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Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 4.2
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
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
4-1
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 4.2
SOLUTION:
Based on the cross section geometry, calculate
the location of the section centroid and
moment of inertia.
Area, mm 2
y , mm
yA, mm3
1 20 × 90 = 1800
50
90 × 103
2 40 × 30 = 1200
∑ A = 3000
20
24 × 103
3
∑ yA = 114 × 10
Y =
3
∑ yA 114 × 10
=
= 38 mm
3000
∑A
(
) (121 bh3 + A d 2 )
1 90 × 203 + 1800 × 12 2 ) + ( 1 30 × 403 + 1200 × 182 )
= (12
12
I x′ = ∑ I + A d 2 = ∑
I = 868 × 103 mm 4 = 868 × 10-9 m 4
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
4-2
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Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 4.2
• 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 =
=
I
868 × 10−9 m 4
3 kN ⋅ m × 0.038 m σ = −131.3 MPa
M cB
B
σB = −
=−
I
868 × 10−9 m 4
σm =
• Calculate the curvature
1
ρ
=
M
EI
=
(165 GPa )(868 ×10-9 m 4 )
3 kN ⋅ m
1
= 20.95 × 10 −3 m -1
ρ
ρ = 47.7 m
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
4-3
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Bending of Members Made of Several Materials
• Consider a composite beam formed from
two materials with E1 and E2.
• Normal strain varies linearly.
εx = −
y
ρ
• Piecewise linear normal stress variation.
σ 1 = E1ε x = −
E1 y
ρ
σ 2 = E2ε x = −
E2 y
ρ
Neutral axis does not pass through
section centroid of composite section.
• Elemental forces on the section are
Ey
E y
dF1 = σ 1dA = − 1 dA dF2 = σ 2 dA = − 2 dA
ρ
My
σx = −
I
σ1 = σ x
σ 2 = nσ x
ρ
• Define a transformed section such that
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
dF2 = −
(nE1 ) y dA = − E1 y (n dA)
ρ
ρ
E
n= 2
E1
4-4
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Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
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.
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.
• 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.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
Fourth
Edition
MECHANICS OF MATERIALS
4-5
Beer • Johnston • DeWolf
Example 4.03
SOLUTION:
• Transform the bar to an equivalent cross section
made entirely of brass.
E
29 × 106 psi
n= s =
= 1.933
Eb 15 × 106 psi
bT = 0.4 in + 1.933 × 0.75 in + 0.4 in = 2.25 in
• Evaluate the transformed cross sectional properties
1 b h3 = 1 (2.25 in.)(3 in.)3
I = 12
T
12
= 5.063 in.4
• Calculate the maximum stresses
σm =
Mc (40 kip ⋅ in.)(1.5 in.)
=
= 11.85 ksi
I
5.063 in.4
(σ b )max = σ m
(σ s )max = nσ m = 1.933 ×11.85 ksi
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
(σ b )max = 11.85 ksi
(σ s )max = 22.9 ksi
4-6
3
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
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 As (d − x ) = 0
2
1 b x2
2
+ n As x − n As d = 0
• The normal stress in the concrete and steel
σx = −
My
I
σc = σ x
σ s = nσ x
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
Fourth
Edition
MECHANICS OF MATERIALS
4-7
Beer • Johnston • DeWolf
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.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
4-8
4
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 4.4
SOLUTION:
• Transform to a section made entirely of concrete.
E
29 × 106 psi
= 8.06
n= s =
Ec 3.6 × 106 psi
( )
2
nAs = 8.06 × 2⎡π4 85 in ⎤ = 4.95 in 2
⎢⎣
⎥⎦
• Evaluate the geometric properties of the
transformed section.
⎛ x⎞
12 x⎜ ⎟ − 4.95(4 − x ) = 0
⎝ 2⎠
(
x = 1.450 in
)
I = 13 (12 in )(1.45 in )3 + 4.95 in 2 (2.55 in )2 = 44.4 in 4
• Calculate the maximum stresses.
Mc1 40 kip ⋅ in × 1.45 in
=
I
44.4 in 4
Mc
40 kip ⋅ in × 2.55 in
σ s = n 2 = 8.06
I
44.4 in 4
σc =
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
σ c = 1.306 ksi
σ s = 18.52 ksi
4-9
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Stress Concentrations
Stress concentrations may occur:
• in the vicinity of points where the
loads are applied
σm = K
Mc
I
• in the vicinity of abrupt changes
in cross section
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4 - 10
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Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Plastic Deformations
• For any member subjected to pure bending
y
c
ε x = − εm
strain varies linearly across the section
• If the member is made of a linearly elastic material,
the neutral axis passes through the section centroid
and
σx = −
My
I
• For a material with a nonlinear stress-strain curve,
the neutral axis location is found by satisfying
Fx = ∫ σ x dA = 0
M = ∫ − yσ x dA
• 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.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
4 - 11
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Plastic Deformations
• 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.
• The modulus of rupture in bending, RB, is found
from an experimentally determined value of MU
and a fictitious linear stress distribution.
RB =
MU c
I
• 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.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
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Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Members Made of an Elastoplastic Material
• Rectangular beam made of an elastoplastic material
Mc
I
σ x ≤ σY
σm =
σ m = σY
I
M Y = σ Y = maximum elastic moment
c
• If the moment is increased beyond the maximum
elastic moment, plastic zones develop around an
elastic core.
⎛
y2 ⎞
M = 32 M Y ⎜1 − 13 Y2 ⎟
⎜
c ⎟⎠
⎝
yY = elastic core half - thickness
• In the limit as the moment is increased further, the
elastic core thickness goes to zero, corresponding to a
fully plastic deformation.
M p = 32 M Y = plastic moment
Mp
= shape factor (depends only on cross section shape)
k=
MY
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