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MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Sample Problem 4.2SOLUTION:
• Based on the cross section geometry, calculate the location of the section centroid and moment of inertia.
( )∑ +∑ 2dAIIAyY ( )∑ +=∑∑= ′
2dAIIAyY x
• Apply the elastic flexural formula to find the maximum tensile and compressive stresses.
Mc
A cast-iron machine part is acted upon by a 3 kN-m couple. Knowing E = 165
IMc
m =σ
• Calculate the curvatureby a 3 kN m couple. Knowing E 165 GPa and neglecting the effects of fillets, determine (a) the maximum tensile and compressive stresses (b)
Calculate the curvature
EIM
=ρ1
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 1
tensile and compressive stresses, (b) the radius of curvature.
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Sample Problem 4.2SOLUTION:
Based on the cross section geometry, calculate the location of the section centroid and moment of inertia.
32 mmmmmmArea Ayy
×=××=×
3
3
3
104220120030402109050180090201
mm,mm,mm Area, Ayy
10114 3×∑ Ay
∑ ×==∑ 3101143000 AyA
mm383000
10114=
×=
∑∑=
AAyY
( ) ( )2312( ) ( )( ) ( )23
12123
121
231212
18120040301218002090 ×+×+×+×=
+=+= ∑∑′ dAbhdAIIx
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 2
49-43 m10868mm10868 ×=×=I
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Sample Problem 4.2
• Apply the elastic flexural formula to find the maximum tensile and compressive stresses.
m022.0mkN 3 ×⋅
=
cMI
Mc
A
m
σ
σ
MPa076+=Aσ
49
49
m10868m038.0mkN 3
m10868
−
−
××⋅
−=−=
×==
IcM
I
BB
AA
σ
σ MPa0.76+Aσ
MPa3.131−=Bσm10868×I
• Calculate the curvature
mkN 3
1
⋅
=EIM
ρ
m1095201 1-3×= −
( )( )49- m10868GPa 165 ×=
m 7.47
m1095.20
=
×=
ρρ
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 3
MECHANICS OF MATERIALS
FourthEdition
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.
ρε y
x −=
• Piecewise linear normal stress variation.
ρεσ
ρεσ yEEyEE xx
222
111 −==−==
ρρNeutral axis does not pass through section centroid of composite section.
• Elemental forces on the section are
dAyEdAdFdAyEdAdFρ
σρ
σ 222
111 −==−==
ρρ
( ) ( ) 211 EdAyEdAynEdF
• Define a transformed section such thatx
nI
My
σσσσ
σ
==
−=
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 4
( ) ( )1
2112 E
ndAnydAydF =−=−=ρρ
xx nσσσσ == 21
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Example 4.03SOLUTION:
• Transform the bar to an equivalent cross qsection made entirely of brass
• Evaluate the cross sectional properties of• Evaluate the cross sectional properties of the transformed section
C l l t th i t i th• Calculate the maximum stress in the transformed section. This is the correct maximum stress for the brass pieces of
Bar is made from bonded pieces of steel (Es = 29x106 psi) and brass
the bar.
• Determine the maximum stress in the ( s p )(Eb = 15x106 psi). Determine the maximum stress in the steel and brass when a moment of 40 kip*in
steel portion of the bar by multiplying the maximum stress for the transformed section by the ratio of the moduli of
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 5
brass when a moment of 40 kip in is applied.
section by the ratio of the moduli of elasticity.
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Example 4.03SOLUTION:• Transform the bar to an equivalent cross section
made entirely of brassmade entirely of brass.
933.1psi1015psi1029
6
6=
×
×==
b
sEEn
• Evaluate the transformed cross sectional properties
in 25.2in 4.0in 75.0933.1in 4.0 =+×+=Tb
( )( )4
31213
121
in.063.5
in. 3in. 25.2
=
== hbI T
• Calculate the maximum stresses( )( ) ksi85.11in.5.1in.kip 40
4 =⋅
==Mc
mσ in.5.063 4Im
( )( ) k i85119331
max = mb σσ ( )( ) k i22 9
ksi 85.11max =bσ
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 6
( ) ksi85.11933.1max ×== ms nσσ ( ) ksi22.9max =sσ
MECHANICS OF MATERIALS
FourthEdition
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
concrete beam carries the compressive load.
nAs where n = Es/Ec.
