Fourth Edition
CHAPTER
8
MECHANICS OF
MATERIALS
Ferdinand P. Beer
E. Russell Johnston, Jr.
John T. DeWolf
Lecture Notes:
J. Walt Oler
Texas Tech University
Principle Stresses
Under a Given
Loading
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Principle Stresses Under a Given Loading
Introduction
Principle Stresses in a Beam
Sample Problem 8.1
Sample Problem 8.2
Design of a Transmission Shaft
Sample Problem 8.3
Stresses Under Combined Loadings
Sample Problem 8.5
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Introduction
• In Chapter 1 and 2, you learned how to determine the normal stress due to
centric loads.
• In Chapter 3, you analyzed the distribution of shearing stresses in a circular
member due to a twisting couple.
• In Chapter 4, you determined the normal stresses caused by bending couples.
• In Chapters 5 and 6, you evaluated the shearing stresses due to transverse
loads.
• In Chapter 7, you learned how the components of stress are transformed by a
rotation of the coordinate axes and how to determine the principal planes,
principal stresses, and maximum shearing stress at a point.
• In Chapter 8, you will learn how to determine the stress in a structural member
or machine element due to a combination of loads and how to find the
corresponding principal stresses and maximum shearing stress.
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Principle Stresses in a Beam
• Prismatic beam subjected to transverse
loading
My
Mc
σm =
I
I
VQ
VQ
τ xy = −
τm =
It
It
σx = −
• Principal stresses determined from methods
of Chapter 7
• Can the maximum normal stress within
the cross-section be larger than
σm =
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Mc
I
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Principle Stresses in a Beam
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Principle Stresses in a Beam
• Cross-section shape results in large values of τxy
near the surface where σx is also large.
• σmax may be greater than σm
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Sample Problem 8.1
SOLUTION:
• Determine shear and bending
moment in Section A-A’
• Calculate the normal stress at top
surface and at flange-web junction.
A 160-kN force is applied at the end
of a W200x52 rolled-steel beam.
• Evaluate the shear stress at flangeweb junction.
Neglecting the effects of fillets and
of stress concentrations, determine
whether the normal stresses satisfy a
design specification that they be
equal to or less than 150 MPa at
section A-A’.
• Calculate the principal stress at
flange-web junction
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Beer • Johnston • DeWolf
Sample Problem 8.1
SOLUTION:
• Determine shear and bending moment in
Section A-A’
M A = (160 kN )( 0.375 m ) = 60 kN - m
V A = 160 kN
• Calculate the normal stress at top surface
and at flange-web junction.
MA
60 kN ⋅ m
=
S
512 × 10−6 m3
= 117.2 MPa
y
90.4 mm
σb = σ a b = (117.2 MPa )
c
103 mm
= 102.9 MPa
σa =
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Sample Problem 8.1
• Evaluate shear stress at flange-web junction.
Q = ( 204 × 12.6 ) 96.7 = 248.6 × 103 mm3
= 248.6 × 10−6 m3
(
)
V AQ (160 kN ) 248.6 × 10−6 m3
τb =
=
It
52.7 × 10− 6 m 4 ( 0.0079 m )
= 95.5 MPa
(
)
• Calculate the principal stress at
flange-web junction
σ max = 12 σ b +
( 12 σ b )2 + τ b2
2
102.9
 102.9 
2
=
+ 
 + ( 95.5)
2
 2 
= 159.9 MPa ( > 150 MPa )
Design specification is not satisfied.
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Sample Problem 8.2
SOLUTION:
• Determine reactions at A and D.
• Determine maximum shear and
bending moment from shear and
bending moment diagrams.
The overhanging beam supports a
uniformly distributed load and a
concentrated load. Knowing that for
the grade of steel to used σall = 24 ksi
and τall = 14.5 ksi, select the wideflange beam which should be used.
• Calculate required section modulus
and select appropriate beam section.
• Find maximum normal stress.
• Find maximum shearing stress.
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Sample Problem 8.2
SOLUTION:
• Determine reactions at A and D.
∑ M A = 0 ⇒ RD = 59 kips
∑ M D = 0 ⇒ R A = 41kips
• Determine maximum shear and bending
moment from shear and bending moment
diagrams.
M max = 239.4 kip ⋅ in
V max = 43 kips
with V = 12.2 kips
• Calculate required section modulus
and select appropriate beam section.
M max 24 kip ⋅ in
S min =
=
= 119.7 in 3
σ all
24 ksi
select W21× 62 beam section
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Sample Problem 8.2
• Find maximum shearing stress.
Assuming uniform shearing stress in web,
V
43 kips
τ max = max =
= 5.12 ksi < 14.5 ksi
Aweb 8.40 in 2
• Find maximum normal stress.
σa =
M max
60 kip ⋅ in
= 2873
= 22.6 ksi
3
S
127in
y
9.88
σb = σ a b = ( 22.6 ksi )
= 21.3 ksi
c
10.5
τb =
V
12.2 kips
=
= 1.45 ksii
Aweb 8.40 in 2
2
21.3 ksi
 21.3 ksi 
2
σ max =
+ 
 + (1.45 ksi )
2
 2 
= 21.4 ksi < 24 ksi
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Beer • Johnston • DeWolf
Design of a Transmission Shaft
• If power is transferred to and from the
shaft by gears or sprocket wheels, the
shaft is subjected to transverse loading
as well as shear loading.
• Normal stresses due to transverse loads
may be large and should be included in
determination of maximum shearing
stress.
• Shearing stresses due to transverse
loads are usually small and
contribution to maximum shear stress
may be neglected.
