bending moment

V0

S 0.25 in 2

0.1963 in 2 P 160 lb A 0.1963 in 2

A Sc 2

• Normal stress due to a centric load

815 8475 V 0 V m

Vc

815 8475

V0 Vm

Vt

Vc

Vt 7660 psi

9260 psi

• Maximum tensile and compressive stresses

Example 4.07

y0

y0

0

4 - 33

3.068 u103 in 4 815 psi 105 lb in 0.0240 in

P I AM

P My0 A I

• Neutral axis location

MECHANICS OF MATERIALS

3 u103 m 2

0.038 m

868 u10 9 m 4

A

Y

I

From Sample Problem 2.4,

Sample Problem 4.8

4 - 34

• The largest allowable load is the smallest of the two critical loads.

• Evaluate the critical loads for the allowable tensile and compressive stresses.

• Superpose the stress due to a centric load and the stress due to bending.

• Determine an equivalent centric load and bending moment.

SOLUTION:

The largest allowable stresses for the cast iron link are 30 MPa in tension and 120 MPa in compression. Determine the largest force P which can be applied to the link.

MECHANICS OF MATERIALS

P Mc A A I P Mc A A I

3 u10

3 u10 P

P

3

3

868 u10

9

868 u10 0.028 P 0.022

9

0.028 P 0.022

1559 P

377 P

1559 P

VB

120 MPa

30 MPa

P

P

• The largest allowable load

377 P

VA

79.6 kN

79.6 kN

P

4 - 35

77.0 kN

• Evaluate critical loads for allowable stresses.

VB

VA

• Superpose stresses due to centric and bending loads

d 0.038 0.010 0.028 m P centric load M Pd 0.028 P bending moment

• Determine an equivalent centric and bending loads.

Sample Problem 4.8

MECHANICS OF MATERIALS

Unsymmetric Bending

4 - 36

• In general, the neutral axis of the section will not coincide with the axis of the couple.

• Cannot assume that the member will bend in the plane of the couples.

• Will now consider situations in which the bending couples do not act in a plane of symmetry.

• The neutral axis of the cross section coincides with the axis of the couple

• Members remain symmetric and bend in the plane of symmetry.

• Analysis of pure bending has been limited to members subjected to bending couples acting in a plane of symmetry.

MECHANICS OF MATERIALS

Fx

0 M y Mz

M

applied couple

• The resultant force and moment from the distribution of elementary forces in the section must satisfy

Wish to determine the conditions under which the neutral axis of a cross section of arbitrary shape coincides with the axis of the couple as shown.

Unsymmetric Bending ³ y dA

y © c

¹

· § dA V V ¨ ³ x ³ m ¸ dA

§ y · ³ y¨ V m ¸dA ¹ © c ım I I I z moment of inertia c

Mz

³ yz dA

§ y · ³ z ¨ V m ¸dA © c ¹ I yz product of inertia ³ zV x dA

4 - 37

couple vector must be directed along a principal centroidal axis

or 0

• 0 My

defines stress distribution

or M

• M

neutral axis passes through centroid

or 0

• 0 Fx

MECHANICS OF MATERIALS

M cosT My

M sin T

tan I

Vx 0 y z

Iz tan T Iy

Mzy Myy Iy Iz

• Along the neutral axis,

Vx

Mzy Myy Iz Iy

Iz

Iy

4 - 38

M cosT y M sin T y

• Superpose the component stress distributions

Mz

• Resolve the couple vector into components along the principle centroidal axes.

Superposition is applied to determine stresses in the most general case of unsymmetric bending.

Unsymmetric Bending

MECHANICS OF MATERIALS

M cosT

My

M sin T

Vx

Mzy Myy Iz Iy

• Combine the stresses from the component stress distributions.

Mz

• Resolve the couple vector into components along the principle centroidal axes and calculate the corresponding maximum stresses.

SOLUTION:

4 - 39

A 1600 lb-in couple is applied to a rectangular wooden beam in a plane • Determine the angle of the neutral forming an angle of 30 deg. with the axis. vertical. Determine (a) the maximum y Iz tan I tan T stress in the beam, (b) the angle that the z Iy neutral axis forms with the horizontal plane.

Example 4.08

MECHANICS OF MATERIALS

Example 4.08

1 1.5 in 3.5 in 3 12 1 3.5 in 1.5 in 3 12

1600 lb in cos 30 1600 lb in sin 30 0.9844 in 4

5.359 in 4

800 lb in

1386 lb in

Mzy Iz 5.359 in

4

1386 lb in 1.75 in

452.6 psi

Iy

0.9844 in

4

800 lb in 0.75 in

609.5 psi

V max

V1 V 2

452.6 609.5

V max

4 - 40

1062 psi

• The largest tensile stress due to the combined loading occurs at A.

V2

M yz

The largest tensile stress due to M z occurs along AD

V1

The largest tensile stress due to M z occurs along AB

Iy

Iz

My

Mz

• Resolve the couple vector into components and calculate the corresponding maximum stresses.

MECHANICS OF MATERIALS

Example 4.08

I

3.143 72.4o

tan I

Iz tan T Iy 0.9844 in

4

5.359 in 4 tan 30

• Determine the angle of the neutral axis.

MECHANICS OF MATERIALS

4 - 41

Pa

Mz

centric force Pb

My Mz y z Iz Iy

P A 4 - 42

• If the neutral axis lies on the section, it may be found from

Vx

P M z y M yz A Iz Iy

• By the principle of superposition, the combined stress distribution is

My

P

• The eccentric force is equivalent to the system of a centric force and two couples.

• Consider a straight member subject to equal and opposite eccentric forces.

General Case of Eccentric Axial Loading

MECHANICS OF MATERIALS