Soluções Resistência dos Materiais HIBBELER 7ª Edição 1

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Stamp
CD1
4/13 US
 ISM fornull
Problem 1-1
Determine the resultant internal normal force acting on the cross section through point A in each
column. In (a), segment BC weighs 300 kg/m and segment CD weighs 400kg/m. In (b), the column
has a mass of 200 kg/m.
a( ) Given: g 9.81
m
s2
:= wBC 300
kg
m
:=
LBC 3m:= wCA 400
kg
m
:=
FB 5kN:= LCA 1.2m:=
FC 3kN:=
Solution:
+↑Σ Fy = 0; FA wBC g⋅( ) LBC⋅− wCA g⋅( ) LCA⋅− FB− 2FC− 0=
FA wBC g⋅( ) LBC⋅ wCA g⋅( ) LCA⋅+ FB+ 2FC+:=
FA 24.5 kN= Ans
b( ) Given: g 9.81
m
s2
:= w 200 kg
m
:=
L 3m:= F1 6kN:=
FB 8kN:= F2 4.5kN:=
Solution:
+↑Σ Fy = 0; FA w L⋅( ) g⋅− FB− 2F1− 2F2− 0=
FA w L⋅( ) g⋅ FB+ 2F1+ 2F2+:=
FA 34.89 kN= Ans
Problem 1-2
Determine the resultant internal torque acting on the cross sections through points C and D of the
shaft. The shaft is fixed at B.
Given: TA 250N m⋅:=
TCD 400N m⋅:=
TDB 300N m⋅:=
Solution:
Equations of equilibrium:
+ TA TC− 0=
TC TA:=
TC 250 N m⋅= Ans
+ TA TCD− TD+ 0=
TD TCD TA−:=
TD 150 N m⋅= Ans
Problem 1-3
Determine the resultant internal torque acting on the cross sections through points B and C.
Given: TD 500N m⋅:=
TBC 350N m⋅:=
TAB 600N m⋅:=
Solution:
Equations of equilibrium:
Σ Mx = 0; TB TBC TD−+ 0=
TB TBC− TD+:=
TB 150 N m⋅= Ans
Σ Mx = 0; TC TD− 0=
TC TD:=
TC 500 N m⋅= Ans
Problem 1-4
A force of 80 N is supported by the bracket as shown. Determine the resultant internal loadings acting
on the section through point A.
Given: P 80N:=
θ 30deg:= φ 45deg:=
a 0.3m:= b 0.1m:=
Solution:
Equations of equilibrium:
 + ΣFx'=0; NA P cos φ θ−( )⋅− 0=
NA P cos φ θ−( )⋅:=
NA 77.27 N= Ans
+ ΣFy'=0; VA P sin φ θ−( )⋅− 0=
VA P sin φ θ−( )⋅:=
VA 20.71 N= Ans
+ ΣΜA=0; MA P cos φ( )⋅ a⋅ cos θ( )+ P sin φ( )⋅ b a sin θ( )+( )⋅− 0=
MA P− cos φ( )⋅ a⋅ cos θ( ) P sin φ( )⋅ b a sin θ( )+( )⋅+:=
MA 0.555− N m⋅= Ans
Note: Negative sign indicates that MA acts in the opposite direction to that shown on FBD.
Problem 1-5
Determine the resultant internal loadings acting on the cross section through point D of member AB.
Given: ME 70N m⋅:=
a 0.05m:= b 0.3m:=
Solution:
Segment AB: Support Reactions
+ ΣΜA=0; −ME By 2 a⋅ b+( )⋅− 0=
By
ME−
2a b+:= By 175− N=
+At B: Bx By
150
200
⎛⎜⎝
⎞
⎠⋅:= Bx 131.25− N=
Segment DB: NB Bx−:= VB By−:=
+ ΣFx=0; ND NB+ 0=
ND NB−:= ND 131.25− N= Ans
+ ΣFy=0; VD VB+ 0=
VD VB−:= VD 175− N= Ans
+ ΣΜD=0; MD− ME− By a b+( )⋅− 0=
MD ME− By a b+( )⋅−:=
MD 8.75− N m⋅= Ans
Problem 1-6
The beam AB is pin supported at A and supported by a cable BC. Determine the resultant internal
loadings acting on the cross section at point D.
Given: P 5000N:=
a 0.8m:= b 1.2m:= c 0.6m:= d 1.6m:=
e 0.6m:=
Solution:
θ atan b
d
⎛⎜⎝
⎞
⎠:= θ 36.87 deg=
φ atan a b+
d
⎛⎜⎝
⎞
⎠ θ−:= φ 14.47 deg=
Member AB: 
+ ΣΜA=0; FBC sin φ( )⋅ a b+( )⋅ P b( )⋅− 0=
FBC
P b( )⋅
sin φ( ) a b+( )⋅:=
FBC 12.01 kN=
Segment BD: 
 + ΣFx=0; ND− FBC cos φ( )⋅− P cos θ( )⋅− 0=
ND FBC− cos φ( )⋅ P cos θ( )⋅−:=
ND 15.63− kN= Ans
+ ΣFy=0; VD FBC sin φ( )⋅+ P sin θ( )⋅− 0=
VD FBC− sin φ( )⋅ P sin θ( )⋅+:=
VD 0 kN= Ans
+ ΣΜD=0; FBC sin φ( )⋅ P sin θ( )⋅−( ) d c−sin θ( )⋅ MD− 0=
MD FBC sin φ( )⋅ P sin θ( )⋅−( ) d c−sin θ( )⋅:=
MD 0 kN m⋅= Ans
Note: Member AB is the two-force member. Therefore the shear force and moment are zero.
Problem 1-7
Solve Prob. 1-6 for the resultant internal loadings acting at point E.
