9789810679941 sm 09

Prévia do material em texto

Problem 9-1
Prove that the sum of the normal stresses σx + σy = σx' + σy' is constant. See Figs. 9-2a and 9-2b.
Solution:
Stress Transformation Equations: Applying Eqs. 9-1 and 9-3 of the text.
σx'
σx σy+
2
σx σy−
2
cos 2θ( )⋅+ τxy sin 2θ( )⋅+= (1)
σy'
σx σy+
2
σx σy−
2
cos 2θ( )⋅− τxy sin 2θ( )⋅−= (2)
(1) + (2) :
LHS σx' σy'+=
RHS
σx σy+
2
σx σy+
2
+= RHS σx σy+=
Hence,
σx' σy'+ σx σy+= (Q.E.D.)
Problem 9-2
The state of stress at a point in a member is shown on the element. Determine the stress components
acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec.
9.1.
Given: σx 3MPa:= σy 5MPa:= τxy 8− MPa:= φ 40deg:=
Solution: Set ∆A m2:= θ 180deg φ+:=
Force Equilibrium: For the sectioned element,
∆Ax ∆A cos φ( )⋅:= ∆Ay ∆A sin φ( )⋅:=
Fxx σx ∆Ax⋅:= Fxy τxy ∆Ax⋅:=
Fyy σy ∆Ay⋅:= Fyx τxy ∆Ay⋅:=
Given
+ ΣFx'=0; ∆Fx' Fxy sin θ( )⋅+ Fxx cos θ( )⋅+ Fyx cos θ( )⋅+ Fyy sin θ( )⋅+ 0=
+ ΣFy'=0; ∆Fy' Fxy cos θ( )⋅+ Fxx sin θ( )⋅− Fyx sin θ( )⋅− Fyy cos θ( )⋅+ 0=
Guess ∆Fx' 1kN:= ∆Fy' 1kN:=
∆Fx'
∆Fy'
⎛⎜⎜⎝
⎞
⎠
Find ∆Fx' ∆Fy',( ):= ∆Fx'∆Fy'
⎛⎜⎜⎝
⎞
⎠
4052.11−
404.38−
⎛⎜⎝
⎞
⎠ kN=
Normal and Shear Stress: σ
0A
F
A
⎛⎜⎝
⎞
⎠lim→=
σx'
∆Fx'
∆A
:= σx' 4.052− MPa= Ans
τx'y'
∆Fy'
∆A
:= τx'y' 0.404− MPa= Ans
The negative signs indicate that the sense of σx' and τx'y' are opposite to that shown in FBD.
Problem 9-3
The state of stress at a point in a member is shown on the element. Determine the stress components
acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec.
9.1.
Given: σx 0.200− MPa:= σy 0.350MPa:=
φ 50deg:= τxy 0MPa:=
Solution: Set ∆A m2:= θ 180deg φ+:=
Force Equilibrium: For the sectioned element,
∆Ax ∆A cos φ( )⋅:= ∆Ay ∆A sin φ( )⋅:=
Fxx σx ∆Ax⋅:= Fxy τxy ∆Ax⋅:= Fxy 0.00=
Fyy σy ∆Ay⋅:= Fyx τxy ∆Ay⋅:= Fyx 0.00=
Given
+ ΣFx'=0; ∆Fx' Fxy sin θ( )⋅+ Fxx cos θ( )⋅+ Fyx cos θ( )⋅+ Fyy sin θ( )⋅+ 0=
+ ΣFy'=0; ∆Fy' Fxy cos θ( )⋅+ Fxx sin θ( )⋅− Fyx sin φ( )⋅− Fyy cos θ( )⋅+ 0=
Guess ∆Fx' 1kN:= ∆Fy' 1kN:=
∆Fx'
∆Fy'
⎛⎜⎜⎝
⎞
⎠
Find ∆Fx' ∆Fy',( ):= ∆Fx'∆Fy'
⎛⎜⎜⎝
⎞
⎠
122.75
270.82
⎛⎜⎝
⎞
⎠ kN=
Normal and Shear Stress: σ
0A
F
A
⎛⎜⎝
⎞
⎠lim→=
σx'
∆Fx'
∆A
:= σx' 0.123 MPa= Ans
τx'y'
∆Fy'
∆A
:= τx'y' 0.271 MPa= Ans
Problem 9-4
The state of stress at a point in a member is shown on the element. Determine the stress components
acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec.
9.1.
Given: σx 0.650− MPa:= σy 0.400MPa:=
φ 60deg:= τxy 0MPa:=
Solution: Set ∆A m2:= θ 90deg φ+( )−:=
Force Equilibrium: For the sectioned element,
∆Ax ∆A sin φ( )⋅:= ∆Ay ∆A cos φ( )⋅:=
Fxx σx ∆Ax⋅:= Fxy τxy ∆Ax⋅:= Fxy 0.00=
Fyy σy ∆Ay⋅:= Fyx τxy ∆Ay⋅:= Fyx 0.00=
Given
 + ΣFx'=0; ∆Fx' Fxy sin θ( )⋅+ Fxx cos θ( )⋅+ Fyx cos θ( )⋅+ Fyy sin θ( )⋅+ 0=
+ ΣFy'=0; ∆Fy' Fxy cos θ( )⋅+ Fxx sin θ( )⋅− Fyx sin θ( )⋅− Fyy cos θ( )⋅+ 0=
Guess ∆Fx' 1kN:= ∆Fy' 1kN:=
∆Fx'
∆Fy'
⎛⎜⎜⎝
⎞
⎠
Find ∆Fx' ∆Fy',( ):= ∆Fx'∆Fy'
⎛⎜⎜⎝
⎞
⎠
387.50−
454.66
⎛⎜⎝
⎞
⎠ kN=
Normal and Shear Stress: σ
0A
F
A
⎛⎜⎝
⎞
⎠lim→=
σx'
∆Fx'
∆A
:= σx' 0.387− MPa= Ans
τx'y'
∆Fy'
∆A
:= τx'y' 0.455 MPa= Ans
The negative signs indicate that the sense of σx' is opposite to that shown in FBD.
Problem 9-5
The state of stress at a point in a member is shown on the element. Determine the stress components
acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec.
9.1.
Given: σx 60− MPa:= σy 50− MPa:=
φ 30deg:= τxy 28MPa:=
Solution: Set ∆A m2:= θ 180deg φ+:=
Force Equilibrium: For the sectioned element,
∆Ax ∆A cos φ( )⋅:= ∆Ay ∆A sin φ( )⋅:=
Fxx σx ∆Ax⋅:= Fxy τxy ∆Ax⋅:= Fxy 24248.71 kN=
Fyy σy ∆Ay⋅:= Fyx τxy ∆Ay⋅:= Fyx 14000.00 kN=
Given
+ ΣFx'=0; ∆Fx' Fxy sin θ( )⋅+ Fxx cos θ( )⋅+ Fyx cos θ( )⋅+ Fyy sin θ( )⋅+ 0=
 + ΣFy'=0; ∆Fy' Fxy cos θ( )⋅+ Fxx sin θ( )⋅− Fyx sin θ( )⋅− Fyy cos θ( )⋅+ 0=
Guess ∆Fx' 1kN:= ∆Fy' 1kN:=
∆Fx'
∆Fy'
⎛⎜⎜⎝
⎞
⎠
Find ∆Fx' ∆Fy',( ):= ∆Fx'∆Fy'
⎛⎜⎜⎝
⎞
⎠
33251.29−
18330.13
⎛⎜⎝
⎞
⎠ kN=
Normal and Shear Stress: σ
0A
F
A
⎛⎜⎝
⎞
⎠lim→=
σx'
∆Fx'
∆A
:= σx' 33.251− MPa= Ans
τx'y'
∆Fy'
∆A
:= τx'y' 18.330 MPa= Ans
The negative signs indicate that the sense of σx' is opposite to that shown in FBD.
Problem 9-6
The state of stress at a point in a member is shown on the element. Determine the stress components
acting on the inclined plane AB. Solve the problem using the method of equilibrium described in Sec.
9.1.
Given: σx 90MPa:= σy 50MPa:=
τxy 35− MPa:= φ 60deg:=
Solution: Set ∆A m2:= θ 90deg φ+( )−:=
Force Equilibrium: For the sectioned element,
∆Ax ∆A sin φ( )⋅:= ∆Ay ∆A cos φ( )⋅:=
Fxx σx ∆Ax⋅:= Fxy τxy ∆Ax⋅:=
Fyy σy ∆Ay⋅:= Fyx τxy ∆Ay⋅:=
Given
+ ΣFx'=0; ∆Fx' Fxy sin θ( )⋅+ Fxx cos θ( )⋅+ Fyx cos θ( )⋅+ Fyy sin θ( )⋅+ 0=
 + ΣFy'=0; ∆Fy' Fxy cos θ( )⋅+ Fxx sin θ( )⋅− Fyx sin θ( )⋅− Fyy cos θ( )⋅+ 0=
Guess ∆Fx' 1kN:= ∆Fy' 1kN:=
∆Fx'
∆Fy'
⎛⎜⎜⎝
⎞
⎠
Find ∆Fx' ∆Fy',( ):= ∆Fx'∆Fy'
⎛⎜⎜⎝
⎞
⎠
49689.11
34820.51−
⎛⎜⎝
⎞
⎠ kN=
Normal and Shear Stress: σ
0A
F
A
⎛⎜⎝
⎞
⎠lim→=
σx'
∆Fx'
∆A
:= σx' 49.69 MPa= Ans
τx'y'
∆Fy'
∆A
:= τx'y' 34.82− MPa= Ans
The negative signs indicate that the sense of τx'y' isopposite to that shown in FBD.
Problem 9-7
Solve Prob. 9-2 using the stress-transformation equations developed in Sec. 9.2.
Given: σx 5MPa:= σy 3MPa:= τxy 8MPa:= φ 40deg:=
Solution: θ 90deg φ+:=
Normal Stress:
σx'
σx σy+
2
σx σy−
2
cos 2θ( )⋅+ τxy sin 2θ( )⋅+:=
σx' 4.05− MPa= Ans
The negative signs indicate that the sense of σx' is a compressive stress.
Shear Stress:
τx'y'
σx σy−
2
− sin 2θ( )⋅ τxy cos 2θ( )⋅+:=
τx'y' 0.404− MPa= Ans
The negative signs indicate that the sense of τx'y' is in the -y' direction.
Problem 9-8
Solve Prob. 9-4 using the stress-transformation equations developed in Sec. 9.2.
Given: σx 0.650− MPa:= σy 0.400MPa:=
φ 60deg:= τxy 0MPa:=
Solution: θ 90deg φ−:=
Normal Stress:
σx'
σx σy+
2
σx σy−
2
cos 2θ( )⋅+ τxy sin 2θ( )⋅+:=
σx' 0.387− MPa= Ans
The negative signs indicate that the sense of σx' is a compressive stress.
Shear Stress:
τx'y'
σx σy−
2
− sin 2θ( )⋅ τxy cos 2θ( )⋅+:=
τx'y' 0.455 MPa= Ans
Problem 9-9
Solve Prob. 9-6 using the stress-transformation equations developed in Sec. 9.2. Show the result on a
sketch.
Given: σx 90MPa:= σy 50MPa:=
τxy 35− MPa:= φ 60deg:=
Solution: θ 90deg φ+( )−:=
Normal Stress:
σx'
σx σy+
2
σx σy−
2
cos 2θ( )⋅+ τxy sin 2θ( )⋅+:=
σx' 49.69 MPa= Ans
Shear Stress:
τx'y'
σx σy−
2
− sin 2θ( )⋅ τxy cos 2θ( )⋅+:=
τx'y' 34.82− MPa= Ans
The negative signs indicate that the sense of τx'y' is in the -y' direction.
Problem 9-10
Determine the equivalent state of stress on an element if the element is oriented 30° counterclockwise
from the element shown. Use the stress-transformation equations.
Unit Used: kPa 1000Pa:=
Given: σx 0kPa:= σy 300− kPa:=
θ 30deg:= τxy 950kPa:=
Solution:
Normal Stress:
σx'
σx σy+
2
σx σy−
2
cos 2θ( )⋅+ τxy sin 2θ( )⋅+:=
σx' 747.7 kPa= Ans
σy'
σx σy+
2
σx σy−
2
cos 2θ( )⋅− τxy sin 2θ( )⋅−:=
σy' 1047.7− kPa= Ans
Shear Stress:
τx'y'
σx σy−
2
− sin 2θ( )⋅ τxy cos 2θ( )⋅+:=
τx'y' 345.1 kPa= Ans
Problem 9-11
Determine the equivalent state of stress on an element if the element is oriented 60° clockwise from the
element shown.
Given: σx 0.300MPa:= σy 0MPa:=
θ 60− deg:= τxy 0.120MPa:=
Solution:
Normal Stress:
σx'
σx σy+
2
σx σy−
2
cos 2θ( )⋅+ τxy sin 2θ( )⋅+:=
σx' 0.0289− MPa= Ans
σy'
σx σy+
2
σx σy−
2
cos 2θ( )⋅− τxy sin 2θ( )⋅−:=
σy'0.329 MPa= Ans
Shear Stress:
τx'y'
σx σy−
2
− sin 2θ( )⋅ τxy cos 2θ( )⋅+:=
τx'y' 0.0699 MPa= Ans
Problem 9-12
Solve Prob. 9-6 using the stress-transformation equations.
Given: σx 90MPa:= σy 50MPa:=
φ 60deg:= τxy 35− MPa:=
Solution: θ 90deg φ+( )−:=
Normal Stress:
σx'
σx σy+
2
σx σy−
2
cos 2θ( )⋅+ τxy sin 2θ( )⋅+:=
σx' 49.69 MPa= Ans
Shear Stress:
τx'y'
σx σy−
2
− sin 2θ( )⋅ τxy cos 2θ( )⋅+:=
τx'y' 34.82− MPa= Ans
Problem 9-13
The state of stress at a point is shown on the element. Determine (a) the principal stresses and (b) the
maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the
element in each case.
Given: σx 45MPa:= σy 60− MPa:= τxy 30MPa:=
Solution:
(a) Principal Stress:
σ1
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2++:= σ1 52.97 MPa=
Ans
σ2
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+−:= σ2 67.97− MPa=
Ans
Orientation of Principal Stress:
tan 2θp( ) 2τxyσx σy−= θp
1
2
atan
2τxy
σx σy−
⎛⎜⎝
⎞
⎠
:= θ'p θp 90deg−:=
θp 14.87 deg= θ'p 75.13− deg=
Use Eq. 9-1 to determine the principal plane of σ1 and σ2.
