4. Combined Stress and
Failure Theories
330:148 (g)
Machine Design
• When parts have multiple types of
loading or more than one type of stress
from a single load
Nageswara Rao Posinasetti
1
2
Grouping Types of Stresses
Objectives
• Group
stresses by type, separating the stresses into
bending and axial versus shear and torsional stresses.
• Combine like types of stresses in an appropriate
manner.
• Combine different types of stresses, using appropriate
combined stress theories.
• Axial and Bending
• Torsion and Shear
Appendix 1
p 466
• Gain further understanding into how these combined
stresses should be compared to the stress allowables
for the materials being used in the design.
3
Combined Axial and Bending
stresses
4
Example Problem 4-1:
Design of a Short Column with Eccentric Load
• Determine the stress in the 2-inch diameter vertical column shown.
• Summation of stresses taking the
directions into account
– Column with an eccentric load
• S = ± Saxial ± Sbending = ± F ± M
A
Z
– Tensile +
– Compressive 5
6
1
Example Problem 4-1:
Design of a Short Column with Eccentric Load
Example Problem 4-1:
Design a Short Column with Eccentric Load (cont’d.)
• Determine the stress in the 2-inch diameter vertical column shown.
– Bending stress:
Fe
M π D4
S=
=
32
Z
= 10,000lb( 2in)
π (2 in) 4
= 25,460 lb / in 2
32
S = ± S axial ± S bending
(4-1)
S = −3183 lb / in 2 − 25,460 lb / in 2
S = 28,647 lb / in 2
• First, determine stresses.
– Axial stress
S =
• The bending stress and axial stress add on the inner side of the
column.
F
10,000 lb
lb
=
= 3183 2
A
in
π (2 in)2
4
• Stresses subtract on the outer side so we are primarily concerned
about the inner surface.
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8
Example Problem 4-2: Coplanar Shear
Coplanar Shear Stresses
• Determine the stress in the 1/2-inch diameter pin.
• Use vectorial addition
• As the bell crank is free to rotate, both forces create shear stresses in the pin.
– Adding forces vectorially:
Ft
=
Ft
=
(600 lb)2 + (800 lb)2
Ft
S =
9
Shear and Torsion
(perpendicular forces)
(F1 2 + F2 2)
=
1000 lb
1000 lb
F
=
A
π (1/2 in)2
4
= 5093 lb/in2
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Example Problem 4-3: Combined Torsion and Shear
• Use vectorial addition since the stresses
are of the same type
• S = Storsion + Sshear
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• A roller chain system transmits 50 hp at a speed of 300 rpm.
• If the chain sprocket has an effective pitch diameter of 10 inches, calculate the combined
stress in the 1-inch diameter shaft.
12
2
Example Problem 4-3: Combined Torsion and Shear
Example Problem 4-3: Combined Torsion and Shear (cont’d.)
• A roller chain system transmits 50 hp at a speed of 300 rpm.
• If the chain sprocket has an effective pitch diameter of 10 inches, calculate the combined
stress in the 1-inch diameter shaft.
• Driving force in chain:
(2-5)
F =
(2-6)
T
10,500 in-lb
=
r
10 in
2
as: T = Fr
F = 2100 lb
63,000 (hp)
T =
n
T =
• Calculate shear and torsional stresses.
63,000 (50 hp)
300 rpm
− Shear in shaft:
T = 10,500 in-lb
Ss =
F
2100 lb
=
A
π (1 in)2
4
Ss = 2674 lb/in2
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Example Problem 4-3: Combined Torsion and Shear (cont’d)
•
Torsional stress:
Ss =
T
Z'
(Appendix 3)
Z' =
Z' =
π (1 in)3
16
Ss =
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Normal and Shear Stresses
• Mohr’s Circle
• σ = equivalent combined normal stress
• S = normal stress from bending or axial
loads
• SS = shear or torsional stress
π D3
16
= .196 in3
10,500 in-lb
.196 in3
Ss = 53,570 lb/in2
(4-3)
Stotal = Storsion + Sshear
2
S ⎡ 2 ⎛S⎞ ⎤
σ = ± ⎢SS + ⎜ ⎟ ⎥
2 ⎣⎢
⎝ 2 ⎠ ⎦⎥
Stotal = 53,570 lb/in2 + 2,674 lb/in2
Stotal = 56,250 lb/in2
• This value would be compared to shear stress allowable for the shaft
material.
1/2
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Normal and Shear Stresses
•
•
•
•
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Example Problem 4-4: Combined Normal and Shear Stress
• A center mounted chain drive system transmits 20 hp at a speed of 500 rpm.
• If the sprocket has a pitch diameter of 8 inches, would this be an acceptable design if the
shaft is made of hot rolled AISI 1020 steel and a safety factor of 2 based on yield is
desired?
S = F / A for axial loads
S = M / Z for bending loads
SS = F / A for shear loads
SS = T c / J for torsion shear
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3
Example Problem 4-4: Combined Normal and Shear Stress
Example Problem 4-4: Combined Normal and Shear Stress
(cont’d.)
• A center mounted chain drive system transmits 20 hp at a speed of 500 rpm.
• If the sprocket has a pitch diameter of 8 inches, would this be an acceptable design if the
shaft is made of hot rolled AISI 1020 steel and a safety factor of 2 based on yield is
desired?
– Torque on shaft:
63,000 hp
n
63,000 (20 hp )
T =
500 rpm
T =
• Force in chain:
F =
T
2520 in-lb
=
= 630 lb
r
8 in
2
(2-6)
− Bending moment in shaft:
T = 2,520 in − lb
(Appendix 2)
Mm =
Mm =
FL
4
630 lb 12 in
4
Mm = 1,890 in-lb
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Example Problem 4-4: Combined Normal and Shear Stress (cont’d.)
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Example Problem 4-4: Combined Normal and Shear Stress (cont’d.)
• Combining stresses:
• Calculating torsional and bending stresses
(4-4)
– Torsional shear stress:
Ss =
T
Z'
S
⎛
⎛ S ⎞2 ⎞ ½
σ = 2 ± ⎜ Ss2 + ⎜ 2 ⎟ ⎟
⎝
⎝ ⎠ ⎠
(Z' from Appendix 3)
Ss =
Ss =
2
2
9860 lb/in2
⎛
⎛9860 lb/in ⎞ ⎞ ½
+ ⎜(6570 lb/in2) + ⎜
⎟ ⎟
2
2
⎝
⎠ ⎠
⎝
2
2520 in-lb
π D3
16
σ =
σ = 13,150 lb/in2
2520 in-lb
π (1.25 in) 3
(Appendix 4)
16
Ss = 6570 in-lb
– Bending (normal) stress:
S =
M
1890 in-lb
=
Z
π D3
32
S =
(Z from Appendix 3)
Sy = 30,000 lb/in2
Sy
= 15,000 lb/in2
2
• This is greater than the combined stress of 13,150 lb/in2, so it is acceptable.
1890 in-lb
π (1.25 in)3
32
S = 9860 lb/in2
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Combined maximum shear stress
• τ = Maximum combined shear stress
• S = normal stress
• SS = shear stress
2
⎡
⎛S⎞ ⎤
τ = ⎢SS 2 + ⎜ ⎟ ⎥
⎝ 2 ⎠ ⎦⎥
⎣⎢
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Distortion energy theory
• Also called as von Mises theory
• Closely duplicate the failure of ductile
materials under static, repeated and
combined stresses
• To use this determine the two principal
stresses using Mohr’s circle or other means
1/2
2
2
σ' = σ1 + σ 2 − σ1 σ 2
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