Given: P 5000N:=
a 0.8m:= b 1.2m:= c 0.6m:= d 1.6m:=
e 0.6m:=
Solution:
θ atan b
d
⎛⎜⎝
⎞
⎠:= θ 36.87 deg=
φ atan a b+
d
⎛⎜⎝
⎞
⎠ θ−:= φ 14.47 deg=
Member AB: 
+ ΣΜA=0; FBC sin φ( )⋅ a b+( )⋅ P b( )⋅− 0=
FBC
P b( )⋅
sin φ( ) a b+( )⋅:=
FBC 12.01 kN=
Segment BE: 
 + ΣFx=0; NE− FBC cos φ( )⋅− P cos θ( )⋅− 0=
NE FBC− cos φ( )⋅ P cos θ( )⋅−:=
NE 15.63− kN= Ans
+ ΣFy=0; VE FBC sin φ( )⋅+ P sin θ( )⋅− 0=
VE FBC− sin φ( )⋅ P sin θ( )⋅+:=
VE 0 kN= Ans
+ ΣΜE=0; FBC sin φ( )⋅ P sin θ( )⋅−( ) e⋅ ME− 0=
ME FBC sin φ( )⋅ P sin θ( )⋅−( ) e⋅:=
ME 0 kN m⋅= Ans
Note: Member AB is the two-force member. Therefore the shear force and moment are zero.
Problem 1-8
The boom DF of the jib crane and the column DE have a uniform weight of 750 N/m. If the hoist and
load weigh 1500 N, determine the resultant internal loadings in the crane on cross sections through
points A, B, and C.
Given: P 1500N:= w 750 N
m
:=
a 2.1m:= b 1.5m:=
c 0.6m:= d 2.4m:= e 0.9m:=
Solution:
Equations of Equilibrium: For point A
+ ΣFx=0; NA 0:= Ans
+
VA w e⋅− P− 0=ΣFy=0;
VA w e⋅ P+:= VA 2.17 kN= Ans
+ ΣΜA=0; MA− w e⋅( ) 0.5 e⋅( )⋅− P e( )⋅− 0=
MA w e⋅( )− 0.5 e⋅( )⋅ P e( )⋅−:= MA 1.654− kN m⋅= Ans
Note: Negative sign indicates that MA acts in the opposite direction to that shown on FBD.
Equations of Equilibrium: For point B
+ ΣFx=0; NB 0:= Ans
+
VB w d e+( )⋅− P− 0=ΣFy=0;
VB w d e+( )⋅ P+:= VB 3.98 kN= Ans
+ ΣΜB=0; −MB w d e+( )⋅[ ] 0.5 d e+( )⋅[ ]⋅− P d e+( )⋅− 0=
MB w d e+( )⋅[ ]− 0.5 d e+( )⋅[ ]⋅ P d e+( )⋅−:=
MB 9.034− kN m⋅= Ans
Note: Negative sign indicates that MB acts in the opposite direction to that shown on FBD.
Equations of Equilibrium: For point C
+ ΣFx=0; VC 0:= Ans
+
NC− w b c+ d+ e+( )⋅− P− 0=ΣFy=0;
NC w− b c+ d+ e+( )⋅ P−:=
NC 5.55− kN= Ans
+ ΣΜB=0; MC− w c d+ e+( )⋅[ ] 0.5 c d+ e+( )⋅[ ]⋅− P c d+ e+( )⋅− 0=
MC w c d+ e+( )⋅[ ]− 0.5 c d+ e+( )⋅[ ]⋅ P c d+ e+( )⋅−:=
MC 11.554− kN m⋅= Ans
Note: Negative sign indicates that NC and MC acts in the opposite direction to that shown
on FBD.
Problem 1-9
The force F = 400 N acts on the gear tooth. Determine the resultant internal loadings on the root of the
tooth, i.e., at the centroid point A of section a-a.
Given: P 400N:=
θ 30deg:= φ 45deg:=
a 4mm:= b 5.75mm:=
Solution: α φ θ−:=
Equations of equilibrium: For section a -a
 + ΣFx'=0; VA P cos α( )⋅− 0=
VA P cos α( )⋅:=
VA 386.37 N= Ans
+ ΣFy'=0; NA sin α( )− 0=
NA P sin α( )⋅:=
NA 103.53 N= Ans
+ ΣΜA=0; MA− P sin α( )⋅ a⋅− P cos α( )⋅ b⋅+ 0=
MA P− sin α( )⋅ a⋅ P cos α( )⋅ b⋅+:=
MA 1.808 N m⋅= Ans
Problem 1-10
The beam supports the distributed load shown. Determine the resultant internal loadings on the cross
section through point C. Assume the reactions at the supports A and B are vertical.
Given: w1 4.5
kN
m
:= w2 6.0
kN
m
:=
a 1.8m:= b 1.8m:= c 2.4m:=
d 1.35m:= e 1.35m:=
Solution: L1 a b+ c+:=
L2 d e+:=
Support Reactions: 
+ ΣΜA=0; By L1⋅ w1 L1⋅( ) 0.5 L1⋅( )− 0.5w2 L2⋅( ) L1 L23+⎛⎜⎝
⎞
⎠⋅− 0=
By w1 L1⋅( ) 0.5( )⋅ 0.5w2 L2⋅( ) 1 L23 L1⋅+
⎛⎜⎝
⎞
⎠
⋅+:= By 22.82 kN=
+ ΣFy=0; Ay By+ w1 L1⋅− 0.5w2 L2⋅− 0=
Ay By− w1 L1⋅+ 0.5w2 L2⋅+:= Ay 12.29 kN=
Equations of Equilibrium: For point C 
+ ΣFx=0; NC 0:= Ans
+ ΣFy=0; Ay w1 a b+( )⋅−⎡⎣ ⎤⎦ VC− 0=
VC Ay w1 a b+( )⋅−⎡⎣ ⎤⎦−:= VC 3.92 kN= Ans
+ ΣΜC=0; MC w1 a b+( )⋅⎡⎣ ⎤⎦ 0.5⋅ a b+( )⋅+ Ay a b+( )⋅− 0=
MC w1 a b+( )⋅⎡⎣ ⎤⎦− 0.5⋅ a b+( )⋅ Ay a b+( )⋅+:=
MC 15.07 kN m⋅= Ans
Note: Negative sign indicates that VC acts in the opposite direction to that shown on FBD.
Problem 1-11
The beam supports the distributed load shown. Determine the resultant internal loadings on the cross
sections through points D and E. Assume the reactions at the supports A and B are vertical.