σx'
σx σy+
2
σx σy−
2
cos 2θp( )⋅+ τxy sin 2θp( )⋅+:=
σx' 52.97 MPa=
Therefore, θp1 θp:= θp1 14.87 deg= Ans
θp2 θ'p:= θp2 75.13− deg= Ans
(b)
τmax
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+:= τmax 60.47 MPa= Ans
σavg
σx σy+
2
:= σavg 7.50− MPa= Ans
Orientation of Maximum In-plane Shear Stress:
tan 2θs( ) σx σy−2τxy−= θs
1
2
atan
σx σy−
2τxy
−⎛⎜⎝
⎞
⎠
:= θ's θs 90deg+:=
θs 30.13− deg= θ's 59.87 deg=
By observation, in order to preserve equilibrium along AB, τmax has to act in the
direction shown in the figure.
Problem 9-14
The state of stress at a point is shown on the element. Determine (a) the principal stresses and (b) the
maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the
element in each case.
Given: σx 180MPa:= σy 0MPa:= τxy 150− MPa:=
Solution:
(a) Principal Stress:
σ1
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2++:= σ1 264.93 MPa= Ans
σ2
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+−:= σ2 84.93− MPa= Ans
Orientation of Principal Stress:
tan 2θp( ) 2τxyσx σy−= θp
1
2
atan
2τxy
σx σy−
⎛⎜⎝
⎞
⎠
:= θ'p θp 90deg+:=
θp 29.52− deg= θ'p 60.48 deg=
Use Eq. 9-1 to determine the principal plane of σ1 and σ2.
σx'
σx σy+
2
σx σy−
2
cos 2θp( )⋅+ τxy sin 2θp( )⋅+:=
σx' 264.93 MPa=
Therefore, θp1 θp:= θp1 29.52− deg= Ans
θp2 θ'p:= θp2 60.48 deg= Ans
(b)
τmax
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+:= τmax 174.93 MPa= Ans
σavg
σx σy+
2
:= σavg 90.00 MPa= Ans
Orientation of Maximum In-plane Shear Stress:
tan 2θs( ) σx σy−2τxy−= θs
1
2
atan
σx σy−
2τxy
−⎛⎜⎝
⎞
⎠
:= θ's θs 90deg−:=
θs 15.48 deg= θ's 74.52− deg=
By observation, in order to preserve equilibrium along AB, τmax has to act in the
direction shown in the figure.
Problem 9-15
The state of stress at a point is shown on the element. Determine (a) the principal stresses and (b) the
maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the
element in each case.
Given: σx 30− MPa:= σy 0MPa:= τxy 12− MPa:=
Solution:
(a) Principal Stress:
σ1
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2++:= σ1 4.21 MPa= Ans
σ2
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+−:= σ2 34.21− MPa= Ans
Orientation of Principal Stress:
tan 2θp( ) 2τxyσx σy−= θp
1
2
atan
2τxy
σx σy−
⎛⎜⎝
⎞
⎠
:= θ'p θp 90deg−:=
θp 19.33 deg= θ'p 70.67− deg=
Use Eq. 9-1 to determine the principal plane of σ1 and σ2.
σx'
σx σy+
2
σx σy−
2
cos 2θp( )⋅+ τxy sin 2θp( )⋅+:=
σx' 34.21− MPa=
Therefore, θp1 θ'p:= θp1 70.67− deg= Ans
θp2 θp:= θp2 19.33 deg= Ans
(b)
τmax
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+:= τmax 19.21 MPa= Ans
σavg
σx σy+
2
:= σavg 15.00− MPa= Ans
Orientation of Maximum In-plane Shear Stress:
tan 2θs( ) σx σy−2τxy−= θs
1
2
atan
σx σy−
2τxy
−⎛⎜⎝
⎞
⎠
:= θ's θs 90deg+:=
θs 25.67− deg= θ's 64.33 deg=
By observation, in order to preserve equilibrium along AB, τmax has to act in the
direction shown in the figure.
Problem 9-16
The state of stress at a point is shown on the element. Determine (a) the principal stresses and (b) the
maximum in-plane shear stress and average normal stress at the point. Specify the orientation of the
element in each case.
Given: σx 200− MPa:= σy 250MPa:= τxy 175MPa:=
Solution:
(a) Principal Stress:
σ1
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2++:= σ1 310.04 MPa=
Ans
σ2
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+−:= σ2 260.04− MPa=
Ans
Orientation of Principal Stress:
tan 2θp( ) 2τxyσx σy−= θp
1
2
atan
2τxy
σx σy−
⎛⎜⎝
⎞
⎠
:= θ'p θp 90deg+:=
θp 18.94− deg= θ'p 71.06 deg=
Use Eq. 9-1 to determine the principal plane of σ1 and σ2.
σx'
σx σy+
2
σx σy−
2
cos 2θp( )⋅+ τxy sin 2θp( )⋅+:=
σx' 260.04− MPa=
Therefore, θp1 θ'p:= θp1 71.06 deg= Ans
θp2 θp:= θp2 18.94− deg= Ans
(b)
τmax
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+:= τmax 285.04 MPa= Ans
σavg
σx σy+
2
:= σavg 25.00 MPa= Ans
Orientation of Maximum In-plane Shear Stress:
tan 2θs( ) σx σy−2τxy−= θs
1
2
atan
σx σy−
2τxy
−⎛⎜⎝
⎞
⎠
:= θ's θs 90deg−:=
θs 26.06 deg= θ's 63.94− deg=
By observation, in order to preserve equilibrium along AB, τmax has to act in the
direction shown in the figure.
Problem 9-17
A point on a thin plate is subjected to the two successive states of stress shown. Determine the
resultant state of stress represented on the element oriented as shown on the right.
Given:
(a) σx'_a 200− MPa:= τx'y'_a 0MPa:=
σy'_a 350− MPa:= θa 30− deg:=
(a) σx'_b 0:= τx'y'_b 58MPa:=
σy'_b 0:= θb 25deg:=
Solution:
Stress Transformation Equations: Applying Eqs. 9-1, 9-2, and 9-3 of the text.
For element (a):
σx_a
σx'_a σy'_a+
2
σx'_a σy'_a−
2
cos 2θa( )⋅+ τx'y'_a sin 2θa( )⋅+⎛⎜⎝
⎞
⎠:= σx_a 237.50− MPa=
σy_a
σx'_a σy'_a+
2
σx'_a σy'_a−
2
cos 2θa( )⋅− τx'y'_a sin 2θa( )⋅−⎛⎜⎝
⎞
⎠:= σy_a 312.50− MPa=
τxy_a
σx'_a σy'_a−
2
− sin 2θa( )⋅ τx'y'_a cos 2θa( )⋅+⎛⎜⎝
⎞
⎠:= τxy_a 64.95 MPa=
For element (b):
σx_b
σx'_b σy'_b+
2
σx'_b σy'_b−
2
cos 2θb( )⋅+ τx'y'_b sin 2θb( )⋅+⎛⎜⎝
⎞
⎠:= σx_b 44.43 MPa=
σy_b
σx'_b σy'_b+
2
σx'_b σy'_b−
2
cos 2θb( )⋅− τx'y'_b sin 2θb( )⋅−⎛⎜⎝
⎞
⎠:= σy_b 44.43− MPa=
τxy_b
σx'_b σy'_b−
2
− sin 2θb( )⋅ τx'y'_b cos 2θb( )⋅+⎛⎜⎝
⎞
⎠:= τxy_b 37.28 MPa=
Combining the stress componenets of two elements yields
σx σx_a σx_b+:= σx 193.1− MPa= Ans
σy σy_a σy_b+:= σy 356.9− MPa= Ans
τxy τxy_a τxy_b+:= τxy 102.2 MPa= Ans
Problem 9-18
The steel bar has a thickness of 12 mm and is subjected to the edge loading shown. Determine the
principal stresses developed in the bar.
Given: d 50mm:= t 12mm:=
q 4
kN
m
:= L 500mm:=
Solution:
Normal and Shear Stress:
In accordance with the established sign convention.
σx 0MPa:= σy 0MPa:=
τxy
q
t
:= τxy 0.333 MPa=
In-plane Principal Stress: Apply Eq. 9-5.
σ1
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2++:= σ1 0.333 MPa= Ans
σ2
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+−:= σ2 0.333− MPa= Ans
Problem 9-19
The steel plate has a thickness of 10 mm and is subjected to the edge loading shown. Determine the
maximum in-plane shear stress and the average normal stress developed in the steel.
Given: ax 300mm:= ay 100mm:= t 10mm:=
qx 30
kN
m
:= qy 40
kN
m
:=
Solution:
Normal and Shear Stress:
In accordance with the established sign convention.
σx
qx
t
:= σx 3.00 MPa=
σy
qy
t
:= σy 4.00 MPa=
τxy 0:=
Maximum In-plane Shear Stress: Apply Eq. 9-7.
τmax
σx σy−2
⎛⎜⎝
⎞
⎠
2
τxy2+:= τmax 0.500 MPa= Ans
Average Normal Stress: Apply Eq. 9-8.
σavg
σx σy+
2
:= σavg 3.50 MPa= Ans
Problem 9-20
The stress acting on two planes at a point is indicated. Determine the shear stress on plane a-a and the
principal stresses at the point.
Given: σa 80MPa:= σb 60MPa:=
θ 45deg:= β 60deg:=
Solution:
σx σb sin β( )⋅:=
τxy σb cos β( )⋅:=
Given
σa
σx σy+
2
σx σy−
2
cos 2θ( )⋅+ τxy sin 2θ( )⋅+=
τa
σx σy−
2
− sin 2θ( )⋅ τxy cos 2θ( )⋅+=
Guess σy 1MPa:= τa 1MPa:=
σy
τa
⎛⎜⎜⎝
⎞
⎠
Find σy τa,( ):= σyτa
⎛⎜⎜⎝
⎞
⎠
48.04
1.96−
⎛⎜⎝
⎞
⎠ MPa= Ans
Principal Stress:
σ1
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2++:= σ1 80.06 MPa= Ans
σ2
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+−:= σ2 19.94 MPa= Ans
Problem 9-21
The stress acting on two planes at a point is indicated. Determine the normal stress σb and the principal
stresses at the point.
Given: σa 4MPa:= τx'y' 2− MPa:=
φbb 45deg:= β 60deg:=
Solution:
Stress Transformation Equations :
Applying Eqs. 9-3 and 9-1 with θ φbb− 90deg−:=
σy σa sin β( )⋅:= τxy σa cos β( )⋅:= σx' σb=
Given
σx'
σx σy+
2
σx σy−
2
cos 2θ( )⋅+ τxy sin 2θ( )⋅+=
τx'y'
σx σy−
2
− sin 2θ( )⋅ τxy cos 2θ( )⋅+=
Guess σx 1MPa:= σx' 1MPa:=
σx
σx'
⎛⎜⎜⎝
⎞
⎠
Find σx σx',( ):= σxσx'
⎛⎜⎜⎝
⎞
⎠
7.464
7.464
⎛⎜⎝
⎞
⎠ MPa=
σb σx':=
σb 7.464 MPa= Ans
In-plane Principal Stress: Applying Eq. 9-5,
σ1
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2++:= σ1 8.29 MPa= Ans
σ2
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+−:= σ2 2.64 MPa= Ans
Problem 9-22
The clamp bears down on the smooth surface at E by tightening the bolt. If the tensile force in the bolt
is 40 kN, determine the principal stresses at points A and B and show the results on elements located at
each of these points. The cross-sectional area at A and B is shown in the adjacent figure.
Given: h 50mm:= t 30mm:= P 40kN:=
a 100mm:= b 200mm:=
Solution: L 3a b+:=
Support Reactions :
+ ΣΜO=0; E' L⋅ P a b+( )⋅− 0=
E'
P a b+( )⋅
L
:= E' 24 kN=
Internal Force and Moment : At Section A-B:
+ ΣFx=0; E V+ 0= V E'−:=
+ ΣΜO=0; M E a( )⋅− 0= M E' a⋅:=
Section Property :
A h t⋅:= A 1500 mm2=
I
1
12
t⋅ h3⋅:= I 312500 mm4=
QB 0.5 h⋅ t⋅( ) 0.25 h⋅( ):= QB 9375 mm3=
QA 0:= (since A' = 0)
Normal Stress: σ M c⋅
I
=
cA 0.5− h:= σA
M cA⋅
I
:= σA 192.00− MPa=
cB 0:= σB
M cB⋅
I
:= σB 0 MPa=
Shear Stress : τ V Q⋅
I t⋅=
τA
V QA⋅
I t⋅:= τA 0.00 MPa=
τB
V QB⋅
I t⋅:= τB 24.00− MPa=
In-plane Principal Stress:
At A: σxA σA:= σyA 0:= τxy τA:=
Since no shear stress acts upon the element,
σA1 σyA:= σA1 0 MPa= Ans
σA2 σxA:= σA2 192− MPa= Ans
At B: σxB σB:= σyB 0:= τxy τB:=
2
2
2.1 22 xy
yxyx τσσσσσ +⎟⎟⎠
⎞
⎜⎜⎝
⎛ −±+=
σB1 τxy−:= σB1 24 MPa= Ans
σB2 τxy:= σB2 24− MPa= Ans
Orientation of Principal Plane: Applying Eq. 9-4 for point B,
tan 2θp( ) 2τBσxB σyB−−= θp
1
2
90deg( ):= θ'p θp 90deg−:=
θp 45 deg= θ'p 45− deg=
Use Eq. 9-1 to determine the principal plane of σ1 and σ2.
σx'_B
σxB σyB+
2
σxB σyB−
2
cos 2θp( )⋅+ τB sin 2θp( )⋅+:=
σx'_B 24.00− MPa=
Therefore, θp1 θ'p:= θp1 45.00− deg= Ans
θp2 θp:= θp2 45.00 deg= Ans
Problem 9-23
Solve Prob. 9-22 for points C and D.
