1. MECHANICS OF MATERIALS
Third Edition
MECHANICS OF MATERIALS
CHAPTER
Ferdinand P. Beer
E. Russell Johnston, Jr. John T. DeWolf
Lecture Notes:
J. Walt Oler
Texas Tech University
Pure Bending
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2. MECHANICS OF MATERIALS
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Beer • Johnston • DeWolf
Pure Bending
Pure Bending
Other Loading Types
Symmetric Member in Pure Bending Bending Deformations
Strain Due to Bending
Beam Section Properties
Properties of American Standard Shapes Deformations in a Transverse Cross Section Sample Problem 4.2
Bending of Members Made of Several Materials
Example 4.03
Reinforced Concrete Beams Sample Problem 4.4
Stress Concentrations Plastic Deformations
Members Made of an Elastoplastic Material
Example 4.03
Reinforced Concrete Beams Sample Problem 4.4
Stress Concentrations Plastic Deformations
Members Made of an Elastoplastic Material Plastic Deformations of Members With a Single
Plane of S...
Residual Stresses Example 4.05, 4.06
Eccentric Axial Loading in a Plane of Symmetry Example 4.07
Sample Problem 4.8 Unsymmetric Bending Example 4.08
General Case of Eccentric Axial Loading
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3. MECHANICS OF MATERIALS
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Pure Bending
Pure Bending: Prismatic members subjected to equal and opposite couples acting in the same longitudinal plane
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4. MECHANICS OF MATERIALS
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Other Loading Types
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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5. MECHANICS OF MATERIALS
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Symmetric Member in Pure Bending
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.
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
M y = ∫ zσ x dA = 0
M z = ∫ − yσ x dA = M
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6. MECHANICS OF MATERIALS
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Bending Deformations
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
stresses and strains are negative (compressive) above the neutral plane and positive (tension) below it
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7. MECHANICS OF MATERIALS
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Strain Due to Bending
Consider a beam segment of length L. After deformation, the length of the neutral
surface remains L. At other sections,
ε x = − c εm
(strain varies linearly) y
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8. MECHANICS OF MATERIALS
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Stress Due to Bending
For a linearly elastic material, My
σ x = − I
First moment with respect to neutral plane is zero. Therefore, the neutral surface must pass through the section centroid.
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9. MECHANICS OF MATERIALS
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Beam Section Properties
The maximum normal stress due to bending,
= Mc = M
I S
σ m
I = section moment of inertia
S = I = section modulus c
A beam section with a larger section modulus
will have a lower maximum stress
Consider a rectangular beam cross section, 1 3 I
12 bh
S = =
c h 2
= 6 bh = 6 Ah
1 1 3
Between two beams with the same cross sectional area, the beam with the greater depth will be more effective in resisting bending.
Structural steel beams are designed to have a large section modulus.
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10. MECHANICS OF MATERIALS
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Third
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Properties of American Standard Shapes
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11. 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
1 = εm = σ m = 1 Mc
ρ c Ec E`c I
= M EI
Expansion above the neutral surface and contraction below it cause an in-plane curvature,
1 = ν = anticlastic curvature ρ′ ρ
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12. MECHANICS OF MATERIALS
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Sample Problem 4.2
SOLUTION:
Based on the cross section geometry, calculate the location of the section centroid and moment of inertia.
I ′ = ∑ (I + A d 2 )
Y = ∑ yA x
∑ A
Apply the elastic flexural formula to find the maximum tensile and compressive stresses.
Mc I
Calculate the curvature
σ m =
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.
1 M =
ρ EI
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13. MECHANICS OF MATERIALS
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Sample Problem 4.2
SOLUTION:
Based on the cross section geometry, calculate the location of the section centroid and moment of inertia.
Area, mm2
y, mm
yA, mm3 1 2
20 × 90 = 1800
40 × 30 = 1200 50 20
90 ×103
24 ×103
∑ A = 3000
∑ yA = 114 ×103 3
Y = ∑ yA = 114 ×10 = 38 mm
∑ A 3000
I x′ = ∑ (I + A d 2 )= ∑ (1 bh3 + A d 2 ) 12
= (1 90 × ×122 )+ (1 30 × ×182 )
12 12
I = 868×103 mm = 868×10-9 m4
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14. MECHANICS OF MATERIALS
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Sample Problem 4.2
Apply the elastic flexural formula to find the maximum tensile and compressive stresses.
= Mc I σ m
= M cA = 3 kN ⋅ m ×0.022 m σ
σ A = MPa A I
868×10−9 mm4
= − M cB = − 3 kN ⋅ m ×0.038 m σ
σ B = −131.3 MPa B I
868×10−9 mm4
Calculate the curvature
1 = M
ρ EI
3 kN ⋅ m
(165 GPa)(868×10-9 m4 ) =
1 = 20.95×10−3 m-1 ρ
ρ = 47.7 m
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15. MECHANICS OF MATERIALS
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Bending of Members Made of Several Materials
Consider a composite beam formed from two materials with E1 and E2.
Normal strain varies linearly.
ε x = − ρ
Piecewise linear normal stress variation. y
= − E1y = − E2 y
σ1 = E1ε x
σ = E ε
2 2 x
ρ ρ
Neutral axis does not pass through section centroid of composite section.
Elemental forces on the section are
dF = σ dA = − E1y dA dF = σ dA = − E2 y dA
1 1
2 2
ρ ρ My
σ x = −
Define a transformed section such that I
= − (nE1)
σ1 = σ x
σ 2 = nσ x
( ) dF
y dA = − E y n dA n = E 1 2 2 ρ ρ E1
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Example 4.03
SOLUTION:
Transform the bar to an equivalent cross section made entirely of brass
Evaluate the cross sectional properties of the transformed section
Calculate the maximum stress in the transformed section. This is the correct maximum stress for the brass pieces of the bar.