Given: w1 4.5
kN
m
:= w2 6.0
kN
m
:=
a 1.8m:= b 1.8m:= c 2.4m:=
d 1.35m:= e 1.35m:=
Solution: L1 a b+ c+:=
L2 d e+:=
Support Reactions: 
+ ΣΜA=0; By L1⋅ w1 L1⋅( ) 0.5 L1⋅( )− 0.5w2 L2⋅( ) L1 L23+⎛⎜⎝
⎞
⎠⋅− 0=
By w1 L1⋅( ) 0.5( )⋅ 0.5w2 L2⋅( ) 1 L23 L1⋅+
⎛⎜⎝
⎞
⎠
⋅+:= By 22.82 kN=
+ ΣFy=0; Ay By+ w1 L1⋅− 0.5w2 L2⋅− 0=
Ay By− w1 L1⋅+ 0.5w2 L2⋅+:= Ay 12.29 kN=
Equations of Equilibrium: For point D 
+ ΣFx=0; ND 0:= Ans
+ ΣFy=0; Ay w1 a( )⋅−⎡⎣ ⎤⎦ VD− 0=
VD Ay w1 a( )⋅−:= VD 4.18 kN= Ans
+ ΣΜD=0; MD w1 a( )⋅⎡⎣ ⎤⎦ 0.5⋅ a( )⋅+
Ay a( )⋅− 0=
MD w1 a( )⋅⎡⎣ ⎤⎦− 0.5⋅ a( )⋅ Ay a( )⋅+:= MD 14.823 kN m⋅= Ans
Equations of Equilibrium: For point E 
+ ΣFx=0; NE 0:= Ans
+ ΣFy=0; VE 0.5w2 0.5 e⋅( )⋅− 0=
VE 0.5w2 0.5 e⋅( )⋅:= VE 2.03 kN= Ans
+ ΣΜD=0; ME− 0.5w2 0.5 e⋅( )⋅⎡⎣ ⎤⎦ e3
⎛⎜⎝
⎞
⎠⋅− 0=
ME 0.5− w2 0.5 e⋅( )⋅⎡⎣ ⎤⎦ e3
⎛⎜⎝
⎞
⎠⋅:= ME 0.911− kN m⋅= Ans
Note: Negative sign indicates that ME acts in the opposite direction to that shown on FBD.
Problem 1-12
Determine the resultant internal loadings acting on (a) section a-a and (b) section b-b. Each section is
located through the centroid, point C.
Given: w 9
kN
m
:= θ 45deg:=
a 1.2m:= b 2.4m:=
Solution: L a b+:=
Support Reactions: 
+ ΣΜA=0; Bx− L⋅ sin θ( )⋅ w L⋅( ) 0.5 L⋅( )+ 0=
Bx
w L⋅( ) 0.5 L⋅( )⋅
L sin θ( )⋅:= Bx 22.91 kN=
+ ΣFy=0; Ay w L⋅ sin θ( )⋅− 0=
Ay w L⋅ sin θ( )⋅:= Ay 22.91 kN=
+ ΣFx=0; Bx w L⋅ cos θ( )⋅− Ax+ 0=
Ax w L⋅ cos θ( )⋅ Bx−:= Ax 0− kN=
(a) Equations of equilibrium: For Section a - a : 
+ ΣFx=0; NC Ay sin θ( )⋅+ 0=
NC Ay− sin θ( )⋅:=
 + ΣFy=0; VC Ay cos θ( )⋅+ w a⋅− 0=
VC Ay− cos θ( )⋅ w a⋅+:= VC 5.4− kN= Ans
+ ΣΜA=0; MC− w a⋅( ) 0.5 a⋅( )⋅− Ay cos θ( ) a⋅+ 0=
MC w a⋅( ) 0.5 a⋅( )⋅ Ay cos θ( ) a⋅−:= MC 12.96− kN m⋅= Ans
(b) Equations of equilibrium: For Section b - b : 
+ ΣFx=0; NC w a⋅ cos θ( )⋅+ 0=
NC w− a⋅ cos θ( )⋅:= NC 7.64− kN= Ans
+ ΣFy=0; VC w a⋅ sin θ( )⋅− Ay+ 0=
VC w a⋅ sin θ( )⋅ Ay−:= VC 15.27− kN= Ans
+ ΣΜA=0; MC− w a⋅( ) 0.5 a⋅( )⋅− Ay cos θ( ) a⋅+ 0=
MC w a⋅( ) 0.5 a⋅( )⋅ Ay cos θ( ) a⋅−:= MC 12.96− kN m⋅= Ans
NC 16.2− kN= Ans
Problem 1-13
Determine the resultant internal normal and shear forces in the member at (a) section a-a and (b)
section b-b, each of which passes through point A. Take θ = 60 degree. The 650-N load is applied
along the centroidal axis of the member.
Given: P 650N:= θ 60deg:=
(a) Equations of equilibrium: For Section a - a : 
+ ΣFy=0; P Na_a− 0=
Na_a P:=
Na_a 650 N= Ans
+ ΣFx=0; Va_a 0:= Ans
(b) Equations of equilibrium: For Section b - b : 
 + ΣFy=0; Vb_b− P cos 90deg θ−( )⋅+ 0=
Vb_b P cos 90deg θ−( )⋅:=
Vb_b 562.92 N= Ans
+ ΣFx=0; Nb_b P sin 90deg θ−( )⋅− 0=
Nb_b P sin 90deg θ−( )⋅:=
Nb_b 325 N= Ans
Problem 1-14
Determine the resultant internal normal and shear forces in the member at section b-b, each as a
function of θ. Plot these results for 0o θ≤ 90o≤ . The 650-N load is applied along the centroidal axis
of the member.
Given: P 650N:= θ 0:=
 Equations of equilibrium: For Section b - b : 
+ ΣFx0; Nb_b P cos θ( )⋅− 0=
Nb_b P cos θ( )⋅:= Ans
 + ΣFy=0; Vb_b− P cos θ( )⋅+ 0=
Vb_b P− cos θ( )⋅:= Ans
Problem 1-15
The 4000-N load is being hoisted at a constant speed using the motor M, which has a weight of 450 N.
Determine the resultant internal loadings acting on the cross section through point B in the beam. The
beam has a weight of 600 N/m and is fixed to the wall at A.