Given: h 50mm:= t 30mm:= P 40kN:=
a 100mm:= b 200mm:= hD 10mm:=
Solution: L 3a b+:=
Support Reactions :
+ ΣΜE=0; P 2a( )⋅ R L⋅− 0=
R
2P a⋅
L
:= R 16 kN=
Internal Force and Moment : At Section C-D:
+ ΣFx=0; R V− 0= V R:=
+ ΣΜO=0; M R b( )⋅− 0= M R b( )⋅:=
Section Property :
A h t⋅:= A 1500 mm2=
I
1
12
t⋅ h3⋅:= I 312500 mm4=
QD hD t⋅( ) 0.5 h⋅ 0.5 hD⋅−( ):= QD 6000 mm3=
QC 0:= (since A' = 0)
Normal Stress: σ M c⋅
I
=
cC 0.5h:= σC
M cC⋅
I
:= σC 256.00 MPa=
cD 0.5− h⋅ hD+:= σD
M cD⋅
I
:= σD 153.6− MPa=
Shear Stress : τ V Q⋅
I t⋅=
τC
V QC⋅
I t⋅:= τC 0.00 MPa=
τD
V QD⋅
I t⋅:= τD 10.24 MPa=
In-plane Principal Stress:
At C: σxC 0:= σyC σC:= τxy τC:=
Since no shear stress acts upon the element,
σC1 σyC:= σC1 256 MPa= Ans
σC2 σxC:= σC2 0 MPa= Ans
At D: σxD 0:= σyD σD:= τxy τD:=
σD1
σxD σyD+
2
σxD σyD−
2
⎛⎜⎝
⎞
⎠
2
τD2++:= σD1 0.680 MPa= Ans
σD2
σxD σyD+
2
σxD σyD−
2
⎛⎜⎝
⎞
⎠
2
τD2+−:= σD2 154.28− MPa= Ans
Orientation of Principal Plane: Applying Eq. 9-4 for point D,
tan 2θp( ) 2τDσxD σyD−= θp
1
2
atan
2τD
σxD σyD−
⎛⎜⎝
⎞
⎠
:= θ'p θp 90deg−:=
θp 3.797 deg= θ'p 86.203− deg=
Use Eq. 9-1 to determine the principal plane of σ1 and σ2.
σx'_D
σxD σyD+
2
σxD σyD−
2
cos 2θp( )⋅+ τD sin 2θp( )⋅+:=
σx'_D 0.680 MPa=
Therefore, θp1 θp:= θp1 3.80 deg= Ans
θp2 θ'p:= θp2 86.20− deg= Ans
Problem 9-24
The grains of wood in the board make an angle of 20° with the horizontal as shown. Determine the
normal and shear stress that act perpendicular to the grains if the board is subjected to an axial load of
250 N.
Unit Used: kPa 1000Pa:=
Given: P 250N:= φ 20deg:=
h 60mm:= t 25mm:=
Solution:
σx
P
h t⋅:= σx 0.1667 MPa=
σy 0:=
τxy 0:=
θ 90deg φ+:=
σx'
σx σy+
2
σx σy−
2
cos 2θ( )⋅+ τxy sin 2θ( )⋅+:= σx' 19.50 kPa= Ans
τx'y'
σx σy−
2
− sin 2θ( )⋅ τxy cos 2θ( )⋅+:= τx'y' 53.57 kPa= Ans
Problem 9-25
The wooden block will fail if the shear stress acting along the grain is 3.85 MPa. If the normal stress
σx = 2.8 MPa, determine the necessary compressive stress σy that will cause failure.
Given: σx 2.8MPa:= τxy 0MPa:=
θgrain 58deg:= τx'y' 3.85MPa:=
Solution: θ θgrain 90deg+:=
Shear Stress:
τx'y'
σx σy−
2
− sin 2θ( )⋅ τxy cos 2θ( )⋅+=
τx'y'
σx σy−
2
− sin 2θ( )⋅=
σy
2τx'y'
sin 2θ( ) σx+:=
σy 5.767− MPa= Ans
Problem 9-26
The T-beam is subjected to the distributed loading that is applied along its centerline. Determine the
principal stresses at points A and B and show the results on elements located at each of these points.
Given: bf 150mm:= tf 20mm:=
dw 150mm:= tw 20mm:=
a 2m:= b 1m:=
hB 50mm:= w 12 kN
m
:=
Solution:
Internal Force and Moment : At Section A-B:
+ ΣFy=0; V w a⋅− 0= V w a⋅:=
+ ΣΜA=0; M w a⋅ 0.5a b+( )⋅− 0= M w a⋅ 0.5a b+( )⋅:=
Section Property : D dw tf+:=
yc
0.5tf bf tf⋅( ) 0.5dw tf+( ) dw tw⋅( )+
bf tf⋅ dw tw⋅+
:= yc 52.50 mm=
I1
1
12
bf⋅ tf3⋅ bf tf⋅( ) 0.5tf yc−( )2⋅+:=
I2
1
12
tw⋅ dw3⋅ dw tw⋅( ) 0.5dw tf+ yc−( )2⋅+:=
I I1 I2+:= I 16562500.00 mm4=
QA 0:= (since A' = 0)
QB hB tw⋅( ) D yc− 0.5 hB⋅−( ):= QB 92500 mm3=
Normal Stress: σ M c⋅
I
=
cA yc:= σA
M cA⋅
I
:= σA 152.15 MPa=
cB D yc− hB−( )−:= σB M cB⋅I:= σB 195.62− MPa=
Shear Stress : τ V Q⋅
I t⋅= τA
V QA⋅
I bf⋅
:= τA 0.00 MPa=
τB
V QB⋅
I tw⋅
:= τB 6.702 MPa=
In-plane Principal Stress:
At A: σxA σA:= σyA 0:= τxy τA:=
Since no shear stress acts upon the element,
σA1 σxA:= σA1 152.15 MPa= Ans
σA2 σyA:= σA2 0 MPa= Ans
At B: σxB σB:= σyB 0:= τxy τB:=
σB1
σxB σyB+
2
σxB σyB−
2
⎛⎜⎝
⎞
⎠
2
τB2++:= σB1 0.229 MPa= Ans
σB2
σxB σyB+
2
σxB σyB−
2
⎛⎜⎝
⎞
⎠
2
τB2+−:= σB2 195.852− MPa= Ans
Orientation of Principal Plane: Applying Eq. 9-4 for point B,
tan 2θp( ) 2τBσxB σyB−−= θp
1
2
atan
2− τB
σxB σyB−
⎛⎜⎝
⎞
⎠
:= θ'p θp 90deg−:=
θp 1.96 deg= θ'p 88.04− deg=
Use Eq. 9-1 to determine the principal plane of σ1 and σ2.
σx'_B
σxB σyB+
2
σxB σyB−
2
cos 2θp( )⋅+ τB sin 2θp( )⋅+:=
σx'_B 194.94− MPa=
Therefore, θp1 θ'p:= θp1 88.04− deg= Ans
θp2 θp:= θp2 1.96 deg= Ans
Problem 9-27
The bent rod has a diameter of 15 mm and is subjected to the force of 600 N. Determine the principal
stresses and the maximum in-plane shear stress that are developed at point A and point B. Show the
results on properly oriented elements located at these points.Given: do 15mm:= a 50mm:= P 0.6kN:=
Solution:
Internal Force and Moment : At Section A-B:
+ ΣFx=0; N P− 0= N P:=
+ ΣΜO=0; M P a( )⋅− 0= M P a⋅:=
Section Property :
A
π do2⋅
4
:= A 176.71 mm2=
I
π do4⋅
64
:= I 2485.05 mm4=
I
Mc
A
N ±=2.1σNormal Stress:
cA 0.5do:= σA
N
A
M cA⋅
I
−:= σA 87.15− MPa=
cB 0.5do:= σB
N
A
M cB⋅
I
+:= σB 93.94 MPa=
In-plane Principal Stress:
At A: σxA σA:= σyA 0:= τxy 0:=
Since no shear stress acts upon the element,
σA1 σyA:= σA1 0 MPa= Ans
σA2 σxA:= σA2 87.15− MPa= Ans
At B: σxB σB:= σyB 0:= τxy 0:=
Since no shear stress acts upon the element,
σB1 σxB:= σB1 93.94 MPa= Ans
σB2 σyB:= σB2 0 MPa= Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7
τA.max
σxA σyA−
2
⎛⎜⎝
⎞
⎠
2
τxy2+:= τA.max 43.6 MPa= Ans
τB.max
σxB σyB−
2
⎛⎜⎝
⎞
⎠
2
τxy2+:= τB.max 47.0 MPa= Ans
Orientation of Plane for Maximum In-plane Shear Stress: Applying Eq. 9-6
∞=)θ2tan( S_A θs_A 45deg:= Anstan 2θs_A( ) σxA σyA−2τxy−= Ansθ's_A θs_A 90deg−:= θ's_A 45− deg= Ans
∞=)θ2tan( S_Btan 2θs_B( ) σxB σyB−2τxy−= θs_B 45− deg:= Ans
Ansθ's_B θs_B 90deg+:= θ's_B 45 deg= Ans
By observation, in order to preserve equilibrium along AB, τmax has to act in the
direction shown in the figure.
Average Normal Stress: Applying Eq. 9-8
σavg_A
σxA σyA+
2
:= σavg_A 43.57− MPa=
σavg_B
σxB σyB+
2
:= σavg_B 46.97 MPa=
Problem 9-28
The simply supported beam is subjected to the traction stress τ0 on its top surface. Determine the
principal stresses at points A and B.
Problem 9-29
The beam has a rectangular cross section and is subjected to the loadings shown. Determine the
principal stresses and the maximum in-plane shear stress that are developed at point A and point B.
These points are just to the left of the 10-kN load. Show the results on properly oriented elements
located at these points.
Given: b 150mm:= d 375mm:=
F 10kN:= P 5kN:= L 1.2m:=
Solution:
Support Reactions : By symmetry, R1=R ; R2= R
+ ΣFy=0; 2R F− 0= R 0.5F:=
+ ΣFx=0; H1 P− 0= H1 P:=
Internal Force and Moment : At Section A-B:
+ ΣFx=0; H1 N+ 0= N H1−:=
+ ΣFy=0; R V+ 0= V R−:=
+ ΣΜO=0; M R 0.5L( )⋅− 0= M 0.5R L⋅:=
Section Property :
A b d⋅:= I 1
12
b⋅ d3⋅:=
QA 0:= QB 0:= (since A' = 0)
Normal Stress: σ N
A
M c⋅
I
+=
cA 0.5− d:= σA
N
A
M cA⋅
I
+:= σA 0.942− MPa=
cB 0.5d:= σB
N
A
M cB⋅
I
+:= σB 0.764 MPa=
Shear Stress : Since QA = QB = 0, τA 0:= τB 0:=
In-plane Principal Stress:
At A: σxA σA:= σyA 0:= τxy 0:=
Since no shear stress acts upon the element,
σA1 σyA:= σA1 0 MPa= Ans
σA2 σxA:= σA2 0.942− MPa= Ans
At B: σxB σB:= σyB 0:= τxy 0:=
Since no shear stress acts upon the element,
σB1 σxB:= σB1 0.764 MPa= Ans
σB2 σyB:= σB2 0 MPa= Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
τmax_A
σxA σyA−
2
⎛⎜⎝
⎞
⎠
2
τxy2+:= τmax_A 0.471 MPa= Ans
τmax_B
σxB σyB−
2
⎛⎜⎝
⎞
⎠
2
τxy2+:= τmax_B 0.382 MPa= Ans
Orientation of Plane for Maximum In-plane Shear Stress: Applying Eq. 9-6
∞=)θ2tan( S_A θs_A 45deg:= Anstan 2θs_A( ) σxA σyA−2τxy−= Ansθ's_A θs_A 90deg−:= θ's_A 45− deg= Ans
∞=)θ2tan( S_Btan 2θs_B( ) σxB σyB−2τxy−= θs_B 45− deg:= Ans
Ansθ's_B θs_B 90deg+:= θ's_B 45 deg= Ans
By observation, in order to preserve equilibrium along AB, τmax has to act in the
direction shown in the figure.
Average Normal Stress: Applying Eq. 9-8,
σavg_A
σxA σyA+
2
:= σavg_A 0.471− MPa= Ans
σavg_B
σxB σyB+
2
:= σavg_B 0.382 MPa= Ans
Problem 9-30
The wide-flange beam is subjected to the loading shown. Determine the principal stress in the beam at
point A and at point B. These points are located at the top and bottom of the web, respectively.
Although it is not very accurate, use the shear formula to compute the shear stress.
Given: bf 200mm:= tf 10mm:=
tw 10mm:= dw 200mm:=
P 25kN:= θ 30deg:=
a 3m:= w 8 kN
m
:=
Solution:
Internal Force and Moment : At Section A-B:
+ ΣFx=0; P cos θ( )⋅ N− 0= N P cos θ( )⋅:=
+ ΣFy=0; V P sin θ( )⋅− w a⋅− 0=
V P sin θ( )⋅ w a⋅+:=
+ ΣΜO=0; M P sin θ( )⋅ a( )⋅− w a⋅( ) 0.5a( )⋅− 0=
M P a⋅ sin θ( )⋅ 0.5w a2⋅+:=
Section Property : D dw 2tf+:=
A bf D⋅ bf tw−( ) dw⋅−:= A 6000 mm2=
I
1
12
bf D
3⋅ bf tw−( ) dw3⋅−⎡⎣ ⎤⎦⋅:= I 50.80 10 6−× m4=
QA bf tf⋅( ) D2
tf
2
−⎛⎜⎝
⎞
⎠⋅:= QA 210000 mm
3= QB QA:=
Normal Stress: σ N
A
M c⋅
I
+=
cA 0.5dw:= σA
N
A
M cA⋅
I
+:= σA 148.293 MPa=
cB 0.5− dw:= σB
N
A
M cB⋅
I
+:= σB 141.077− MPa=
Shear Stress : τ V Q⋅
I t⋅=
τA
V QA⋅
I tw⋅
:= τA 15.09 MPa=
τB
V QB⋅
I tw⋅
:= τB 15.09 MPa=
In-plane Principal Stress:
At A: σxA σA:= σyA 0:= τxy τA:=
σA1
σxA σyA+
2
σxA σyA−
2
⎛⎜⎝
⎞
⎠
2
τA2++:= σA1 149.8 MPa= Ans
σA2
σxA σyA+
2
σxA σyA−
2
⎛⎜⎝
⎞
⎠
2
τA2+−:= σA2 1.52− MPa= Ans
At B: σxB σB:= σyB 0:= τxy τB:=
σB1
σxB σyB+
2
σxB σyB−
2
⎛⎜⎝
⎞
⎠
2
τB2++:= σB1 1.60 MPa= Ans
σB2
σxB σyB+
2
σxB σyB−
2
⎛⎜⎝
⎞
⎠
2
τB2+−:= σB2 142.7− MPa= Ans
Problem 9-31
The shaft has a diameter d and is subjected to the loadings shown. Determine the principal stresses and
the maximum in-plane shear stress that is developed anywhere on the surface of the shaft.