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Design of a Transmission Shaft
• At any section,
Mc
I
Tc
τm =
J
σm =
where M 2 = M y2 + M z2
• Maximum shearing stress,
2
2
σ 
 Mc   Tc 
τ max =  m  + (τ m ) 2 = 
 + 
 2 
 2I   J 
2
for a circular or annular cross - section, 2 I = J
τ max =
c
M2 +T2
J
• Shaft section requirement,
 M 2 + T 2 

 max
J
=
 
τ all
 c  min
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Sample Problem 8.3
SOLUTION:
• Determine the gear torques and
corresponding tangential forces.
• Find reactions at A and B.
• Identify critical shaft section from
torque and bending moment diagrams.
Solid shaft rotates at 480 rpm and
transmits 30 kW from the motor to
gears G and H; 20 kW is taken off at
gear G and 10 kW at gear H.
Knowing that σall = 50 MPa, determine
the smallest permissible diameter for
the shaft.
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• Calculate minimum allowable shaft
diameter.
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Sample Problem 8.3
SOLUTION:
• Determine the gear torques and corresponding
tangential forces.
P
30 kW
=
= 597 N ⋅ m
2πf 2π ( 8 Hz )
T
597 N ⋅ m
FE = E =
= 3.73 kN
rE
0.16 m
TE =
20 kW
= 398 N ⋅ m
2π ( 8 Hz )
10 kW
TD =
= 199 N ⋅ m
2π ( 8 Hz )
TC =
FC = 6.63 kN
FD = 2.49 kN
• Find reactions at A and B.
Ay = 0.932 kN
Az = 6.22 kN
B y = 2.80 kN
Bz = 2.90 kN
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Sample Problem 8.3
• Identify critical shaft section from torque and
bending moment diagrams.
(
M y2 + M z2 + T 2
)
max
= 11602 + 3732 + 597 2
= 1357 N ⋅ m
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Sample Problem 8.3
• Calculate minimum allowable shaft diameter.
M y2 + M z2 + T 2
J
=
c
τ all
=
1357 N ⋅ m
= 27.14 × 10− 6 m3
50 MPa
For a solid circular shaft,
J π 3
= c = 27.14 × 10− 6 m3
c 2
c = 0.02585 m = 25.85 m
d = 2c = 51.7 mm
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Stresses Under Combined Loadings
• Wish to determine stresses in slender
structural members subjected to
arbitrary loadings.
• Pass section through points of interest.
Determine force-couple system at
centroid of section required to maintain
equilibrium.
• System of internal forces consist of
three force components and three
couple vectors.
• Determine stress distribution by
applying the superposition principle.
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Stresses Under Combined Loadings
• Axial force and in-plane couple vectors
contribute to normal stress distribution
in the section.
• Shear force components and twisting
couple contribute to shearing stress
distribution in the section.
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Stresses Under Combined Loadings
• Normal and shearing stresses are used to
determine principal stresses, maximum
shearing stress and orientation of principal
planes.
• Analysis is valid only to extent that
conditions of applicability of superposition
principle and Saint-Venant’s principle are
met.
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Sample Problem 8.5
SOLUTION:
• Determine internal forces in Section
EFG.
• Evaluate normal stress at H.
• Evaluate shearing stress at H.
Three forces are applied to a short
steel post as shown. Determine the
principle stresses, principal planes and
maximum shearing stress at point H.
• Calculate principal stresses and
maximum shearing stress.
Determine principal planes.
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Sample Problem 8.5
SOLUTION:
• Determine internal forces in Section EFG.
Vx = −30 kN P = 50 kN Vz = −75 kN
M x = ( 50 kN )( 0.130 m ) − ( 75 kN )( 0.200 m )
= −8.5 kN ⋅ m
M y = 0 M z = ( 30 kN )( 0.100 m ) = 3 kN ⋅ m
Note: Section properties,
A = ( 0.040 m )( 0.140 m ) = 5.6 × 10−3 m 2
1 ( 0.040 m )( 0.140 m ) 3 = 9.15 × 10− 6 m 4
I x = 12
1 ( 0.140 m )( 0.040 m ) 3 = 0.747 × 10 − 6 m 4
I z = 12
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Sample Problem 8.5
• Evaluate normal stress at H.
σy =+
=
P Mz a Mx b
+
−
A
Iz
Ix
( 3 kN ⋅ m )( 0.020 m )
+
5.6 × 10-3 m 2
0.747 × 10− 6 m 4
50 kN
−
( 8.5 kN ⋅ m )( 0.025 m )
9.15 × 10−6 m 4
= ( 8.93 + 80.3 − 23.2 ) MPa = 66.0 MPa
• Evaluate shearing stress at H.
Q = A1 y1 = [ ( 0.040 m )( 0.045 m ) ] ( 0.0475 m )
= 85.5 × 10−6 m3
(
)
(
Vz Q
75 kN ) 85.5 × 10− 6 m3
τ yz =
=
I xt
9.15 × 10 −6 m 4 ( 0.040 m )
(
)
= 17.52 MPa
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Sample Problem 8.5
• Calculate principal stresses and maximum
shearing stress.
Determine principal planes.
τ max = R = 33.02 + 17.522 = 37.4 MPa
σ max = OC + R = 33.0 + 37.4 = 70.4 MPa
σ min = OC − R = 33.0 − 37.4 = −7.4 MPa
tan 2θ p =
CY 17.52
=
2θ p = 27.96°
CD 33.0
θ p = 13.98°
τ max = 37.4 MPa
σ max = 70.4 MPa
σ min = −7.4 MPa
θ p = 13.98°
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