Given:
W1 4000N:= w 600
N
m
:=
W2 450N:=
a 1.2m:= b 1.2m:= c 0.9m:=
d 0.9m:= e 1.2m:=
f 0.45m:= r 0.075m:=
Solution:
Tension in rope: T
W1
2
:=
T 2.00 kN=
Equations of Equilibrium: For point B 
+ ΣFx=0; NB− T−( ) 0=
NB T−:= NB 2− kN= Ans
+ ΣFy=0; VB w e( )⋅− W1− 0=
VB w e( )⋅ W1+:= VB 4.72 kN= Ans
+ ΣΜB=0; MB− w e( )⋅[ ] 0.5⋅ e( )⋅− W1 e r+( )⋅− T f( )⋅+ 0=
MB w e( )⋅[ ]− 0.5⋅ e( )⋅ W1 e r+( )⋅− T f( )⋅+:=
MB 4.632− kN m⋅= Ans
Problem 1-16
Determine the resultant internal loadings acting on the cross section through points C and D of the
beam in Prob. 1-15.
Given: W1 4000N:=
w 600
N
m
:=
W2 450N:=
a 1.2m:= b 1.2m:= c 0.9m:=
d 0.9m:= e 1.2m:=
f 0.45m:= r 0.075m:=
Solution:
Tension in rope: T
W1
2
:= T 2.00 kN=
Equations of Equilibrium: For point C LC d e+:=
+ ΣFx=0; NC− T−( ) 0=
NC T−:= NC 2− kN= Ans
+ ΣFy=0; VC w LC( )⋅− W1− 0=
VC w LC( )⋅ W1+:= VC 5.26 kN= Ans
+ ΣΜC=0; MC− w LC( )⋅⎡⎣ ⎤⎦ 0.5⋅ LC( )⋅− W1 LC r+( )⋅− T f( )⋅+ 0=
MC w LC( )⋅⎡⎣ ⎤⎦− 0.5⋅ LC( )⋅ W1 LC r+( )⋅− T f( )⋅+:=
MC 9.123− kN m⋅= Ans
Equations of Equilibrium: For point D LD b c+ d+ e+:=
+ ΣFx=0; ND 0:= ND 0 kN= Ans
+ ΣFy=0; VD w LD( )⋅− W1− W2− 0=
VD w LD( )⋅ W1+ W2+:= VD 6.97 kN= Ans
+ ΣΜC=0; MD− w LD( )⋅⎡⎣ ⎤⎦ 0.5⋅ LD( )⋅− W1 LD r+( )⋅− W2 b( )⋅− 0=
MD w LD( )⋅⎡⎣ ⎤⎦− 0.5⋅ LD( )⋅ W1 LD r+( )⋅− W2 b( )⋅−:=
MD 22.932− kN m⋅= Ans
Problem 1-17
Determine the resultant internal loadings acting on the cross section at point B.
Given: w 900
kN
m
:=
a 1m:= b 4m:=
Solution: L a b+:=
Equations of Equilibrium: For point B 
+ ΣFx=0; NB 0:=
NB 0 kN= Ans
+ ΣFy=0; VB 0.5 w
b
L
⎛⎜⎝
⎞
⎠⋅
⎡⎢⎣
⎤⎥⎦⋅ b( )⋅− 0=
VB 0.5 w
b
L
⎛⎜⎝
⎞
⎠⋅
⎡⎢⎣
⎤⎥⎦⋅ b( )⋅:=
VB 1440 kN= Ans
+ ΣΜB=0; MB− 0.5 w
b
L
⎛⎜⎝
⎞
⎠⋅
⎡⎢⎣
⎤⎥⎦⋅ b( )⋅
b
3
⎛⎜⎝
⎞
⎠⋅− 0=
MB 0.5− w
b
L
⎛⎜⎝
⎞
⎠⋅
⎡⎢⎣
⎤⎥⎦⋅ b( )⋅
b
3
⋅:=
MB 1920− kN m⋅= Ans
Problem 1-18
The beam supports the distributed load shown. Determine the resultant internal loadings acting on the
cross section through point C. Assume the reactions at the supports A and B are vertical.
Given: w1 0.5
kN
m
:= a 3m:=
w2 1.5
kN
m
:=
Solution: L 3 a⋅:= w w2 w1−:=
Support Reactions: 
+ ΣΜA=0; By L⋅ w1 L⋅( ) 0.5 L⋅( )− 0.5 w( ) L⋅[ ] 2L3⎛⎜⎝ ⎞⎠⋅− 0=
By w1 L⋅( ) 0.5( )⋅ 0.5 w( ) L⋅[ ] 23⎛⎜⎝ ⎞⎠⋅+:=
By 5.25 kN=
+ ΣFy=0; Ay By+ w1 L⋅− 0.5 w( ) L⋅− 0=
Ay By− w1 L⋅+ 0.5 w( ) L⋅+:=
Ay 3.75 kN=
Equations of Equilibrium: For point C 
+ ΣFx=0; NC 0:= NC 0 kN= Ans
+ ΣFy=0; VC w1 a⋅+ 0.5 w
a
L
⎛⎜⎝
⎞
⎠⋅
⎡⎢⎣
⎤⎥⎦⋅ a( )⋅+ Ay− 0=
VC w1− a⋅ 0.5 w
a
L
⎛⎜⎝
⎞
⎠⋅
⎡⎢⎣
⎤⎥⎦⋅ a( )⋅− Ay+:=
VC 1.75 kN= Ans
+ ΣΜC=0; MC w1 a⋅( ) 0.5 a⋅( )+ 0.5 w aL⎛⎜⎝ ⎞⎠⋅⎡⎢⎣ ⎤⎥⎦⋅ a( )⋅ a3⎛⎜⎝ ⎞⎠⋅+ Ay a⋅− 0=
MC w1− a⋅( ) 0.5 a⋅( ) 0.5 w aL⎛⎜⎝ ⎞⎠⋅⎡⎢⎣ ⎤⎥⎦⋅ a( )⋅ a3⎛⎜⎝ ⎞⎠⋅− Ay a⋅+:=
MC 8.5 kN m⋅= Ans
Problem 1-19
Determine the resultant internal loadings acting on the cross section through point D in Prob. 1-18.