Problem 9-32
A paper tube is formed by rolling a paper strip in a spiral and then gluing the edges together as shown.
Determine the shear stress acting along the seam, which is at 30° from the vertical, when the tube is
subjected to an axial force of 10 N. The paper is 1 mm thick and the tube has an outer diameter of 30
mm.
Unit used: kPa 1000Pa:=
Given: do 30mm:= t 1mm:=
θ 30deg:= P 10N:=
Solution:
Section Property : di do 2t−:=
A
π
4
do
2 di
2−⎛⎝ ⎞⎠⋅:= A 91.11 mm2=
Normal Stress:
σx
P
A
:= σx 109.76 kPa=
σy 0:=
τxy 0:=
Shear stress along the seam:
τx'y'
σx σy−
2
− sin 2θ( )⋅ τxy cos 2θ( )⋅+:=
τx'y' 47.53− kPa= Ans
Problem 9-33
Solve Prob. 9-32 for the normal stress acting perpendicular to the seam.
Unit used: kPa 1000Pa:=
Given: do 30mm:= t 1mm:=
θ 30deg:= P 10N:=
Solution:
Section Property : di do 2t−:=
A
π
4
do
2 di
2−⎛⎝ ⎞⎠⋅:= A 91.11 mm2=
Normal Stress:
σx
P
A
:= σx 109.76 kPa=
σy 0:=
τxy 0:=
Normal stress perpendicular to the seam:
σx'
σx σy+
2
σx σy−
2
cos 2θ( )⋅+ τxy sin 2θ( )⋅+:=
σx' 82.32 kPa= Ans
Problem 9-34
The shaft has a diameter d and is subjected to the loadings shown. Determine the principal stresses and
the maximum in-plane shear stress that is developed at point A. The bearings only support vertical
reactions.
Problem 9-35
The drill pipe has an outer diameter of 75 mm, a wall thickness of 6 mm, and a weight of 0.8 kN/m. If
it is subjected to a torque and axial load as shown, determine (a) the principal stresses and (b) the
maximum in-plane shear stress at a point on its surface at section a.
Given: do 75mm:= t 6mm:= L 6m:=
P 7.5kN:= Mx 1.2kN m⋅:= w 0.8
kN
m
:=
Solution:
Internal Force and Moment : At section a:
ΣFx=0; N P+ w L⋅+ 0= N P− w L⋅−:=
ΣΜx=0; T Mx− 0= T Mx:=
Section Property : di do 2t−:=
A
π
4
do
2 di
2−⎛⎝ ⎞⎠⋅:= J π32 do
4 di
4−⎛⎝ ⎞⎠⋅:=
Normal Stress: σ N
A
:= σ 9.457− MPa=
Shear Stress :
c 0.5do:= τ
T c⋅
J
:= τ 28.850 MPa=
a) In-plane Principal Stresses:
σx 0:= σy σ:= τxy τ:=
for any point on the shaft's surface. Applying Eq. 9-5,
σ1
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2++:= σ1 24.51 MPa= Ans
σ2
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+−:= σ2 33.96− MPa= Ans
b) Maximum In-plane Shear Stress: Applying Eq. 9-7,
τmax
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+:= τmax 29.24 MPa= Ans
Problem 9-36
The internal loadings at a section of the beam are shown. Determine the principal stresses at point A.
Also computethe maximum in-plane shear stress at this point.
Given: bf 200mm:= tf 50mm:=
tw 50mm:= dw 200mm:=
Px 500− kN:= My 30− kN m⋅:=
Py 800− kN:= Mz 40kN m⋅:=
Solution:
Section Property : D dw 2tf+:=
A bf D⋅ bf tw−( ) dw⋅−:= A 30000 mm2=
Iz
1
12
bf D
3⋅ bf tw−( ) dw3⋅−⎡⎣ ⎤⎦⋅:= Iz 350.00 10 6−× m4=
Iy
1
12
2tf bf
3⋅ dw tw3⋅+⎛⎝ ⎞⎠⋅:= Iy 68.75 10 6−× m4=
QA 0:= (since A' = 0)
Normal Stress: yA 0.5D:= zA 0.5bf:=
σA
Px
A
Mz yA⋅
Iz
−
My zA⋅
Iy
+:= σA 77.446− MPa=
Shear Stress : Since QA = 0, τA 0:=
In-plane Principal Stress:
σx σA:= σy 0:= τxy 0:=
Since no shear stress acts upon the element,
σ1 σy:= σ1 0 MPa= Ans
σ2 σx:= σ2 77.45− MPa= Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
τmax
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+:= τmax 38.72 MPa= Ans
Problem 9-37
Solve Prob. 9-36 for point B.
Given: bf 200mm:= tf 50mm:=
tw 50mm:= dw 200mm:=
Px 500− kN:= My 30− kN m⋅:=
Py 800− kN:= Mz 40kN m⋅:=
Solution:
Section Property : D dw 2tf+:=
A bf D⋅ bf tw−( ) dw⋅−:= A 30000 mm2=
Iz
1
12
bf D
3⋅ bf tw−( ) dw3⋅−⎡⎣ ⎤⎦⋅:= Iz 350.00 10 6−× m4=
Iy
1
12
2tf bf
3⋅ dw tw3⋅+⎛⎝ ⎞⎠⋅:= Iy 68.75 10 6−× m4=
QB 0:= (since A' = 0)
Normal Stress: yB 0.5− D:= zB 0.5− bf:=
σB
Px
A
Mz yB⋅
Iz
−
My zB⋅
Iy
+:= σB 44.113 MPa=
Shear Stress : Since QB = 0, τB 0:=
In-plane Principal Stress:
σx σB:= σy 0:= τxy 0:=
Since no shear stress acts upon the element,
σ1 σx:= σ1 44.113 MPa= Ans
σ2 σy:= σ2 0.00 MPa= Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
τmax
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+:= τmax 22.06 MPa= Ans
Problem 9-38
Solve Prob. 9-36 for point C, located in the center on the bottom of the web.
Given: bf 200mm:= tf 50mm:=
tw 50mm:= dw 200mm:=
Px 500− kN:= My 30− kN m⋅:=
Py 800− kN:= Mz 40kN m⋅:=
Solution:
Section Property : D dw 2tf+:=
A bf D⋅ bf tw−( ) dw⋅−:= A 30000 mm2=
Iz
1
12
bf D
3⋅ bf tw−( ) dw3⋅−⎡⎣ ⎤⎦⋅:= Iz 350.00 10 6−× m4=
Iy
1
12
2tf bf
3⋅ dw tw3⋅+⎛⎝ ⎞⎠⋅:= Iy 68.75 10 6−× m4=
QC bf tf⋅( ) D2
tf
2
−⎛⎜⎝
⎞
⎠⋅:= QC 1250000 mm
3=
Normal Stress: yC 0.5− dw:= zC 0:=
σC
Px
A
Mz yC⋅
Iz
−
My zC⋅
Iy
+:= σC 5.238− MPa=
Shear Stress : τ V Q⋅
I t⋅=
τC
Py QC⋅
Iz tw⋅
:= τC 57.14− MPa=
In-plane Principal Stress:
σx σC:= σy 0:= τxy τC:=
σ1
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2++:= σ1 54.58 MPa= Ans
σ2
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+−:= σ2 59.82− MPa= Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
τmax
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+:= τmax 57.20 MPa= Ans
Problem 9-39
The wide-flange beam is subjected to the 50-kN force. Determine the principal stresses in the beam at
point A located on the web at the bottom of the upper flange. Although it is not very accurate, use the
shear formula to calculate the shear stress.
Given: bf 200mm:= tf 12mm:=
tw 10mm:= dw 250mm:=
P 50kN:= a 3m:=
Solution:
Internal Force and Moment : At Section A-B:
+ ΣFy=0; V P− 0= V P:=
+ ΣΜO=0; M P a( )⋅− 0= M P a⋅:=
Section Property : D dw 2tf+:=
A bf D⋅ bf tw−( ) dw⋅−:= A 7300 mm2=
I
1
12
bf D
3⋅ bf tw−( ) dw3⋅−⎡⎣ ⎤⎦⋅:= I 95.45 10 6−× m4=
QA bf tf⋅( ) D2
tf
2
−⎛⎜⎝
⎞
⎠⋅:= QA 314400 mm
3=
Normal Stress: σ M c⋅
I
=
cA 0.5dw:= σA
M cA⋅
I
:= σA 196.435 MPa=
Shear Stress : τ V Q⋅
I t⋅=
τA
V QA⋅
I tw⋅
:= τA 16.47 MPa=
In-plane Principal Stress:
σx σA:= σy 0:= τxy τA:=
σ1
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2++:= σ1 197.81 MPa= Ans
σ2
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+−:= σ2 1.37− MPa= Ans
Problem 9-40
Solve Prob. 9-39 for point B located on the web at the top of the bottom flange.
Given: bf 200mm:= tf 12mm:=
tw 10mm:= dw 250mm:=
P 50kN:= a 3m:=
Solution:
Internal Force and Moment : At Section A-B:
+ ΣFy=0; V P− 0= V P:=
+ ΣΜO=0; M P a( )⋅− 0= M P a⋅:=
Section Property : D dw 2tf+:=
A bf D⋅ bf tw−( ) dw⋅−:= A 7300 mm2=
I
1
12
bf D
3⋅ bf tw−( ) dw3⋅−⎡⎣ ⎤⎦⋅:= I 95.45 10 6−× m4=
QB bf tf⋅( ) D2
tf
2
−⎛⎜⎝
⎞
⎠⋅:= QB 314400 mm
3=
Normal Stress: σ M c⋅
I
=
cB 0.5− dw:= σB
M cB⋅
I
:= σB 196.435− MPa=
Shear Stress : τ V Q⋅
I t⋅=
τB
V QB⋅
I tw⋅
:= τB 16.47 MPa=
In-plane Principal Stress:
σx σB:= σy 0:= τxy τB:=
σ1
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2++:= σ1 1.37 MPa= Ans
σ2
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+−:= σ2 197.81− MPa= Ans
Problem 9-41
The bolt is fixed to its support at C. If a force of 90 N is applied to the wrench to tighten it, determine
the principal stresses developed in the bolt shank at point A. Represent the results on an element located
at this point. The shank has a diameter of 6 mm.
Given: do 6mm:= a 150mm:= L 50mm:=
P 90N:=
Solution:
Internal Force and Moment : At section AB:
Mx P L⋅:=
Ty P a⋅:=
Section Property :
I
π
64
do
4⋅:= J π
32
do
4⋅:= cA 0.5do:=
Normal Stress: σA
Mx cA⋅
I
:= σA 212.21 MPa=
Shear Stress : τA
Ty cA⋅
J
:= τA 318.31 MPa=
In-plane Principal Stresses: Applying Eq. 9-5,
σx σA:= σy 0:= τxy τA:=
σ1
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2++:= σ1 441.63 MPa= Ans
σ2
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+−:= σ2 229.42− MPa= Ans
Orientation of Principal Stress:
tan 2θp( ) 2τxyσx σy−= θp
1
2
atan
2τxy
σx σy−
⎛⎜⎝
⎞
⎠
:= θ'p θp 90deg−:=
θp 35.783 deg= θ'p 54.217− deg=
Use Eq. 9-1 to determine the principal plane of σ1 and σ2.
σx'
σx σy+
2
σx σy−
2
cos 2θp( )⋅+ τxy sin 2θp( )⋅+:=
σx' 441.63 MPa=
Therefore, θp1 θp:= θp1 35.78 deg= Ans
θp2 θ'p:= θp2 54.22− deg= Ans
Problem 9-42
Solve Prob. 9-41 for point B.
Given: do 6mm:= a 150mm:= L 50mm:=
P 90N:=
Solution:
Internal Force and Moment : At section AB:
Mx P L⋅:=
Ty P a⋅:=
Vz P:=
Section Property : ro 0.5do:=
I
π
64
do
4⋅:= J π
32
do
4⋅:=
QB
4ro
3π
π ro2⋅
2
⎛⎜⎝
⎞
⎠⋅:=
Normal Stress: cBσ 0:= σB
Mx cBσ⋅
I
:= σB 0 MPa=
Shear Stress : bB do:= cBτ ro:=
τB
Vz QB⋅
I bB⋅
Ty cBτ⋅
J
−:= τB 314.07− MPa=
In-plane Principal Stresses: Applying Eq. 9-5,
σx σB:= σy 0:= τxy τB:=
σ1
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2++:= σ1 314.07 MPa= Ans
σ2
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+−:= σ2 314.07− MPa= Ans
Orientation of Principal Stress:
∞=)θ2tan( Ptan 2θp( ) 2τxyσx σy−= θp1 45deg:= Ans
θp2 θp1 90deg−:= θp2 45− deg= Ans
Problem 9-43
The beam has a rectangular cross section and is subjected to the loadings shown. Determine the
principal stresses that are developed at point A and point B, which are located just to the left of the
20-kN load. Show the results on elements located at these points.
Given: b 100mm:= d 200mm:=
F 20kN:= P 10kN:=
L 4m:=
Solution:
Support Reactions : By symmetry, R1=R ; R2= R
+ ΣFy=0; 2R F− 0= R 0.5F:=
+ ΣFx=0; H1 P− 0= H1 P:=
Internal Force and Moment : At Section A-B:
+ ΣFx=0; H1 N+ 0= N H1−:=
+ ΣFy=0; R V+ 0= V R−:=
+ ΣΜO=0; M R 0.5L( )⋅− 0= M 0.5R L⋅:=
Section Property :
A b d⋅:= I 1
12
b⋅ d3⋅:=
QA 0:= (since A' = 0) QB b 0.5d( )⋅ 0.25d( )⋅:=
Normal Stress: σ N
A
M c⋅
I
+=
cA 0.5− d:= σA
N
A
M cA⋅
I
+:= σA 30.5− MPa=
cB 0:= σB
N
A
M cB⋅
I
+:= σB 0.5− MPa=
Shear Stress :
Since QA = 0, τA 0:=
τB
V QB⋅
I b⋅:= τB 0.75− MPa=
In-plane Principal Stress: Applying Eq. 9-5
At A: σxA σA:= σyA 0:= τxy 0:=
Since no shear stress acts upon the element,
σA1 σyA:= σA1 0 MPa= Ans
σA2 σxA:= σA2 30.50− MPa= Ans
At B: σxB σB:= σyB 0:= τxy τB:=
σB1
σxB σyB+
2
σxB σyB−
2
⎛⎜⎝
⎞
⎠
2
τB2++:= σB1 0.541 MPa= Ans
σB2
σxB σyB+
2
σxB σyB−
2
⎛⎜⎝
⎞
⎠
2
τB2+−:= σB2 1.041− MPa= Ans
Orientation of Principal Plane: Applying Eq. 9-4 for point B,
tan 2θp( ) 2τBσxBσyB−= θp
1
2
atan
2τB
σxB σyB−
⎛⎜⎝
⎞
⎠
:= θp 35.783 deg=
θ'p θp 90deg−:= θ'p 54.217− deg=
Use Eq. 9-1 to determine the principal plane of σ1 and σ2.
