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CHAPTER

4

MECHANICS OF MATERIALS
Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf David F. Mazurek Lecture Notes: Brock E. Barry U.S. Military Academy

Pure Bending

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Pure Bending

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

Fig. 4.2 (a) Free-body diagram of the barbell pictured in the chapter opening photo and (b) Free-body diagram of the center bar portion showing pure bending.

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Other Loading Types

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Fig. 4.3 (a) Free-body diagram of a clamp, (b) freebody diagram of the upper portion of the clamp.

• Eccentric Loading: Axial loading which does not pass through section centroid produces internal forces equivalent to an axial force and a couple • Transverse Loading: Concentrated or distributed transverse load produces internal forces equivalent to a shear force and a couple • 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.
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Fig. 4.4 (a) Cantilevered beam with end loading. (b) As portion AC shows, beam is not in pure bending.

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Symmetric Member in Pure Bending

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• Internal forces in any cross section are equivalent to a couple. The moment of the couple is the section bending moment. • From statics, a couple M consists of two equal and opposite forces.
Fig. 4.5 (a) A member in a state of pure bending. (b) Any intermediate portion of AB will also be in pure bending.

• The sum of the components of the forces in any direction is zero. • The moment is the same about any axis perpendicular to the plane of the couple and zero about any axis contained in the plane. • These requirements may be applied to the sums of the components and moments of the statically indeterminate elementary internal forces.
Fx = ∫ σ x dA = 0

Fig. 4.6 Summation of the infinitesimal stress elements must produce the equivalent pure-bending moment.

M y = ∫ zσ x dA = 0 M z = ∫ − yσ x dA = M
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Bending Deformations

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Beam with a plane of symmetry in pure bending:
• member remains symmetric • bends uniformly to form a circular arc • cross-sectional plane passes through arc center and remains planar • length of top decreases and length of bottom increases • a neutral surface must exist that is parallel to the upper and lower surfaces and for which the length does not change
Fig. 4.9 Member subject to pure bending shown in two views. (a) Longitudinal, vertical view (plane of symmetry) and (b) Longitudinal, horizontal view.

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

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Consider a beam segment of length L. After deformation, the length of the neutral surface remains L. At other sections,
L′ = (ρ − y )θ δ = L′ − L = (ρ − y )θ − ρθ = − yθ δ yθ y εx = = − =− (strain varies linearly) L ρθ ρ c c εm = or ρ =

ρ

εm

εx = − εm
Fig. 4.10 Kinematic definitions for pure bending. (a) Longitudinalvertical view and (b) Transverse section at origin.

y c

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Stress Due to Bending
• For a linearly elastic and homogeneous material, σ x = Eε x = − Eε m y = − σ m (stress varies linearly) c y c

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• For static equilibrium, y Fx = 0 = ∫ σ x dA = ∫ − σ m dA c

Fig. 4.11 Bending stresses vary linearly with distance from the neutral axis.

• For static equilibrium, y M = ∫ (− yσ x dA) = ∫ (− y ) − σ m dA c σ σ I M = m ∫ y 2 dA = m c c Mc M = σm = I S y Substituting σ x = − σ m c My σx = − I
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σ 0 = − m ∫ y dA c

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

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Beam Section Properties σm =

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• The maximum normal stress due to bending,
Mc M = I S I = section moment of inertia I S = = section modulus c

A beam section with a larger section modulus will have a lower maximum stress
Fig. 4.12 Wood beam cross sections.

• Consider a rectangular beam cross section,
3 1 I 12 bh S= = = 1 bh3 = 1 Ah 6 6 c h2

Between two beams with the same cross sectional area, the beam with the larger depth h will be more effective in resisting bending.
Fig. 4.13 Two type of steel beam cross sections. (a) S-beam and (b) W-beam

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

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Properties of American Standard Shapes

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MECHANICS OF MATERIALS

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Deformations in a Transverse Cross Section
• Deformation due to bending moment M is quantified by the curvature of the neutral surface ε σ 1 Mc = m = m = c Ec Ec I ρ M = EI
1

• Although transverse cross sectional planes remain planar when subjected to bending moments, inplane deformations are nonzero, ε y = −νε x = νy ρ ε z = −νε x = νy ρ

• Expansion above the neutral surface and contraction below it cause an in-plane curvature,
1 ν = = anticlastic curvature ′ ρ ρ
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Fig. 4.16 Deformation of a transverse cross section.

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Sample Problem 4.2

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SOLUTION: • Based on the cross section geometry, calculate the location of the section centroid and moment of inertia.
Y = ∑ yA ∑A I x′ = ∑ I + A d 2

(

)

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

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.

Mc I

• Calculate the curvature
1

ρ

=

M EI

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Sample Problem 4.2
SOLUTION:

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Based on the cross section geometry, calculate the location of the section centroid and moment of inertia.
Area, mm 2 1 20 × 90 = 1800 2 40 × 30 = 1200 ∑ A = 3000 y , mm 50 20 yA, mm3 90 × 103 24 × 103 3 ∑ yA = 114 × 10

Fig. 1 Composite areas for calculating centroid.