Given: w1 0.5
kN
m
:= a 3m:=
w2 1.5
kN
m
:=
Solution: L 3 a⋅:= w w2 w1−:=
Support Reactions: 
+ ΣΜA=0; By L⋅ w1 L⋅( ) 0.5 L⋅( )− 0.5 w( ) L⋅[ ] 2L3⎛⎜⎝ ⎞⎠⋅− 0=
By w1 L⋅( ) 0.5( )⋅ 0.5 w( ) L⋅[ ] 23⎛⎜⎝ ⎞⎠⋅+:=
By 5.25 kN=
+ ΣFy=0; Ay By+ w1 L⋅− 0.5 w( ) L⋅− 0=
Ay By− w1 L⋅+ 0.5 w( ) L⋅+:=
Ay 3.75 kN=
Equations of Equilibrium: For point D 
+ ΣFx=0; ND 0:= ND 0 kN= Ans
+ ΣFy=0; VD w1 2a( )⋅+ 0.5 w
2a
L
⎛⎜⎝
⎞
⎠⋅
⎡⎢⎣
⎤⎥⎦⋅ 2a( )⋅+ Ay− 0=
VD w1− 2a( )⋅ 0.5 w
2 a⋅
L
⎛⎜⎝
⎞
⎠⋅
⎡⎢⎣
⎤⎥⎦⋅ 2a( )⋅− Ay+:=
VD 1.25− kN= Ans
+ ΣΜD=0; MD w1 2a( )⋅⎡⎣ ⎤⎦ a( )+ 0.5 w 2 a⋅L
⎛⎜⎝
⎞
⎠⋅
⎡⎢⎣
⎤⎥⎦⋅ 2a( )⋅
2a
3
⎛⎜⎝
⎞
⎠⋅+ Ay 2a( )⋅− 0=
MD w1− 2a( )⋅⎡⎣ ⎤⎦ a( ) 0.5 w 2 a⋅L
⎛⎜⎝
⎞
⎠⋅
⎡⎢⎣
⎤⎥⎦⋅ 2a( )⋅
2a
3
⎛⎜⎝
⎞
⎠⋅− Ay 2a( )⋅+:=
MD 9.5 kN m⋅= Ans
Problem 1-20
The wishbone construction of the power pole supports the three lines, each exerting a force of 4 kN
on the bracing struts. If the struts are pin connected at A, B, and C, determine the resultant internal
loadings at cross sections through points D, E, and F.
Given: P 4kN:= a 1.2m:= b 1.8m:=
Solution:
Support Reactions: FBD (a) and (b).
Given
+ ΣΜA=0; By a( )⋅ Bx 0.5 b⋅( )⋅+ P a( )⋅− 0= [1]
+ ΣΜC=0; Bx 0.5 b⋅( )⋅ P a( )⋅+ By a( )⋅− P a( )⋅− 0= [2]
Solving [1] and [2]: Initial guess: Bx 1kN:= By 2kN:=
Bx
By
⎛⎜⎜⎝
⎞
⎠
Find Bx By,( ):= BxBy
⎛⎜⎜⎝
⎞
⎠
2.67
2
⎛⎜⎝
⎞
⎠ kN=
From FBD (a):
+ ΣFx=0; Bx Ax− 0=
Ax Bx:= Ax 2.67 kN=
+ ΣFy=0; Ay P− By− 0=
Ay P By+:= Ay 6 kN=
From FBD (b):
+ ΣFx=0; Cx Bx− 0=
Cx Bx:= Cx 2.67 kN=
+ ΣFy=0; Cy By+ P− P− 0=
Cy 2P By−:= Cy 6 kN=
Equations of Equilibrium: For point D [FBD (c)].
+ ΣFx=0; VD 0:= VD 0 kN= Ans
+ ΣFy=0; ND 0:= ND 0 kN= Ans
+ ΣΜD=0; MD 0:= MD 0 kN m⋅= Ans
For point E [FBD (d)].
+ ΣFx=0; VF Ax Cx−+ 0= VF Ax− Cx+:= VF 0 kN= Ans
+ ΣFy=0; NF Ay Cy−⋅− 0= NF Ay Cy+:= NF 12 kN= Ans
+ ΣΜF=0; MF Ax Cx+( ) 0.5 b⋅( )⋅− 0= MF Ax
Cx+( ) 0.5 b⋅( )⋅:= MF 4.8 kN m⋅= Ans
+ ΣFx=0; Ax VE− 0= VE Ax:=
+ ΣFy=0; NE Ay− 0= NE Ay:=
+ ΣΜE=0; ME Ax 0.5 b⋅( )⋅− 0= ME Ax 0.5 b⋅( )⋅:= ME 2.4 kN m⋅= Ans
For point F [FBD (e)].
VE 2.67 kN= Ans
NE 6 kN= Ans
Problem 1-21
The drum lifter suspends the 2.5-kN drum. The linkage is pin connected to the plate at A and B. The
gripping action on the drum chime is such that only horizontal and vertical forces are exerted on the
drum at G and H. Determine the resultant internal loadings on the cross section through point I.
Given: P 2.5 kN:= θ 60deg:=
a 200mm:= b 125mm:= c 75mm:=
d 125mm:= e 125mm:= f 50mm:=
Solution:
Equations of Equilibrium: Memeber Ac and BD are
two-force members.
ΣFy=0; P 2 F⋅ sin θ( )− 0= [1]
F
P
2 sin θ( )⋅:= [2]
F 1.443 kN=
Equations of Equilibrium: For point I.
+ ΣFx=0; VI F cos θ( )⋅− 0= VI F cos θ( )⋅:= VI 0.722 kN= Ans
+ ΣFy=0; NI− F sin θ( )⋅+ 0= NI F sin θ( )⋅:= NI 1.25 kN= Ans
+ ΣΜI=0; MI− F cos θ( )⋅ a( )⋅+ 0=
MI F cos θ( )⋅ a( )⋅:= MI 0.144 kN m⋅= Ans
Problem 1-22
Determine the resultant internal loadings on the cross sections through points K and J on the drum lifter
in Prob. 1-21.
Given: P 2.5 kN:= θ 60deg:=
a 200mm:= b 125mm:= c 75mm:=
d 125mm:= e 125mm:= f 50mm:=
Solution:
Equations of Equilibrium: Memeber Ac and BD are
two-force members.
ΣFy=0; P 2 F⋅ sin θ( )− 0= [1]
F
P
2 sin θ( )⋅:= [2]
F 1.443 kN=
Equations of Equilibrium: For point J.
+ ΣFy'=0; VI 0:= VI 0 kN= Ans
 + ΣFx'=0; NI F+ 0= NI F−:= NI 1.443− kN= Ans
+ ΣΜJ=0; MJ 0:= MJ 0 kN m⋅= Ans
Note: Negative sign indicates that NJ acts in the opposite direction to that shown on FBD.