σx'_B
σxB σyB+
2
σxB σyB−
2
cos 2θp( )⋅+ τB sin 2θp( )⋅+:=
σx'_B 1.04− MPa=
Therefore, θp1 θ'p:= θp1 54.22− deg= Ans
θp2 θp:= θp2 35.78 deg= Ans
Problem 9-44
The solid propeller shaft on a ship extends outward from the hull. During operation it turns at ω = 15
rad/s when the engine develops 900 kW of power. This causes a thrust of F = 1.23 MN on the shaft.
If the shaft has an outer diameter of 250 mm, determine the principal stresses at any point located on
the surface of the shaft.
Given: do 250mm:= L 0.75m:=
F 1230kN:= P 900kW:= ω 15 rad
s
:=
Solution:
Internal Force and Moment : As shown on FBD
To
P
ω:= To 60.00 kN m⋅=
N F−:=
Section Property :
A
π do2⋅
4
:= A 49087.39 mm2=
J
π do4⋅
32
:= J 383495196.97 mm4=
Normal Stress:
σa
N
A
:= σa 25.06− MPa=
Shear Stress : cmax 0.5 do⋅:=
τo
To cmax⋅
J
:= τo 19.56 MPa=
In-plane Principal Stresses: Applying Eq. 9-5,
σx σa:= σy 0:= τxy τo:=
σ1
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2++:= σ1 10.70 MPa= Ans
σ2
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+−:= σ2 35.75− MPa= Ans
Problem 9-45
The solid propeller shaft on a ship extends outward from the hull. During operation it turns at ω = 15
rad/s when the engine develops 900 kW of power. This causes a thrust of F = 1.23 MN on the shaft.
If the shaft has a diameter of 250 mm, determine the maximum in-plane shear stress at any point
located on the surface of the shaft.
Given: do 250mm:= L 0.75m:=
F 1230kN:= P 900kW:= ω 15 rad
s
:=
Solution:
Internal Force and Moment : As shown on FBD
To
P
ω:= To 60.00 kN m⋅=
N F−:=
Section Property :
A
π do2⋅
4
:= A 49087.39 mm2=
J
π do4⋅
32
:= J 383495196.97 mm4=
Normal Stress:
σa
N
A
:= σa 25.06− MPa=
Shear Stress : cmax 0.5 do⋅:=
τo
To cmax⋅
J
:= τo 19.56 MPa=
Maximum In-plane Shear Stress: Applying Eq. 9-7
σx σa:= σy 0:= τxy τo:=
τmax
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+:= τmax 23.2 MPa= Ans
L 250mm:=Given: do 75mm:= di 68mm:=
At section AB:
P 100N:= a 300mm:=
Solution:
Internal Force and Moment :
Ans
Tx P a⋅:=
My P L⋅:=
Vz P:=
Section Property : ro 0.5do:= ri 0.5di:=
I
π
64
do
4 di
4−⎛⎝ ⎞⎠⋅:= J π32 do
4 di
4−⎛⎝ ⎞⎠⋅:=
QAz
4ro
3π
π ro2⋅
2
⎛⎜⎝
⎞
⎠⋅
4ri
3π
π ri2⋅
2
⎛⎜⎝
⎞
⎠⋅−:=
Normal Stress: cAσ 0:= σA
My cAσ⋅
I
:= σA 0 MPa=
Shear Stress : bA do di−:= cAτ ro:=
τA
Vz QAz⋅
I bA⋅
Tx cAτ⋅
J
−:= τA 0.863− MPa=
In-plane Principal Stresses: Applying Eq. 9-5,
σx σA:= σz 0:= τxz τA:=
σ1
σx σz+
2
σx σz−
2
⎛⎜⎝
⎞
⎠
2
τxz2++:= σ1 0.863 MPa= Ans
σ2
σx σz+
2
σx σz−
2
⎛⎜⎝
⎞
⎠
2
τxz2+−:= σ2 0.863− MPa=
Problem 9-46
The steel pipe has an inner diameter of 68mm and an outer diameter of 75 mm. If it is fixed at C and
subjected to the horizontal 100-N force acting on the handle of the pipe wrench at its end, determine
the principal stresses in the pipe at point A which is located on the surface of the pipe.
L 250mm:=Given: do 75mm:= di 68mm:=
At section AB:
P 100N:= a 300mm:=
Solution:
Internal Force and Moment :
Ans
Tx P a⋅:=
My P L⋅:=
Vz P:=
Section Property : ro 0.5do:= ri 0.5di:=
I
π
64
do
4 di
4−⎛⎝ ⎞⎠⋅:= J π32 do
4 di
4−⎛⎝ ⎞⎠⋅:=
QBz 0:= (Since A'=0)
Normal Stress: cBσ ro:= σB
My cBσ⋅
I
:=
σB 1.862 MPa=
Shear Stress : cBτ ro:= τB
Tx cBτ⋅
J
−:=
τB 1.117− MPa=
In-plane Principal Stresses: Applying Eq. 9-5,
σx σB:= σz 0:= τxz τB:=
σ1
σx σz+
2
σx σz−
2
⎛⎜⎝
⎞
⎠
2
τxz2++:= σ1 2.385 MPa= Ans
σ2
σx σz+
2
σx σz−
2
⎛⎜⎝
⎞
⎠
2
τxz2+−:= σ2 0.523− MPa=
Problem 9-47
Solve Prob. 9-46 for point B, which is located on the surface of the pipe.
Problem 9-48
The cantilevered beam is subjected to the load at its end. Determine the principal stresses in the beam at
points A and B.
Given: b 120mm:= h 150mm:= P 15kN:= L 1.2m:=
yA 45mm:= zA 60mm:= yθ
4−
5
:= zθ
3
5
:=
yB 75mm:= zB 20− mm:=
Solution:
Internal Force and Moment : At Section A-B:
+ ΣFz0; Vz P zθ⋅+ 0= Vz P− zθ⋅:=
+ ΣFy=0; Vy P yθ⋅+ 0= Vy P− yθ⋅:=
Mz P yθ⋅ L⋅:=
My P− zθ⋅ L⋅:=
Section Property :
Iz
1
12
b⋅ h3⋅:= Iz 33.75 10 6−× m4=
Iy
1
12
h⋅ b3⋅:= Iy 21.60 10 6−× m4=
QA.y b 0.5h yA−( )⋅ yA 0.5 0.5h yA−( )+⎡⎣ ⎤⎦⋅:= QA.y 216000 mm3=
QB.z h 0.5b zB−( )⋅ zB 0.5 0.5b zB−( )+⎡⎣ ⎤⎦⋅:= QB.z 240000 mm3=
QA.z 0:= (since A' = 0)
QB.y 0:= (since A' = 0)
Normal Stress:
σA
Mz yA⋅
Iz
−
My zA⋅
Iy
+:= σA 10.8− MPa=
σB
Mz yB⋅
Iz
−
My zB⋅
Iy
+:= σB 42.0 MPa=
Shear Stress : τ V Q⋅
I t⋅=
τA
Vy QA.y⋅
Iz b⋅
:= τA 0.640 MPa=
τB
Vz QB.z⋅
Iy h⋅
:= τB 0.667− MPa=
In-plane Principal Stress: Applying Eq. 9-5
At A: σxA σA:= σyA 0:= τxy 0:=
σA1
σxA σyA+
2
σxA σyA−
2
⎛⎜⎝
⎞
⎠
2
τA2++:= σA1 0.0378 MPa= Ans
σA2
σxA σyA+
2
σxA σyA−
2
⎛⎜⎝
⎞
⎠
2
τA2+−:= σA2 10.84− MPa= Ans
At B: σxB σB:= σzB 0:= τxz τB:=
σB1
σxB σzB+
2
σxB σzB−
2
⎛⎜⎝
⎞
⎠
2
τB2++:= σB1 42.01 MPa= Ans
σB2
σxB σzB+
2
σxB σzB−
2
⎛⎜⎝
⎞
⎠
2
τB2+−:= σB2 0.0106− MPa= Ans
Problem 9-49
The box beam is subjected to the loading shown. Determine the principal stresses in the beam at points
A and B.
Given: bo 200mm:= bi 150mm:=
do 200mm:= di 150mm:=
L1 0.9m:= L2 1.5m:=
P1 4kN:= P2 6kN:=
Solution: L L1 2L2+:=
Support Reactions : Given
+ ΣFy=0; R1 R2+ P1− P2− 0=
+ ΣΜR2=0;
Ans
Guess R1 1kN:= R2 1kN:=
R1
R2
⎛⎜⎜⎝
⎞
⎠
Find R1 R2,( ):= R1R2
⎛⎜⎜⎝
⎞
⎠
8.2
1.8
⎛⎜⎝
⎞
⎠ kN=
Internal Force and Moment : At section A-B:
N 0:=
V P1 R1−:=
M P1 L1 0.5L2+( )⋅ R1 0.5L2( )⋅−:=
Section Property :
I
1
12
bo do
3⋅ bi di3⋅−⎛⎝ ⎞⎠⋅:=
QA 0:= QB 0:= (Since A'=0)
For point A: τA 0:=
cA 0.5 do⋅:= σA
M cA⋅
I
:= σA 0.494 MPa=
σ1 σA:= σ1 0.494 MPa= Ans
σ2 0:= σ2 0.000 MPa= Ans
For point B: τB 0:=
cB 0.5− di⋅:= σB
M cB⋅
I
:= σB 0.37− MPa=
σ1 0:= σ1 0.000 MPa= Ans
σ2 σB:= σ2 0.370− MPa=
P1 L⋅ R1 L L1−( )⋅− P2 L2⋅+ 0=
Problem 9-50
A bar has a circular cross section with a diameter of 25 mm. It is subjected to a torque and a bending
moment. At the point of maximum bending stress the principal stresses are 140 MPa and -70 MPa.
Determine the torque and the bending moment.
Given: do 6mm:=
σ1 140MPa:= σ2 70− MPa:=
Solution:
In-plane Principal Stresses: Applying Eq. 9-5,
Given σy 0:=
σ1
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2++= (1)
σ2
σx σy+
2
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+−= (2)
Solving Eqs. (1) and (2):
Guess σx 2MPa:= τxy 1MPa:=
σx
τxy
⎛⎜⎜⎝
⎞
⎠
Find σx τxy,( ):= σxτxy
⎛⎜⎜⎝
⎞
⎠
70.00
98.99
⎛⎜⎝
⎞
⎠ MPa=
Section Property : ro 0.5do:=
I
π
64
do
4⋅:= J π
32
do
4⋅:=
Normal Stress: σ M c⋅
I
=
c ro:= M
σx I⋅
c
:= M 1.484 N m⋅= Ans
Shear Stress: τ T c⋅
J
=
c ro:= T
τxy J⋅
c
:= T 4.199 N m⋅= Ans
Problem 9-51
The internal loadings at a section of the beam consist of an axial force of 500 N, a shear force of 800
N, and two moment components of 30 N·m and 40 N·m. Determine the principal stresses at point A.
Also calculate the maximum in-plane shear stress at this point.
Unit Used: kPa 1000Pa:=
Given: b 100mm:= h 200mm:= Px 0.5kN:=
yA 100mm:= zA 0mm:= Py 0.8kN:=
My 0.04− kN m⋅:= Mz 0.03kN m⋅:=
Solution:
Section Property :
A b h⋅:= A 20000 mm2=
Iz
1
12
b⋅ h3⋅:= Iz 66.67 10 6−× m4=
Iy
1
12
h⋅ b3⋅:= Iy 16.67 10 6−× m4=
QA.y 0:= (since A' = 0)
Normal Stress:
σA
Px
A
Mz yA⋅
Iz
−
My zA⋅
Iy
+:= σA 20.0− kPa=
Shear Stress : SinceQA = 0, τA 0:=
In-plane Principal Stress:
σx σA:= σy 0:= τxy 0:=
Since no shear stress acts upon the element,
σ1 σy:= σ1 0 kPa= Ans
σ2 σx:= σ2 20− kPa= Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
τmax
σx σy−
2
⎛⎜⎝
⎞
⎠
2
τxy2+:= τmax 10 kPa= Ans
Problem 9-52
The internal loadings at a section of the beam consist of an axial force of 500 N, a shear force of 800
N, and two moment components of 30 N·m and 40 N·m. Determine the principal stresses at point B.
Also calculate the maximum in-plane shear stress at this point.
Unit Used: kPa 1000Pa:=
Given: b 100mm:= h 200mm:= Px 0.5kN:=
yB 0mm:= zB 50− mm:= Py 0.8kN:=
My 0.04− kN m⋅:= Mz 0.03kN m⋅:=
Solution:
Section Property :
A b h⋅:= A 20000 mm2=
Iz
1
12
b⋅ h3⋅:= Iz 66.67 10 6−× m4=
Iy
1
12
h⋅ b3⋅:= Iy 16.67 10 6−× m4=
QB.y b 0.5h( )⋅ 0.25h( )⋅:=
Normal Stress:
σB
Px
A
Mz yB⋅
Iz
−
My zB⋅
Iy
+:= σB 145.0 kPa=
Shear Stress :
τB
Py QB.y⋅
Iz b⋅
:= τB 60.00 kPa=
In-plane Principal Stress:
σxB σB:= σyB 0:= τxy τB:=
σB1
σxB σyB+
2
σxB σyB−
2
⎛⎜⎝
⎞
⎠
2
τB2++:= σB1 166.6 kPa= Ans
σB2
σxB σyB+
2
σxB σyB−
2
⎛⎜⎝
⎞
⎠
2
τB2+−:= σB2 21.61− kPa= Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
τmax
σxB σyB−
2
⎛⎜⎝
⎞
⎠
2
τxy2+:= τmax 94.11 kPa= Ans
Problem 9-53
The internal loadings at a section of the beam consist of an axial force of 500 N, a shear force of 800
N, and two moment components of 30 N·m and 40 N·m. Determine the principal stresses at point C.
Also calculate the maximum in-plane shear stress at this point.