• To determine the location of the neutral axis,( ) ( ) 0dAxb( ) ( )
0
022
21 =−+
=−−
dAnxAnxb
xdAnbx
ss
s
• The normal stress in the concrete and steel
x IMyσ −=
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 7
xsxc nI
σσσσ ==
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Sample Problem 4.4
SOLUTION:
• Transform to a section made entirely• Transform to a section made entirely of concrete.
• Evaluate geometric properties of• Evaluate geometric properties of transformed section.
• Calculate the maximum stresses
A t fl l b i i f d ith
• 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
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 8
stress in the concrete and steel.
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Sample Problem 4.4SOLUTION:• Transform to a section made entirely of concrete.
6
( ) 225
6
6
in954in2068
06.8psi 106.3psi1029
⎥⎤
⎢⎡×
=×
×==
π
c
s
nA
EEn
( )85
4 in95.4in206.8 =⎥⎦⎤
⎢⎣⎡×= π
snA
• Evaluate the geometric properties of the t f d titransformed section.
( )
( )in450.10495.4
212 ==−−⎟
⎠⎞
⎜⎝⎛ xxxx
( )( ) ( )( ) 422331 in4.44in55.2in95.4in45.1in12 =+=I
• Calculate the maximum stresses.
2
41
in55.2inkip40068
in44.4in1.45inkip40
×⋅
×⋅==
Mc
IMc
cσ ksi306.1=cσ
k i5218
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 9
42
in44.4in55.2inkip4006.8==
IMcnsσ ksi52.18=sσ
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Stress Concentrations
Stress concentrations may occur:
• in the vicinity of points where theI
McKm =σ
in the vicinity of points where the loads are applied
• in the vicinity of abrupt changes
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 10
y p gin cross section
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Plastic Deformations• For any member subjected to pure bending
mx cy εε −= strain varies linearly across the sectionc
• If the member is made of a linearly elastic material, the neutral axis passes through the section centroid
IMy
x −=σand
• For a material with a nonlinear stress strain curve• For a material with a nonlinear stress-strain curve, the neutral axis location is found by satisfying
∫ −=∫ == dAyMdAF xxx σσ 0
• For a member with vertical and horizontal planes of symmetry and a material with the same tensile and y ycompressive stress-strain relationship, the neutral axis is located at the section centroid and the stress-strain relationship may be used to map the strain
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 11
strain relationship may be used to map the strain distribution from the stress distribution.
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Plastic Deformations• When the maximum stress is equal to the ultimate
strength of the material, failure occurs and the corresponding moment M is referred to as thecorresponding moment MU is referred to as the ultimate bending moment.
• The modulus of rupture in bending R is found• The modulus of rupture in bending, RB, is found from an experimentally determined value of MU and a fictitious linear stress distribution.
IcMR U
B =
• RB may be used to determine MU of any member made of the same material and with the same cross sectional shape but differentsame cross sectional shape but different dimensions.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 12
MECHANICS OF MATERIALS
FourthEdition
Beer • Johnston • DeWolf
Members Made of an Elastoplastic Material• Rectangular beam made of an elastoplastic material
=≤ mYx IMcσσσ
moment elastic maximum === YYYm cIM σσσ
• If the moment is increased beyond the maximum elastic moment, plastic zones develop around an elastic core.
thickness-half core elastic 1 2
2
31
23 =⎟
⎟⎠
⎞⎜⎜⎝
⎛−= Y
YY y
cyMM
• In the limit as the moment is increased further, the elastic core thickness goes to zero, corresponding to a fully plastic deformationfully plastic deformation.
moment plastic 23 ==
p
Yp
M
MM
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 13
shape)section crosson only (dependsfactor shape==Y
pM
k