Support Reactions: For Member DFH : 
+ ΣΜH=0; FEF c( )⋅ F cos θ( ) a b+ c+( )⋅− F sin θ( )⋅ f( )⋅+ 0=
FEF F cos θ( )⋅ a b+ c+c⎛⎜⎝
⎞
⎠⋅ F sin θ( )⋅
f
c
⎛⎜⎝
⎞
⎠⋅−:=
FEF 3.016 kN=
Equations of Equilibrium: For point K.
+ ΣFx=0; NK FEF+ 0= NK FEF:= NK 3.016 kN= Ans
+ ΣFy=0; VK 0:= VK 0 kN= Ans
+ ΣΜK=0; MK 0:= MK 0 kN m⋅= Ans
Problem 1-23
The pipe has a mass of 12 kg/m. If it is fixed to the wall at A, determine the resultant internal loadings
acting on the cross section at B. Neglect the weight of the wrench CD.
Given: g 9.81
m
s2
:= ρ 12 kg
m
:= P 60N:=
a 0.150m:= b 0.400m:=
c 0.200m:= d 0.300m:=
Solution: w ρ g⋅:=
ΣFx=0; NBx 0N:= Ans
ΣFy=0; VBy 0N:= Ans
ΣFz=0; VBz P− P+ w b c+( )⋅− 0=
VBz P P− w b c+( )⋅+:=
VBz 70.6 N= Ans
ΣΜx=0; TBx P b( )⋅+ P b( )⋅− w b⋅( ) 0.5 b⋅( )⋅− 0=
TBx P− b( )⋅ P b( )⋅+ w b⋅( ) 0.5 b⋅( )⋅+:= TBx 9.42 N m⋅= Ans
ΣΜy=0; MBy P 2a( )⋅− w b⋅( ) c( )⋅+ w c⋅( ) 0.5 c⋅( )⋅+ 0=
MBy P 2 a⋅( )⋅ w b⋅( ) c( )⋅− w c⋅( ) 0.5 c⋅( )⋅−:= MBy 6.23 N m⋅= Ans
ΣΜz=0; MBz 0N m⋅:= Ans
Problem 1-24
The main beam AB supports the load on the wing of the airplane. The loads consist of the wheel
reaction of 175 kN at C, the 6-kN weight of fuel in the tank of the wing, having a center of gravity at
D, and the 2-kN weight of the wing, having a center of gravity at E. If it is fixed to the fuselage at A,
determine the resultant internal loadings on the beam at this point. Assume that the wing does not
transfer any of the loads to the fuselage, except through the beam.
Given: PC 175kN:= PE 2kN:= PD 6kN:=
a 1.8m:= b 1.2m:= e 0.3m:=
c 0.6m:= d 0.45m:=
Solution:
ΣFx=0; VAx 0kN:= Ans
ΣFy=0; NAy 0kN:= Ans
ΣFz=0; VAz PD− PE− PC+ 0=
VAz PD PE PC−+:=
VAz 167− kN= Ans
ΣΜx=0;
MAx PD a( )⋅ PE a b+ c+( )⋅+ PC a b+( )⋅−:= MAx 507− kN m⋅= Ans
ΣΜy=0; TAy PD d( )⋅+ PE e( )⋅− 0=
TAy PD− d( )⋅ PE e( )⋅+:= TAy 2.1− kN m⋅= Ans
ΣΜz=0; MAz 0kN m⋅:= Ans
MAx PD a( )⋅− PE a b+ c+( )⋅− PC a b+( )⋅+ 0=
Problem 1-25
Determine the resultant internal loadings acting on the cross section through point B of the signpost.
The post is fixed to the ground and a uniform pressure of 50 N/m2 acts perpendicular to the face of the
sign.
Given: a 4m:= d 2m:= p 50 N
m2
:=
b 6m:= e 3m:=
c 3m:=
Solution: P p c( )⋅ d e+( )⋅:=
ΣFx=0; VBx P− 0=
VBx P:=
VBx 750 N= Ans
ΣFy=0; VBy 0N:= Ans
ΣFz=0; NBz 0N:= Ans
ΣΜx=0; MBx 0N m⋅:= Ans
ΣΜy=0; MBy P b 0.5 c⋅+( )⋅− 0=
MBy P b 0.5 c⋅+( )⋅:=
MBy 5625 N m⋅= Ans
ΣΜz=0; TBz P e 0.5 d e+( )⋅−[ ]⋅− 0=
TBz P e 0.5 d e+( )⋅−[ ]⋅:=
TBz 375 N m⋅= Ans
Problem 1-26
The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the
pulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section through
point D. The 400-N forces act in the -z direction and the 200-N and 80-N forces act in the +y
direction. The journal bearings at A and B exert only y and z components of force on the shaft.
Given: P1z 400N:= P2y 200N:= P3y 80N:=
a 0.3m:= b 0.4m:= c 0.3m:= d 0.4m:=
Solution: L a b+ c+ d+:=
Support Reactions:
ΣΜz=0; 2P3y d( )⋅ 2P2y c d+( )⋅+ Ay L( )⋅− 0=
Ay 2P3y
d
L
⋅ 2P2y
c d+
L
⋅+:=
Ay 245.71 N=
ΣFy=0; Ay− By− 2 P2y⋅+ 2 P3y⋅+ 0=
By Ay− 2 P2y⋅+ 2 P3y⋅+:= By 314.29 N=
ΣΜy=0; 2P1z b c+ d+( )⋅ Az L( )⋅− 0=
Az 2P1z
b c+ d+
L
⋅:= Az 628.57 N=
ΣFz=0; Bz Az+ 2 P1z⋅− 0=
Bz Az− 2 P1z⋅+:= Bz 171.43 N=
Equations of Equilibrium: For point D.