Unit Used: kPa 1000Pa:=
Given: b 100mm:= h 200mm:= Px 0.5kN:=
yC 50− mm:= zC 0mm:= Py 0.8kN:=
My 0.04− kN m⋅:= Mz 0.03kN m⋅:=
Solution:
Section Property :
A b h⋅:= A 20000 mm2=
Iz
1
12
b⋅ h3⋅:= Iz 66.67 10 6−× m4=
Iy
1
12
h⋅ b3⋅:= Iy 16.67 10 6−× m4=
QC.y b 0.5h yC−( )⋅ yC 0.5 0.5h yC−( )+⎡⎣ ⎤⎦⋅:= QC.y 375000 mm3=
Normal Stress:
σC
Px
A
Mz yC⋅
Iz
−
My zC⋅
Iy
+:= σC 47.5 kPa=
Shear Stress :
τC
Py QC.y⋅
Iz b⋅
:= τC 45.00 kPa=
In-plane Principal Stress:
σxC σC:= σyC 0:= τxy τC:=
σC1
σxC σyC+
2
σxC σyC−
2
⎛⎜⎝
⎞
⎠
2
τC2++:= σC1 74.63 kPa= Ans
σC2
σxC σyC+
2
σxC σyC−
2
⎛⎜⎝
⎞
⎠
2
τC2+−:= σC2 27.13− kPa= Ans
Maximum In-plane Shear Stress: Applying Eq. 9-7,
τmax
σxC σyC−
2
⎛⎜⎝
⎞
⎠
2
τxy2+:= τmax 50.88 kPa= Ans
Problem 9-54
The beam has a rectangular cross section and is subjected to the loads shown. Write a computer
program that can be used to determine the principal stresses at points A, B, C, and D. Show an
application of the program using the values h = 300 mm, b = 200 mm, Nx = 2 kN, Vy = 1.5 kN, Vz = 0,
My = 0, and Mz = -225 kN·m.
Problem 9-55
The member has a rectangular cross section and is subjected to the loading shown. Write a computer
program that can be used to determine the principal stresses at points A, B, and C. Show an application
of the program using the values b = 150 mm, h = 200 mm, P = 1.5 kN, x = 75 mm, z = -50 mm, Vx =
300 N, and Vz = 600 N.
Problem 9-56
Solve Prob. 9-4 using Mohr's circle.
Given: σx 0.650− MPa:= σy 0.400MPa:=
φ' 60deg:= τxy 0MPa:=
Solution: θ 90deg φ'−:= θ 30.00 deg=
Center :
σc
σx σy+
2
:= σc 0.125− MPa=
Radius :
R σx σc−:= R 0.525 MPa=
Coordinates:
A σx 0,( ) B σy 0,( ) C σc 0,( )
Stresses: 
σx' σc R cos 2θ( )⋅−:= σx' 0.387− MPa= Ans
τx'y' R sin 2θ( )⋅:= τx'y' 0.455 MPa= Ans
Problem 9-57
Solve Prob. 9-2 using Mohr's circle.
Given: σx 5MPa:= σy 3MPa:= τxy 8MPa:= φ' 40deg:=
Solution:
Center :
σc
σx σy+
2
:= σc 4 MPa=
Radius :
R σx σc−( )2 τxy2+:= R 8.062 MPa=
Angles: θ 90deg φ'+:= θ 130 deg=
φ atan
τxy
σx σc−
⎛⎜⎝
⎞
⎠
:= φ 82.875 deg=
α 180deg 2θ φ−( )−:= α 2.875 deg=
Stresses: 
σx' σc R cos α( )⋅−:= σx' 4.052− MPa= Ans
τx'y' R− sin α( )⋅:= τx'y' 0.404− MPa= Ans
Problem 9-58
Solve Prob. 9-3 using Mohr's circle.
Given: σx 0.350MPa:= σy 0.200− MPa:=
φ' 50deg:= τxy 0MPa:=
Solution: θ 90deg φ'+:= θ 140 deg=
Center :
σc
σx σy+
2
:= σc 0.075 MPa=
Radius :
R σx σc−( )2 τxy2+:= R 0.275 MPa=
Coordinates:
A σx 0,( ) C σc 0,( )
Angles:
α 360deg 2θ−:= α 80 deg=
Stresses: (represented by coordinates of point P)
σx' σc R cos α( )⋅+:= σx' 0.123 MPa= Ans
τx'y' R sin α( )⋅:= τx'y' 0.271 MPa= Ans
Problem 9-59
Solve Prob. 9-10 using Mohr's circle.
Given: σx 0MPa:= σy 0.300− MPa:=
θ 30deg:= τxy 0.950MPa:=
Solution:
Center :
σc
σx σy+
2
:= σc 0.15− MPa=
Radius :
R σx σc−( )2 τxy2+:= R 0.962 MPa=
Angles:
φ atan
τxy
σx σc−
⎛⎜⎝
⎞
⎠
:= φ 81.027 deg=
α 2θ φ−:= α 21.027− deg=
Stresses: 
σx' σc R cos α( )⋅+:= σx' 0.748 MPa= Ans
σy' σc R cos α( )⋅−:= σy' 1.048− MPa= Ans
τx'y' R− sin α( )⋅:= τx'y' 0.345 MPa= Ans
Problem 9-60
Solve Prob. 9-6 using Mohr's circle.
Given: σx 90MPa:= σy 50MPa:= τxy 35− MPa:=
φ' 60deg:=
Solution:
Center :
σc
σx σy+
2
:= σc 70 MPa=
Radius :
R σx σc−( )2 τxy2+:= R 40.311 MPa=
Angles: θ 90deg φ'−:=
φ atan
τxy
σx σc−
⎛⎜⎝
⎞
⎠
:= φ 60.255− deg=
α 180deg 2θ φ−( )−:= α 59.745 deg=
Stresses: 
σx' σc R cos α( )⋅−:= σx' 49.689 MPa= Ans
τx'y' R− sin α( )⋅:= τx'y' 34.821− MPa= Ans
Problem 9-61
Solve Prob. 9-11 using Mohr's circle.
Given: σx 0.300MPa:= σy 0MPa:=
θ 60− deg:= τxy 0.120MPa:=
Solution:
Center :
σc
σx σy+
2
:= σc 0.15 MPa=
Radius :
R σx σc−( )2 τxy2+:= R 0.1921 MPa=
Coordinates:
A σx τxy,( ) C σc 0,( )
Angles:
φ atan
τxy
σx σc−
⎛⎜⎝
⎞
⎠
:= φ 38.660 deg=
α 180deg 2θ+ φ−:= α 21.340 deg=
Stresses: (represented by coordinates of points P and Q)
σx' σc R cos α( )⋅−:= σx' 0.0289− MPa= Ans
σy' σc R cos α( )⋅+:= σy' 0.329 MPa= Ans
τx'y' R sin α( )⋅:= τx'y' 0.0699 MPa= Ans
Problem 9-62
Solve Prob. 9-13 using Mohr's circle.
Given: σx 45MPa:= σy 60− MPa:= τxy 30MPa:=
Solution:
Center :
σc
σx σy+
2
:= σc 7.5− MPa=
Radius :
R σx σc−( )2 τxy2+:= R 60.467 MPa=
Coordinates:
A σx τxy,( ) C σc 0,( )
Stresses: 
σ1 σc R+:= σ1 52.97 MPa= Ans
σ2 σc R−:= σ2 67.97− MPa= Ans
τmax R:= τmax 60.47 MPa= Ans
σavg σc:= σavg 7.5− MPa= Ans
Angles:
θp1
1
2
atan
τxy
σx σc−
⎛⎜⎝
⎞
⎠
:=
θp1 14.872 deg= (Counter-clockwise) Ans
2θs1 90deg 2θp1−=
θs1 45deg θp1−:=
θs1 30.128 deg= (Clockwise) Ans
Problem 9-63
Solve Prob. 9-14 using Mohr's circle.
Given: σx 180MPa:= σy 0MPa:= τxy 150− MPa:=
Solution:
Center :
σc
σx σy+
2
:= σc 90 MPa=
Radius :
R σx σc−( )2 τxy2+:= R 174.929 MPa=
Coordinates:
A σx τxy,( ) B σy τxy−,( ) C σc 0,( )
Stresses: 
σ1 σc R+:= σ1 264.93 MPa= Ans
σ2 σc R−:= σ2 84.93− MPa= Ans
τmax R:= τmax 174.93 MPa= Ans
σavg σc:= σavg 90 MPa= Ans
Angles:
θp
1
2
atan
τxy
σx σc−
⎛⎜⎝
⎞
⎠
:=
θp 29.518 deg= (Clockwise) Ans
2θs 90deg 2θp−=
θs 45deg θp−:=
θs 15.482 deg= (Counter-clockwise) Ans
Problem 9-64
Solve Prob. 9-16 using Mohr's circle.
Given: σx 200− MPa:= σy 250MPa:= τxy 175MPa:=
Solution:
Center :
σc
σx σy+
2
:= σc 25 MPa=
Radius :
R σx σc−( )2 τxy2+:= R 285.044 MPa=
Coordinates:
A σx τxy,( ) B σy τxy−,( ) C σc 0,( )
Stresses: 
σ1 σc R+:= σ1 310.04 MPa= Ans
σ2 σc R−:= σ2 260.04− MPa= Ans
τmax R:= τmax 285.04 MPa= Ans
σavg σc:= σavg 25 MPa= Ans
Angles:
θp
1
2
atan
τxy
σx σc−
⎛⎜⎝
⎞
⎠
:=
θp 18.937 deg= (Clockwise) Ans
2θs 90deg 2θp−=
θs 45deg θp−:=
θs 26.063 deg= (Counter-clockwise) Ans
Problem 9-65
Solve Prob. 9-15 using Mohr's circle.
Given: σx 30− MPa:= σy 0MPa:= τxy 12− MPa:=
Solution:
Center :
σc
σx σy+
2
:= σc 15− MPa=
Radius :
R σx σc−( )2 τxy2+:= R 19.209 MPa=
Coordinates:
A σx τxy,( ) C σc 0,( )
Stresses: 
σ1 σc R+:= σ1 4.21 MPa=Ans
σ2 σc R−:= σ2 34.21− MPa= Ans
τmax R:= τmax 19.21 MPa= Ans
σavg σc:= σavg 15− MPa= Ans
Angles:
θp2
1
2
atan
τxy
σx σc−
⎛⎜⎝
⎞
⎠
:=
θp2 19.330 deg= (Countr-clockwise) Ans
2θs2 2θp2 90deg+=
θs2 θp2 45deg+:=
θs2 64.330 deg= (Countr-clockwise) Ans
Problem 9-66
Determine the equivalent state of stress if an element is oriented 20° clockwise from the element
shown. Show the result on the element.
Given: σx 3MPa:= σy 2− MPa:=
θ 20− deg:= τxy 4− MPa:=
Solution:
Center :
σc
σx σy+
2
:= σc 0.5 MPa=
Radius :
R σx σc−( )2 τxy2+:= R 4.717 MPa=
Coordinates:
A σx τxy,( ) C σc 0,( )
Angles:
φ atan
τxy
σx σc−
⎛⎜⎝
⎞
⎠
:= φ 57.995 deg=
α φ 2θ+:= α 17.995 deg=
Stresses: (represented by coordinates of points P and Q)
σx' σc R cos α( )⋅+:= σx' 4.986 MPa= Ans
σy' σc R cos α( )⋅−:= σy' 3.986− MPa= Ans
τx'y' R− sin α( )⋅:= τx'y' 1.457− MPa= Ans
Problem 9-67
Determine the equivalent state of stress if an element is oriented 60° counterclockwise from the
element shown.
Given: σx 0.750MPa:= σy 0.800− MPa:=
θ 60deg:= τxy 0.450MPa:=
Solution:
Center :
σc
σx σy+
2
:= σc 0.025− MPa=
Radius :
R σx σc−( )2 τxy2+:= R 0.8962 MPa=
Coordinates:
A σx τxy,( ) B σy τxy−,( ) C σc 0,( )
Angles:
φ atan
τxy
σx σc−
⎛⎜⎝
⎞
⎠
:= φ 30.141 deg=
α 2θ φ−:= α 89.859 deg=
Stresses: 
σx' σc R cos α( )⋅+:= σx' 0.0228− MPa= Ans
σy' σc R cos α( )⋅−:= σy' 0.0272− MPa= Ans
τx'y' R− sin α( )⋅:= τx'y' 0.896− MPa= Ans
Problem 9-68
Determine the equivalent state of stress if an element is oriented 30° clockwise from the element
shown.
Given: σx 350MPa:= σy 230MPa:=
θ 30deg:= τxy 480− MPa:=
Solution:
Center :
σc
σx σy+
2
:= σc 290 MPa=
Radius :
R σx σc−( )2 τxy2+:= R 483.7355 MPa=
Coordinates:
A σx τxy,( ) B σy τxy−,( ) C σc 0,( )
Angles:
φ atan
τxy
σx σc−
⎛⎜⎝
⎞
⎠
:= φ 82.875− deg=
α 2− θ φ−:= α 22.875 deg=
Stresses: 
σx' σc R cos α( )⋅+:= σx' 735.6922 MPa= Ans
σy' σc R cos α( )⋅−:= σy' 155.6922− MPa= Ans
τx'y' R− sin α( )⋅:= τx'y' 188.038− MPa= Ans
Problem 9-69
Determine the equivalent state of stress if an element is oriented 30° clockwise from the element
shown. Show the result on the element.
Given: σx 7MPa:= σy 8MPa:=
θ 30− deg:= τxy 15MPa:=
Solution:
Center :
σc
σx σy+
2
:= σc 7.5 MPa=
Radius :
R σx σc−( )2 τxy2+:= R 15.0083 MPa=
Coordinates:
A σx τxy,( ) C σc 0,( )
Angles:
φ atan
τxy
σx σc−
⎛⎜⎝
⎞
⎠
:= φ 88.091 deg=
α 180deg 2 θ φ+( )−:= α 31.909 deg=
Stresses: (represented by coordinates of points P and Q)
σx' σc R cos α( )⋅−:= σx' 5.240− MPa= Ans
σy' σc R cos α( )⋅+:= σy' 20.240 MPa= Ans
τx'y' R sin α( )⋅:= τx'y' 7.933 MPa= Ans
Problem 9-70
Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal
stress. Specify the orientation of the element in each case.