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.
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.
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17. MECHANICS OF MATERIALS
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Example 4.03
SOLUTION:
Transform the bar to an equivalent cross section made entirely of brass. 6
n = Es = 29 ×10 psi = 1.933 Eb
15×106 psi
bT = 0.4 in × 0.75 in in = 2.25 in
𝐼= 𝑏 ℎ 3 = 1 12 ∗2.25∗ 3 3 =5.06 in4
Evaluate the transformed cross sectional properties
Calculate the maximum stresses
Mc (40 kip ⋅ in)(1.5 in)
σ m = =
= ksi I
5.063 in4
(σ b )max = σ m
(σ s )max = nσ m = 1.933×11.85 ksi
(σ b )max = ksi
(σ s )max = 22.9 ksi
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18. MECHANICS OF MATERIALS
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Reinforced Concrete Beams
Concrete beams subjected to bending moments are reinforced by steel rods.
The steel rods carry the entire tensile load below the neutral surface. The upper part of the concrete beam carries the compressive load.
In the transformed section, the cross sectional area of the steel, As, is replaced by the equivalent area nAs where n = Es/Ec.
To determine the location of the neutral axis,
(bx) x − n A (d − x) = 0 s 2
1 b x2 + n As x − n Asd = 0 2
The normal stress in the concrete and steel My
σ x = − I
σ c = σ x
σ s = nσ x
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19. MECHANICS OF MATERIALS
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Sample Problem 4.4
SOLUTION:
Transform to a section made entirely of concrete.
Evaluate geometric properties of transformed section.
Calculate the maximum stresses in the concrete and steel.
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.
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20. MECHANICS OF MATERIALS
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Beer • Johnston • DeWolf
Sample Problem 4.4
SOLUTION:
Transform to a section made entirely of concrete.
Es 29 ×106 psi
n = E
= = 8.06
3.6 ×106 psi c
nA = 8.06 × 2⎡π (5 in)2 ⎤ = 4.95in2 s
⎢⎣ 4 8 ⎥⎦
Evaluate the geometric properties of the transformed section.
⎛ x ⎞
12x⎜ ⎟ − 4.95(4 − x) = 0
x = 1.450in
⎝ 2 ⎠
I = 1 (12in)(1.45in)3 + (4.95in2 )(2.55in)2 = 44.4in4 3
Calculate the maximum stresses.
= Mc1 = 40 kip ⋅ in ×1.45 in σ
σ c = ksi c I
44.4in4
= n Mc2 = kip ⋅ in × 2.55 in σ
σ s = ksi s I
44.4in4
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21. 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 a pure bending moment
σ x = (σ x )centric + (σ x )bending
= P − My A I
Eccentric loading
F = P M = Pd
Validity requires stresses below proportional limit, deformations have negligible effect on geometry, and stresses not evaluated near points of load application.
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22. MECHANICS OF MATERIALS
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Example 4.07
SOLUTION:
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.
Evaluate the maximum tensile and compressive stresses at the inner and outer edges, respectively, of the superposed stress distribution.
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 is zero.
(a) maximum tensile and compressive stresses, (b) distance between section centroid and neutral axis
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23. MECHANICS OF MATERIALS
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Example 4.07
Normal stress due to a centric load
A = πc2 = π (0.25in)2
= in2
P 160 lb
σ 0 = A = in2
= 815 psi
Normal stress due to bending moment
I = 1 πc4 = 1 π (0.25)4 4 4
= 3.068×10−3 in4
Equivalent centric load and bending moment
P = 160 lb
M = Pd = (160 lb)(0.6in)
= 104 lb ⋅ in
Mc (104 lb ⋅ in )(0.25 in )
σ = = m I
.068×10−3 in4
= 8475 psi
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24. MECHANICS OF MATERIALS
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Third
Beer • Johnston • DeWolf
Example 4.07
Maximum tensile and compressive stresses
Neutral axis location
P My0
0 = −
A I
σ t = σ 0 + σ m =
σ c = σ 0 − σ m
= 815 − 8475
σ t = 9260 psi
P I
3.068×10−3in4
y0 = A M = (815 psi)
105lb ⋅ in
σ c = −7660 psi
y0 = in
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25. MECHANICS OF MATERIALS
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Sample Problem 4.8
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 an equivalent centric load and bending moment.
Superpose the stress due to a centric load and the stress due to bending.
Evaluate the critical loads for the allowable tensile and compressive stresses.
The largest allowable load is the smallest of the two critical loads.
From Sample Problem 2.4,
A = 3×10−3 m2
Y = m
I = 868×10−9 m4
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26. MECHANICS OF MATERIALS
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Beer • Johnston • DeWolf
Sample Problem 4.8
Determine an equivalent centric and bending loads.
d = − = m
P = centric load
M = Pd = P = bending moment
Superpose stresses due to centric and bending loads
P McA P (0.028 P )(0.022)
σ A = − +
= − + = +377 P
A I
P McA P (0.028 P )(0.022)
3×10−3
868×10−9
σ B = − −
= −
− = −1559 P
A I
3×10−3 868×10−9
Evaluate critical loads for allowable stresses.
σ A = +377 P = 30 MPa
σ B = −1559 P = −120 MPa
P = 79.6 kN
P = 77.0 kN
The largest allowable load
P = 77.0 kN
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