ΣFx=0; NDx 0N:= Ans
ΣFy=0; VDy By− 2 P3y⋅+ 0=
VDy By 2 P3y⋅−:=
VDy 154.3 N= Ans
ΣFz=0; VDz Bz+ 0=
VDz Bz−:=
VDz 171.4− N= Ans
ΣΜx=0; TDx 0N m⋅:= Ans
ΣΜy=0; MDy Bz d 0.5 c⋅+( )⋅+ 0=
MDy Bz d 0.5 c⋅+( )⋅⎡⎣ ⎤⎦−:= MDy 94.29− N m⋅= Ans
ΣΜz=0; MDz By d 0.5 c⋅+( )⋅+ 2 P3y⋅ 0.5 c⋅( )⋅− 0=
MDz By− d 0.5 c⋅+( )⋅ 2 P3y⋅ 0.5 c⋅( )⋅+:= MDz 148.86− N m⋅= Ans
Problem 1-27
The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the
pulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section through
point C. The 400-N forces act in the -z direction and the 200-N and 80-N forces act in the +y
direction. The journal bearings at A and B exert only y and z components of force on the shaft.
Given: P1z 400N:= P2y 200N:= P3y 80N:=
a 0.3m:= b 0.4m:= c 0.3m:= d 0.4m:=
Solution: L a b+ c+ d+:=
Support Reactions:
ΣΜz=0; 2P3y d( )⋅ 2P2y c d+( )⋅+ Ay L( )⋅− 0=
Ay 2P3y
d
L
⋅ 2P2y
c d+
L
⋅+:= Ay 245.71 N=
ΣFy=0; Ay− By− 2 P2y⋅+ 2 P3y⋅+ 0=
By Ay− 2 P2y⋅+ 2 P3y⋅+:= By 314.29 N=
ΣΜy=0; 2P1z b c+ d+( )⋅ Az L( )⋅− 0=
Az 2P1z
b c+ d+
L
⋅:= Az 628.57 N=
ΣFz=0; Bz Az+ 2 P1z⋅− 0=
Bz Az− 2 P1z⋅+:= Bz 171.43 N=
Equations of Equilibrium: For point C.
ΣFx=0; NCx 0N:= Ans
ΣFy=0; VCy Ay− 0=
VCy Ay:=
VCy 245.7 N= Ans
ΣFz=0; VCz Az+ 2P1z− 0=
VCz Az− 2P1z+:=
VCz 171.4 N= Ans
ΣΜx=0; TCx 0N m⋅:= Ans
ΣΜy=0; MCy Az a 0.5 b⋅+( )⋅− 2 P1z⋅ 0.5 b⋅( )⋅+ 0=
MCy Az a 0.5 b⋅+( )⋅ 2 P1z⋅ 0.5 b⋅( )⋅−:= MCy 154.29 N m⋅= Ans
ΣΜz=0; MCz Ay a 0.5 b⋅+( )⋅+ 0=
MCz Ay− a 0.5 b⋅+( )⋅:= MCz 122.86− N m⋅= Ans
Problem 1-28
Determine the resultant internal loadings acting on the cross section of the frame at points F and G.
The contact at E is smooth.
Given: a 1.2m:= b 1.5m:= c 0.9m:= P 400N:=
d 0.9m:= e 1.2m:= θ 30deg:=
Solution: L d2 e2+:=
Member DEF : 
+ ΣΜD=0; NE b( )⋅ P a b+( )⋅− 0=
NE P
a b+
b
⋅:= NE 720 N=
Member BCE : 
+ ΣΜB=0; FAC
e
L
⎛⎜⎝
⎞
⎠⋅ d( )⋅ NE sin θ( )⋅ c d+( )⋅− 0=
FAC
L
e d⋅
⎛⎜⎝
⎞
⎠ NE sin θ( )⋅ c d+( )⋅⎡⎣ ⎤⎦⋅:=
FAC 900 N=
+ ΣFx=0; Bx FAC
d
L
⎛⎜⎝
⎞
⎠⋅+ NE cos θ( )⋅− 0=
Bx FAC−
d
L
⎛⎜⎝
⎞
⎠⋅ NE cos θ( )⋅+:=
Bx 83.54 N=
+ ΣFy=0; By− FAC
e
L
⎛⎜⎝
⎞
⎠⋅+ NE sin θ( )⋅− 0=
By FAC
e
L
⎛⎜⎝
⎞
⎠⋅ NE sin θ( )⋅−:=
By 360 N=
Equations of Equilibrium: For point F.
+ ΣFy'=0; NF 0:= NF 0 N= Ans
 + ΣFx'=0; VF P− 0= VF P:= VF 400 N= Ans
+ ΣΜF=0; MF P 0.5 a⋅( )⋅− 0= MF P 0.5 a⋅( )⋅:= MF 240 N m⋅= Ans
Equations of Equilibrium: For point G.
+ ΣFx=0; Bx NG− 0= NG Bx:= NG 83.54 N= Ans
+ ΣFy=0; VG By− 0= VG By:= VG 360 N= Ans
+ ΣΜG=0; MG− By 0.5 d⋅( )⋅+ 0= MG By 0.5 d⋅( )⋅:= MG 162 N m⋅= Ans
Problem
1-29
The bolt shank is subjected to a tension of 400 N. Determine the resultant internal loadings acting on
the cross section at point C. 
Given:
P 400N:=
r 150mm:=
θ 90deg:=
Solution:
Equations of Equilibrium: For segment AC.
+ ΣFx=0; NC P+ 0= NC P:= NC 400 N= Ans
+ ΣFy=0; VC 0:= VC 0 N= Ans
+ ΣΜG=0; MC P r( )⋅+ 0= MC P− r( )⋅:= MC 60− N m⋅= Ans
Problem 1-30
The pipe has a mass of 12 kg/m. If it is fixed to the wall at A, determine the resultant internal loadings
acting on the cross section through B.
Given: P 750N:= MC 800N m⋅:=
ρ 12 kg
m
:= g 9.81 m
s2
:=
a 1m:= b 2m:= c 2m:=
Solution:
Py
4
5
⎛⎜⎝
⎞
⎠ P⋅:= Pz
3
5
− P⋅:=
Equations of Equilibrium: For point B.