Given: σx 350MPa:= σy 200− MPa:= τxy 500MPa:=
Solution:
Center :
σc
σx σy+
2
:= σc 75 MPa=
Radius :
R σx σc−( )2 τxy2+:= R 570.636 MPa=
Coordinates:
A σx τxy,( ) C σc 0,( )
a) Principal Stresses:
σ1 σc R+:= σ1 645.64 MPa= Ans
σ2 σc R−:= σ2 495.64− MPa= Ans
Angles:
θp1
1
2
atan
τxy
σx σc−
⎛⎜⎝
⎞
⎠
:=
θp1 30.595 deg= (Counter-clockwise) Ans
b) Maximum In-plane Shear Stress: (represented by coordinates of point E)
τmax R:= τmax 570.64 MPa= Ans
σavg σc:= σavg 75 MPa= Ans
2θs 90deg 2 θp1⋅−=
θs 45deg θp1−:=
θs 14.405 deg= (Clockwise) Ans
Problem 9-71
Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal
stress. Specify the orientation of the element in each case.
Given: σx 10MPa:= σy 80MPa:= τxy 60− MPa:=
Solution:
Center :
σc
σx σy+
2
:= σc 45 MPa=
Radius :
R σx σc−( )2 τxy2+:= R 69.462 MPa=
Coordinates:
A σx τxy,( ) C σc 0,( )
a) Principal Stresses:
σ1 σc R+:= σ1 114.46 MPa= Ans
σ2 σc R−:= σ2 24.46− MPa= Ans
Angles:
θp2
1
2
atan
τxy
σx σc−
⎛⎜⎝
⎞
⎠
:=
θp1 90deg θp2−:=
θp1 60.128 deg= (Clockwise) Ans
b) Maximum In-plane Shear Stress: (represented by coordinates of point E)
τmax R−:= τmax 69.46− MPa= Ans
σavg σc:= σavg 45 MPa= Ans
2θs2 90deg 2 θp2⋅−=
θs2 45deg θp2−:=
θs2 15.128 deg= (Clockwise) Ans
Problem 9-72
Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal
stress. Specify the orientation of the element in each case.
Given: σx 0MPa:= σy 50MPa:= τxy 30− MPa:=
Solution:
Center :
σc
σx σy+
2
:= σc 25 MPa=
Radius :
R σx σc−( )2 τxy2+:= R 39.051 MPa=
Coordinates:
A σx τxy,( ) B σy τxy−,( ) C σc 0,( )
a) Principal Stresses:
σ1 σc R+:= σ1 64.05 MPa= Ans
σ2 σc R−:= σ2 14.05− MPa= Ans
Angles:
θp
1
2
atan
τxy
σx σc−
⎛⎜⎝
⎞
⎠
:=
θp 25.097 deg= (Clockwise)
b) Maximum In-plane Shear Stress: (represented by coordinates of point E)
τmax R:= τmax 39.05 MPa= Ans
σavg σc:= σavg 25 MPa= Ans
2θs2 90deg 2 θp⋅−=
θs2 45deg θp−:=
θs2 19.903 deg= (Counter-clockwise)
Problem 9-73
Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal
stress. Specify the orientation of the element in each case.
Given: σx 12− MPa:= σy 8− MPa:= τxy 4MPa:=
Solution:
Center :
σc
σx σy+
2
:= σc 10− MPa=
Radius :
R σx σc−( )2 τxy2+:= R 4.472 MPa=
Coordinates:
A σx τxy,( ) B σy τxy−,( ) C σc 0,( )
a) Principal Stresses:
σ1 σc R+:= σ1 5.53− MPa= Ans
σ2 σc R−:= σ2 14.47− MPa= Ans
Angles:
θp
1
2
atan
τxy
σx σc−
⎛⎜⎝
⎞
⎠
:=
θp 31.717 deg= (Clockwise)
b) Maximum In-plane Shear Stress: (represented by coordinates of point E)
τmax R:= τmax 4.47 MPa= Ans
σavg σc:= σavg 10− MPa= Ans
2θs2 90deg 2 θp⋅−=
θs2 45deg θp−:=
θs2 13.283 deg= (Counter-clockwise)
Problem 9-74
Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal
stress. Specify the orientation of the element in each case.
Given: σx 45MPa:= σy 30MPa:= τxy 50− MPa:=
Solution:
Center :
σc
σx σy+
2
:= σc 37.5 MPa=
Radius :
R σx σc−( )2 τxy2+:= R 50.559 MPa=
Coordinates:
A σx τxy,( ) B σy τxy−,( ) C σc 0,( )
a) Principal Stresses:
σ1 σc R+:= σ1 88.06 MPa= Ans
σ2 σc R−:= σ2 13.06− MPa= Ans
Angles:
θp
1
2
atan
τxy
σx σc−
⎛⎜⎝
⎞
⎠
:=
θp 40.735 deg= (Clockwise)
b) Maximum In-plane Shear Stress: (represented by coordinates of point E)
τmax R:= τmax 50.56 MPa= Ans
σavg σc:= σavg 37.5 MPa= Ans
2θs2 90deg 2 θp⋅−=
θs2 45deg θp−:=
θs2 4.265 deg= (Counter-clockwise)
Problem 9-75
The square steel plate has a thickness of 12 mm and is subjected to the edge loading shown. Determine
the principal stresses developed in the steel.
Given: σx 0MPa:= σy 0MPa:= qxy 3.2
kN
m
:=
L 100mm:= t 12mm:=
Solution:
τxy
qxy
t
:= τxy 0.267 MPa=
Center :
σc
σx σy+
2
:= σc 0 MPa=
Radius :
R σx σc−( )2 τxy2+:= R 0.267 MPa=
Coordinates:
A σx τxy,( ) C σc 0,( )
In-plane Principal Stresses: 
The coordinates of points B and D represent σ1 and σ2, respectively.
σ1 σc R+:= σ1 0.267 MPa= Ans
σ2 σc R−:= σ2 0.267− MPa= Ans
σ1 115.60 MPa=
Given: σx 105MPa:= σy 0MPa:= τxy 35− MPa:=
Solution:
Center :
σc
σx σy+
2
:= σc 52.5 MPa=
Radius :
R σx σc−( )2 τxy2+:= R 63.097 MPa=
Coordinates:
A σx τxy,( ) C σc 0,( )
a) In-planePrincipal Stresses:
The coordinates of points B and D represent σ1 and σ2, respectively.
σ1 σc R+:=
Ans
Ans
σ2 σc R−:= σ2 10.60− MPa= Ans
Angles:
θp
1
2
atan
τxy
σx σc−
⎛⎜⎝
⎞
⎠
:=
θp 16.845 deg= (Clockwise)
b) Maximum In-plane Shear Stresses:
Represented by the coordinates of point E on the circle.
τmax R−:= τmax 63.10− MPa= Ans
2θs 90deg 2 θp⋅−=
θs 45deg θp−:=
θs 28.155 deg= (Counter-clockwise)
Problem 9-76Determine (a) the principal stress and (b) the maximum in-plane shear stress and average normal
stress. Specify the orientation of the element in each case.
Problem 9-77
Draw Mohr's circle that describes each of the following states of stress.
Given: σx 30− MPa:= σy 30MPa:=
τxy 0MPa:=
Solution:
Cases (a) and (b) are the same.
Center :
σc
σx σy+
2
:= σc 0 MPa=
Radius :
R σx σc−( )2 τxy2+:= R 30 MPa=
Coordinates:
A σx τxy,( ) C σc 0,( )
Problem 9-78
Draw Mohr's circle that describes each of the following states of stress.
a) Given: σx 0.8MPa:= τxy 0MPa:=
σy 0.6− MPa:=
Solution:
Center : σc
σx σy+
2
:=
σc 0.1 MPa=
Radius : R σx σc−( )2 τxy2+:=
R 0.7 MPa=
Coordinates: A σx τxy,( ) C σc 0,( )
b) Given: σx 0MPa:= τxy 0MPa:=
σy 2− MPa:=
Solution:
Center : σc
σx σy+
2
:=
σc 1− MPa=
Radius : R σx σc−( )2 τxy2+:=
R 1 MPa=
Coordinates: A σx τxy,( ) C σc 0,( )
c) Given: σx 0MPa:= τxy 20MPa:=
σy 0MPa:=
Solution:
Center : σc
σx σy+
2
:=
σc 0 MPa=
Radius : R σx σc−( )2 τxy2+:=
R 20 MPa=
Coordinates: A σx τxy,( ) C σc 0,( )
Problem 9-79
A point on a thin plate is subjected to two successive states of stress as shown. Determine the resulting
state of stress with reference to an element oriented as shown on the bottom.
Unit used: kPa 1000Pa:=
Given: σx_a 50kPa:= σy_a 0:= τxy_a 0:=
σy_b 18− kPa:= σx_b 0:= τxy_b 45− kPa:=
θa 30deg:= βb 50deg:=
Solution:
For element a:
Center : σc_a
σx_a σy_a+
2
:= σc_a 25 kPa=
Radius : Ra σx_a σc_a−( )2 τxy_a2+:= Ra 25 kPa=
Coordinates: A σx_a τxy_a,( ) B σx_a τxy_a−,( ) C σc_a 0,( )
σ'x_a σc_a Ra cos 2θa( )⋅+:=
σ'y_a σc_a Ra cos 2θa( )⋅−:=
τ'xy_a Ra sin 2θa( )⋅:=
For element b: θb 90deg βb−( )−:=
Center : σc_b
σx_b σy_b+
2
:= σc_b 9− kPa=
Radius : Rb σx_b σc_b−( )2 τxy_b2+:= Rb 45.891 kPa=
Coordinates: A σx_b τxy_b,( ) B σx_a τxy_b−,( ) C σc_b 0,( )
Angles : 
φ atan
τxy_b
σx_b σc_b−
⎛⎜⎝
⎞
⎠
:= φ 78.69 deg=
αb 2θb φ+:=
αb 1.310− deg=
σ'x_b σc_b Rb cos αb( )⋅+:=
σ'y_b σc_b Rb cos αb( )⋅−:=
τ'xy_b Rb sin αb( )⋅:=
Resultants:
σ'x σ'x_a σ'x_b+:= σ'x 74.38 kPa= Ans
σ'y σ'y_a σ'y_b+:= σ'y 42.38− kPa= Ans
τ'xy τ'xy_a τ'xy_b+:= τ'xy 22.70 kPa= Ans
Problem 9-80
Mohr's circle for the state of stress in Fig. 9-15a is shown in Fig. 9-15b. Show that finding the
coordinates of point P(σx' , τx'y' ) on the circle gives the same value as the stress-transformation Eqs.
9-1 and 9-2.
Problem 9-81
The cantilevered rectangular bar is subjected to the force of 25 kN. Determine the principal stresses at
point A.
Given: b 80mm:= d 160mm:= cA 40mm:=
P 25kN:= L 0.4m:= bA 40− mm:=
rv
3
5
:= rh
4
5
:=
Solution:
Internal Force and Moment : At Section A-B:
+ ΣFx=0; P− rh⋅ N+ 0= N P rh⋅:=
+ ΣFy=0; P− rv⋅ V+ 0= V P rv⋅:=
+ ΣΜO=0; M P rv⋅( ) L⋅− 0= M P rv⋅( ) L⋅:=
Section Property : dA 0.5d cA−:=
A b d⋅:= I 1
12
b⋅ d3⋅:=
QA b dA⋅ 0.5dA cA+( )⋅:=
Normal Stress:
σA
N
A
M cA⋅
I
+:= σA 10.352 MPa=
Shear Stress :
τA
V QA⋅
I b⋅:= τA 1.318 MPa=
Construction of Mohr's Circle :
σx σA:= σy 0MPa:= τxy τA−:=
Center : σc
σx σy+
2
:= σc 5.176 MPa=
Radius : R σx σc−( )2 τxy2+:=
R 5.341 MPa=
Coordinates: A σx τxy,( ) C σc 0,( )
In-plane Principal Stresses: 
The coordinates of points B and D represent σ1 and σ2, respectively.
σ1 σc R+:= σ1 10.52 MPa= Ans
σ2 σc R−:= σ2 0.165− MPa= Ans
Problem 9-82
Solve Prob. 9-81 for the principal stresses at point B.
Given: b 80mm:= d 160mm:= cB 25− mm:=
P 25kN:= L 0.4m:= bB 25mm:=
rv
3
5
:= rh
4
5
:=
Solution:
Internal Force and Moment : At Section A-B:
+ ΣFx=0; P− rh⋅ N+ 0= N P rh⋅:=
+ ΣFy=0; P− rv⋅ V+ 0= V P rv⋅:=
+ ΣΜO=0; M P rv⋅( ) L⋅− 0= M P rv⋅( ) L⋅:=
Section Property : dB 0.5d cB−:=
A b d⋅:= I 1
12
b⋅ d3⋅:=
QB b dB⋅ 0.5dB cB+( )⋅:=
Normal Stress:
σB
N
A
M cB⋅
I
+:= σB 3.931− MPa=
Shear Stress :
τB
V QB⋅
I b⋅:= τB 1.586 MPa=
Construction of Mohr's Circle :
σx σB:= σy 0MPa:= τxy τB−:=
Center : σc
σx σy+
2
:= σc 1.965− MPa=
Radius : R σx σc−( )2 τxy2+:=
R 2.526 MPa=
Coordinates: A σx τxy,( ) C σc 0,( )
In-plane Principal Stresses: 
The coordinates of points B and D represent σ1 and σ2, respectively.
σ1 σc R+:= σ1 0.560 MPa= Ans
σ2 σc R−:= σ2 4.491− MPa= Ans
Problem 9-83
The stair tread of the escalator is supported on two of its sides by the moving pin at A and the roller at
B. If a man having a weight of 1500 N (~150 kg) stands in the center of the tread, determine the
principal stresses developed in the supporting truck on the cross section at point C. The stairs move at
constant velocity.
Given: b 12mm:= d 50mm:= cC 0mm:=
W 1.5kN:= hC 150mm:= θ 30deg:=
vA 450mm:= hA 375mm:= hB 150mm:=
Solution:
Support Reactions :
+ ΣΜA=0; W hA⋅ By hB⋅− 0= By
W hA⋅
hB
:=
Bx By− tan θ( )⋅:=
Internal Force and Moment : At Section C:
+ ΣFx=0; Bx V+ 0= V Bx−:=
+ ΣFy=0; By N+ 0= N By−:=
+ ΣΜO=0; Bx hC⋅ M− 0= M Bx hC⋅:=
Section Property : dC 0.5d:=
A b d⋅:= I 1
12
b⋅ d3⋅:= QC b dC⋅ 0.5dC( )⋅:=
Normal Stress:
σC
N
A
M cC⋅
I
+:= σC 6.250− MPa=
Shear Stress :
τC
V QC⋅
I b⋅:= τC 5.413 MPa=
Construction of Mohr's Circle :
σx 0MPa:= σy σC:= τxy τC:=
Center : σc
σx σy+
2
:= σc 3.125− MPa=
Radius : R σx σc−( )2 τxy2+:= R 6.25 MPa=
Coordinates: A σx τxy,( ) C σc 0,( )
In-plane Principal Stresses: 
The coordinates of points B and D represent σ1 and σ2, respectively.