∑ yA 114 × 10 Y = = = 38 mm 3000 ∑A

3

I x′ = ∑ I + A d 2 = ∑
Fig. 2 Composite sections for calculating moment of inertia.

1 ) (12 bh3 + A d 2 ) 1 1 = (12 90 × 203 + 1800 × 122 ) + (12 30 × 403 + 1200 × 182 )

(

I = 868 × 103 mm 4 = 868 × 10-9 m 4
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Sample Problem 4.2

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• Apply the elastic flexural formula to find the maximum tensile and compressive stresses.
Mc I M c A 3 kN ⋅ m × 0.022 m σ A = +76.0 MPa σA = = −9 4 I 868 × 10 m M cB 3 kN ⋅ m × 0.038 m σ = −131.3 MPa B σB = − =− −9 4 I 868 × 10 m

σm =

Fig. 3 Deformed radius of curvature is measured to the centroid of the cross sections.

• Calculate the curvature
1

ρ

= =

M EI

(165 GPa )(868 ×10-9 m 4 )

3 kN ⋅ m

1

ρ ρ = 47.7 m

= 20.95 × 10−3 m -1

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MECHANICS OF MATERIALS

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Eccentric Axial Loading in a Plane of Symmetry
• Stress due to eccentric loading found by superposing the uniform stress due to a centric load and linear stress distribution due to a pure bending moment σ x = (σ x )centric + (σ x )bending
Fig. 4.39 (a) Member with eccentric loading. (b) Free-body diagram of a member with internal loads at section C.

=

P My − A I

• Eccentric loading
F=P M = Pd

• Result are valid if stresses do not exceed the proportional limit, deformations have negligible effect on geometry, and stresses are not evaluated near points of load application.

Fig. 4.41 Stress distribution for eccentric loading is obtained by superposing the axial and pure bending distributions.

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Concept Application 4.7
SOLUTION:

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• Find the equivalent centric load and bending moment • Superpose the uniform stress due to the centric load and the linear stress due to the bending moment.
Fig. 4.43 Open chain link under loading.

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

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Concept Application 4.7

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• Normal stress due to a centric load
A = πc 2 = π (0.25 in )2 = 0.1963 in 2 P 160 lb σ0 = = A 0.1963 in 2 = 815 psi
Fig. 4.43 Free-body diagram for section at C to find axial force and moment. Stress at section C is superposed axial and bending stresses.

• Normal stress due to bending moment
I = 1 πc 4 = 1 π (0.25)4 4 4 = 3.068 × 10−3 in 4 Mc (104 lb ⋅ in )(0.25 in ) = = I 3.068 × 10−3 in 4 = 8475 psi
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• Equivalent centric load and bending moment
P = 160 lb M = Pd = (160 lb )(0.65 in ) = 104 lb ⋅ in

σm

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MECHANICS OF MATERIALS
Concept Application 4.7

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Fig. 4.43 (c) Axial stress at section C. (d) Bending stress at C. (e) Superposition of stresses.

(c)

(d)

(e)

• Maximum tensile and compressive stresses σt = σ0 +σm
= 815 + 8475 σc = σ0 −σ m = 815 − 8475

• Neutral axis location
0= P My0 − A I

σ t = 9260 psi σ c = −7660 psi

3.068 × 10−3 in 4 P I = (815 psi ) y0 = 105 lb ⋅ in AM y0 = 0.0240 in

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Sample Problem 4.8

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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. SOLUTION: • Determine equivalent centric load and bending moment. • Superpose the stress due to a centric load and the stress due to bending.
Fig. 1 Section geometry to find centroid location.

From Sample Problem 4.2,
A = 3 × 10−3 m 2 Y = 0.038 m I = 868 × 10−9 m 4

• Evaluate the critical loads for the allowable tensile and compressive stresses. • The largest allowable load is the smallest of the two critical loads.
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Sample Problem 4.8

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• Determine equivalent centric and bending loads. d = 0.038 − 0.010 = 0.028 m P = centric load M = Pd = 0.028 P = bending moment

• Superpose stresses due to centric and bending loads
Fig. 2 Section dimensions for finding location of point D.

σA = −

(0.028 P )(0.022) = +377 P P Mc A P + =− + A I 3 × 10−3 868 × 10−9 (0.028 P )(0.022) = −1559 P P Mc P σB = − − A = − − A I 3 ×10−3 868 ×10−9

• Evaluate critical loads for allowable stresses.
P = 79.6 kN σ A = +377 P = 30 MPa σ B = −1559 P = −120 MPa P = 77.0 kN
Figs. 4 Stress distribution at section C is superposition of axial and bending distributions acting at centroid.

• The largest allowable load

P = 77.0 kN

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