σ1 σc R+:= σ1 3.125 MPa= Ans
σ2 σc R−:= σ2 9.375− MPa= Ans
Problem 9-84
The pedal crank for a bicycle has the cross section shown. If it is fixed to the gear at B and does not
rotate while subjected to a force of 400 N, determine the principal stresses in the material on the cross
section at point C.
Given: b 7.5mm:= d 20mm:= cC 5mm:=
P 0.4kN:= L 100mm:=
Solution:
Internal Force and Moment : At Section C:
+ ΣFy=0; V P− 0= V P:=
+ ΣΜO=0; M P L⋅+ 0= M P− L⋅:=
Section Property : dC 0.5d cC−:=
A b d⋅:= I 1
12
b⋅ d3⋅:=
QC b dC⋅ cC 0.5dC+( )⋅:=
Normal Stress:
σC
M cC⋅
I
−:= σC 40.000 MPa=
Shear Stress :
τC
V QC⋅
I b⋅:= τC 3.000 MPa=
Construction of Mohr's Circle :
σx σC:= σy 0MPa:= τxy τC:=
Center : σc
σx σy+
2
:= σc 20 MPa=
Radius : R σx σc−( )2 τxy2+:= R 20.224 MPa=
Coordinates: A σx τxy,( ) C σc 0,( )
In-plane Principal Stresses: 
The coordinates of points B and D represent σ1 and σ2, respectively.
σ1 σc R+:= σ1 40.224 MPa= Ans
σ2 σc R−:= σ2 0.224− MPa= Ans
Problem 9-85
The frame supports the distributed loading of 200 N/m. Determine the normal and shear stresses at
point D that act perpendicular and parallel, respectively, to the grains. The grains at this point make an
angle of 30° with the horizontal as shown.
Unit Used: kPa 1000Pa:=
Given: b 100mm:= a 1m:= w 0.2 kN
m
:=
h 200mm:= L 2.5m:=
hD 75mm:= θ 60deg:=
Solution:
Support Reactions: Due to symmetry, B=C=R
+ ΣFy=0; 2R w L⋅− 0= R 0.5w L⋅:=
Internal Force and Moment : At Section D:
+ ΣFy=0; V− R+ w a⋅− 0=
V R w a⋅−:=
+ ΣΜO=0; M R a⋅− w a 0.5a( )⋅+ 0=
M R a⋅ w a⋅ 0.5a( )⋅−:=
Section Property : cD 0.5h hD−:=
A b h⋅:= I 1
12
b⋅ h3⋅:=
QD b hD⋅ cD 0.5hD+( )⋅:=
Normal Stress: σD
M cD−( )⋅
I
−:= σD 56.25 kPa=
Shear Stress : τD
V QD⋅
I b⋅:= τD 3.516 kPa=
Construction of Mohr's Circle :
σx σD:= σy 0MPa:= τxy τD:=
Center : σc
σx σy+
2
:= σc 28.125 kPa=
Radius : R σx σc−( )2 τxy2+:= R 28.344 kPa=
Coordinates: A σx τxy−,( ) C σc 0,( )
Angles:
φ atan
τxy
σx σc−
⎛⎜⎝
⎞
⎠
:= φ 7.125 deg=
α 180deg 2 θ φ+( )−:= α 52.875 deg=
Stresses on the Rotated Element: (represented by coordinates of point P)
σx' σc R cos α( )⋅−:= σx' 11.02 kPa= Ans
τx'y' R− sin α( )⋅:= τx'y' 22.60− kPa= Ans
Problem 9-86
The frame supports the distributed loadingof 200 N/m. Determine the normal and shear stresses at
point E that act perpendicular and parallel, respectively, to the grains. The grains at this point make an
angle of 60° with the horizontal as shown.
Unit Used: kPa 1000Pa:=
Given: b 50mm:= w 0.2 kN
m
:=
h 100mm:=
L 2.5m:= θ 60deg:=
Solution:
Support Reactions: Due to symmetry, B=C=R
+ ΣFy=0; 2R w L⋅− 0= R 0.5w L⋅:=
Internal Force: At Section E:
+ ΣFy=0; N− R− 0= N R−:=
Section Property :
A b h⋅:= A 5000 mm2=
Normal Stress: σE
N
A
:= σE 50− kPa=
Shear Stress : τD 0:=
Construction of Mohr's Circle :
σx σE:= σy 0MPa:= τxy τD:=
Center : σc
σx σy+
2
:= σc 25− kPa=
Radius : R σx σc−( )2 τxy2+:= R 25 kPa=
Coordinates: A σx τxy,( ) C σc 0,( )
Angles:
φ atan
τxy
σx σc−
⎛⎜⎝
⎞
⎠
:= φ 0.000 deg=
α 180deg 2 θ φ+( )−:= α 60.000 deg=
Stresses on the Rotated Element: (represented by coordinates of point P)
σx' σc R cos α( )⋅+:= σx' 12.50− kPa= Ans
τx'y' R sin α( )⋅:= τx'y' 21.65 kPa= Ans
Problem 9-87
The bent rod has a diameter of 15 mm and is subjected to the force of 600 N. Determine the principal
stresses and the maximum in-plane shear stress that are developed at point A and point B. Show the
results on elements located at these points.
Given: do 15mm:= P 0.6kN:=
a 50mm:= L 0.1m:=
Solution:
Internal Force: At Section AB:
N P:= N 0.600 kN=
M P a⋅:= M 0.030 kN m⋅=
Section Property :
A
π
4
do
2⋅:= A 176.71 mm2=
I
π
64
do
4⋅:= I 0 m2 mm2=
QA 0:= (since A' = 0)
QB 0:= (since A' = 0)
Normal Stress:
cA 0.5do:= σA
N
A
M cA⋅
I
−:= σA 87.15− MPa=
cB 0.5do:= σB
N
A
M cB⋅
I
+:= σB 93.94 MPa=
Shear Stress :
τA 0:= (since QA = 0)
τB 0:= (since QB = 0)
For A:
Construction of Mohr's Circle :
σx σA:= σy 0MPa:= τxy τA:=
Center : σc
σx σy+
2
:= σc 43.573− MPa=
Radius : R σx σc−( )2 τxy2+:= R 43.573 MPa=
Coordinates: A σx τxy,( ) C σc 0,( )
In-plane Principal Stresses: 
The coordinates of points B and A represent σ1 and σ2, respectively.
σ1 σc R+:= σ1 0 MPa= Ans
σ2 σc R−:= σ2 87.15− MPa= Ans
Maximum In-plane Shear Stresses: 
Represented by the coordinates of point E on the circle.
τmax R:= τmax 43.57 MPa= Ans
2θs 90deg= θs 45deg:= (Counter-clockwise) Ans
For B:
Construction of Mohr's Circle :
σx σB:= σy 0MPa:= τxy τB:=
Center : σc
σx σy+
2
:= σc 46.968 MPa=
Radius : R σx σc−( )2 τxy2+:= R 46.968 MPa=
Coordinates: A σx τxy,( ) C σc 0,( )
In-plane Principal Stresses: 
The coordinates of points A and B represent σ1 and σ2, respectively.
σ1 σc R+:= σ1 93.94 MPa= Ans
σ2 σc R−:= σ2 0 MPa= Ans
Maximum In-plane Shear Stresses: 
Represented by the coordinates of point E on the circle.
τmax R:= τmax 46.97 MPa= Ans
2θs 90deg= θs 45deg:= (Clockwise) Ans
Problem 9-88
Draw the three Mohr's circles that describe each of the following states of stress.
a) σmax 6MPa:= σint 0:=
σmin 0:=
b) σmax 50MPa:= σint 0:=
σmin 40− MPa:=
c) Unit used: kPa 1000Pa:=
σmax 600kPa:= σint 200kPa:=
σmin 100kPa:=
d) σmax 0:= σint 7− MPa:=
σmin 9− MPa:=
e) σmax 30− MPa:= σint 30− MPa:=
σmin 30− MPa:=
Problem 9-89
Draw the three Mohr's circles that describe each of the following states of stress.
a) σ1 15MPa:= σ2 0:=
σ3 15− MPa:=
τmax 15MPa:=
b) σ1 65MPa:= σ2 65− MPa:=
σ3 65− MPa:=
τmax 65MPa:=
Problem 9-90
The stress at a point is shown on the element. Determine the principal stresses and the absolute
maximum shear stress.
Given: σx 100− MPa:= σy 90MPa:= σz 80− MPa:=
τxy 0MPa:= τyz 40MPa:= τxz 0MPa:=
Solution:
Construction of Mohr's Circle in y-z Plane :
Center : σc
σy σz+
2
:= σc 5 MPa=
Radius : R σy σc−( )2 τyz2+:= R 93.941 MPa=
Coordinates: A σy τyz,( ) C σc 0,( )
In-plane Principal Stresses: 
The coordinates of points A and B represent σ1 and σ2, respectively.
σ1 σc R+:= σ1 98.941 MPa=
σ2 σc R−:= σ2 88.941− MPa=
Construction of Three Circles :
From the results obtained above,
σmax σ1:= σint σ2:= σmin σx:=
σmax 98.94 MPa= Ans
σint 88.94− MPa= Ans
σmin 100.00− MPa= Ans
Absolute Maximum Shear Stress : 
From the three Mohr's circles,
τabs.max
σmax σmin−
2
:=
τabs.max 99.47 MPa= Ans
Problem 9-91
The stress at a point is shown on the element. Determine the principal stresses and the absolute
maximum shear stress.
Given: σx 0MPa:= σy 7MPa:= σz 0MPa:=
τxy 0MPa:= τyz 50MPa:= τxz 5MPa:=
Solution:
Construction of Mohr's Circle in x-z Plane :
Center : σc
σx σz+
2
:= σc 0 MPa=
Radius : R σx σc−( )2 τxz2+:= R 5 MPa=
Coordinates: A σy τxz,( ) C σc 0,( )
In-plane Principal Stresses: 
The coordinates of points A and B represent σ1 and σ2, respectively.
σ1 σc R+:= σ1 5 MPa=
σ2 σc R−:= σ2 5− MPa=
Construction of Three Circles :
From the results obtained above,
σmax σy:= σint σ1:= σmin σ2:=
σmax 7 MPa= Ans
σint 5 MPa= Ans
σmin 5− MPa= Ans
Absolute Maximum Shear Stress : 
From the three Mohr's circles,
τmax
σmax σmin−
2
:=
τmax 6 MPa= Ans
Problem 9-92
The stress at a point is shown on the element. Determine the principal stresses and the absolute
maximum shear stress.
Given: σx 150MPa:= σy 0MPa:= σz 90MPa:=
τxy 0MPa:= τyz 80− MPa:= τxz 0MPa:=
Solution:
Construction of Mohr's Circle in y-z Plane :
Center : σc
σy σz+
2
:= σc 45 MPa=
Radius : R σz σc−( )2 τyz2+:= R 91.788 MPa=
Coordinates: A σy τyz,( ) B σz τyz−,( ) C σc 0,( )
In-plane Principal Stresses: 
The coordinates of points A and B represent σ1 and σ2, respectively.
σ1 σc R+:= σ1 136.788 MPa=
σ2 σc R−:= σ2 46.788− MPa=
Construction of Three Circles :
From the results obtained above,
σmax σx:= σint σ1:= σmin σ2:=
σmax 150.00 MPa= Ans
σint 136.79 MPa= Ans
σmin 46.79− MPa= Ans
Absolute Maximum Shear Stress : 
From the three Mohr's circles,
τmax
σmax σmin−
2
:=
τmax 98.39 MPa= Ans
Problem 9-93
The principal stresses acting at a point in a body are shown. Draw the three Mohr's circles that
describe this state of stress, and find the maximum in-plane shear stresses and associated average
normal stresses for the x-y, y-z, and x-z planes. For each case, show the results on the element oriented
in the appropriate direction.
Given: σx 40MPa:= σy 40− MPa:= σz 40− MPa:=
τxy 0MPa:= τyz 0MPa:= τxz 0MPa:=
Solution:
Construction of Three Circles :
σmax σx:= σint σy:= σmin σz:=
σmax 40.00 MPa=
σint 40.00− MPa=
σmin 40.00− MPa=
For x-y Plane :
σavg
σx σy+
2
:= σavg 0 MPa= Ans
τmax
σx σy−
2
:= τmax 40 MPa= Ans
For y-z Plane :
σ'avg
σy σz+
2
:= σ'avg 40− MPa= Ans
τ'max
σy σz−
2
:= τ'max 0 MPa= Ans
For x-y Plane :
σ''avg
σx σz+
2
:= σ''avg 0 MPa= Ans
τ''max
σx σz−
2
:= τ''max 40 MPa= Ans
Problem 9-94
Consider the general case of plane stress as shown. Write a computer program that will show a plot of
the three Mohr's circles for the element, and will also calculate the maximum in-plane shear stress and
the absolute maximum shear stress.
Problem 9-95
The solid shaft is subjected to a torque, bending moment, and shear force as shown. Determine the
principal stresses acting at points A and B and the absolute maximum shear stress.
Given: ro 25mm:= L 450mm:=
P 0.8kN:= To 0.045kN m⋅:= Mo 0.3kN m⋅:=
Solution: ρ ro:=
Internal Force and Moment : At Section AB:
Vy P:= Tx To:= Mz Mo P L⋅−:=
Section Property :
J
π
2
ro
4⋅:=A π ro2⋅:= Iz
π
4
ro
4⋅:=
QA 0:= (Since A'=0) QB
4 ro⋅
3 π⋅ 0.5A( )⋅:=
Normal Stress: cA ro:= cB 0:=
σA
Mz cA⋅
Iz
−:= σA 4.889 MPa=
σB
Mz cB⋅
Iz
−:= σB 0 MPa=
Shear Stress : bB 2 ro⋅:=
τA
Tx ρ⋅
J
:= τA 1.833 MPa=
τB
Vy QB⋅
Iz bB⋅
Tx ρ⋅
J
−:= τB 1.290− MPa=
For Point A:
Construction of Mohr's Circle:
σx σA:= σz 0MPa:= τxz τA−:=
Center : σc
σx σz+
2
:= σc 2.445 MPa=
Radius : R σz σc−( )2 τxz2+:= R 3.056