ΣFx=0; VBx 0kip:= Ans
ΣFy=0; NBy Py+ 0=
NBy Py−:= Ans
NBy 600− N= Ans
ΣFz=0; VBz Pz+ ρ g⋅ c⋅− ρ g⋅ b⋅− 0=
VBz Pz− ρ g⋅ c⋅+ ρ g⋅ b⋅+:=
VBz 920.9 N= Ans
ΣΜx=0; MBx Pz b( )⋅+ ρ g⋅ c⋅ b( )⋅− ρ g⋅ b⋅ 0.5 b⋅( )⋅− 0=
MBx Pz− b( )⋅ ρ g⋅ c⋅ b( )⋅+ ρ g⋅ b⋅ 0.5 b⋅( )⋅+:=
MBx 1606.3 N m⋅= Ans
ΣΜy=0; TBy 0N m⋅:= Ans
ΣΜz=0; MBy Mc+ 0=
MBy MC−:=
MBy 800− N m⋅= Ans
Problem 1-31
The curved rod has a radius r and is fixed to the wall at B. Determine the resultant internal loadings
acting on the cross section through A which is located at an angle θ from the horizontal.
Solution: P kN:= θ deg:=
Equations of Equilibrium: For point A.
+ ΣFx=0; NA− P cos θ( )⋅+ 0=
NA P cos θ( )⋅:= Ans
 + ΣFy=0; VA P sin θ( )⋅− 0=
VA P sin θ( )⋅:= Ans
+ ΣΜA=0; MA P r⋅ 1 cos θ( )−( )⋅− 0=
MA P r⋅ 1 cos θ( )−( )⋅:= r Ans
Problem 1-32
The curved rod AD of radius r has a weight per length of w. If it lies in the horizontal plane, determine
the resultant internal loadings acting on the cross section through point B. Hint: The distance from the
centroid C of segment AB to point O is CO = 0.9745r.
Given: θ 22.5deg:= r m:= a 0.9745r:= w kN
m2
:=
Solution:
Equations of Equilibrium: For point B.
ΣFz=0; VB
π
4
r⋅ w⋅− 0= VB 0.785 w⋅ r⋅:= Ans
ΣFx=0; NB 0:= Ans
ΣΜx=0; TB
π
4
r⋅ w⋅ 0.09968r( )⋅− 0= TB 0.0783w r2⋅:= Ans
ΣΜy=0; MB
π
4
r⋅ w⋅ 0.37293r( )⋅+ 0= MB 0.293− w r2⋅:= Ans
Problem 1-33
A differential element taken from a curved bar is shown in the figure. Show that dN/dθ = V, 
dV/dθ = -N, dM/dθ = -T, and dT/dθ = M, 
Solution:
Problem 1-34
The column is subjected to an axial force of 8 kN, which is applied through the centroid of the
cross-sectional area. Determine the average normal stress acting at section a–a. Show this distribution
of stress acting over the area’s cross section.
Given: P 8kN:=
b 150mm:= d 140mm:= t 10mm:=
Solution:
A 2 b t⋅( ) d t⋅+:= A 4400.00 mm2=
σ P
A
:= σ 1.82 MPa= Ans
Problem 1-35
The anchor shackle supports a cable force of 3.0 kN. If the pin has a diameter of 6 mm, determine the
average shear stress in the pin.
Given: P 3.0kN:= d 6mm:=
Solution:
+ ΣFy=0; 2 V⋅ P− 0=
V 0.5P:=
V 1500 N=
A
π d2⋅
4
:= A 28.2743 mm2=
τavg
V
A
:= τavg 53.05 MPa= Ans
Problem 1-36
While running the foot of a 75-kg man is momentarily subjected to a force which is 5 times his
weight. Determine the average normal stress developed in the tibia T of his leg at the mid section a-a.
The cross section can be assumed circular, having an outer diameter of 45 mm and an inner diameter
of 25 mm. Assume the fibula F does not support a load.
Given: g 9.81
m
s2
=
M 75kg:=
do 45mm:= di 25mm:=
Solution:
A
π
4
do
2 di
2−⎛⎝ ⎞⎠⋅:= A 1099.5574 mm2=
σ 5M g⋅
A
:= σ 3.345 MPa= Ans
Problem 1-37
The thrust bearing is subjected to the loads shown. Determine the average normal stress developed on
cross sections through points B, C, and D. Sketch the results on a differential volume element located
at each section.
Units Used: kPa 103Pa:=
Given: P 500N:= Q 200N:=
dB 65mm:= dC 140mm:= dD 100mm:=
Solution:
AB
π dB2⋅
4
:= AB 3318.3 mm2=
σB
P
AB
:= σB 150.7 kPa= Ans
AC
π dC2⋅
4
:= AC 15393.8 mm2=
σC
P
AC
:= σC 32.5 kPa= Ans
AD
π dD2⋅
4
:= AD 7854.0 mm2=
σD
Q
AD
:= σD 25.5 kPa= Ans
Problem 1-38
The small block has a thickness of 5 mm. If the stress distribution at the support developed by the load
varies as shown, determine the force F applied to the block, and the distance d to where it is applied.
Given: a 60mm:= b 120mm:= t 5mm:=
σ1 0MPa:= σ2 40MPa:= σ3 60MPa:=
Solution:
F Aσ⌠⎮⌡ d=
F 0.5 σ2⋅ a t⋅( )⋅ σ2 b t⋅( )⋅+ 0.5 σ3 σ2−( )⋅ b t⋅( )⋅+:=
F 36.00 kN= Ans
Require:
F d⋅ Ax σ⋅⌠⎮⌡ d=
d
0.5 σ2⋅ a t⋅( )⋅⎡⎣ ⎤⎦ 2a3⋅ σ2 b t⋅( )⋅⎡⎣ ⎤⎦ a 0.5 b⋅+( )⋅+ 0.5 σ3 σ2−( )⋅ b t⋅( )⋅⎡⎣ ⎤⎦ a 2 b⋅3+⎛⎜⎝ ⎞⎠⋅+
F
:=
d 110 mm= Ans
Problem 1-39
The lever is held to the fixed shaft using a tapered pin AB, which has a mean diameter of 6 mm. If a
couple is applied to the lever, determine the average shear stress in the pin between the pin and lever.
Given: a 250mm:= b 12mm:=
d 6mm:= P 20N:=
Solution:
+ ΣΜO=0; V b⋅ P 2a( )⋅− 0=
V P
2a
b
⎛⎜⎝
⎞
⎠⋅:=
V 833.33 N=
A
π d2⋅
4
:= A 28.2743 mm2=
τavg
V
A
:= τavg 29.47 MPa= Ans
Problem 1-40
The cinder block has the dimensions shown. If the material fails when the average normal stress
reaches 0.840 MPa, determine the largest centrally applied vertical load P it can support.
Given: σallow 0.840MPa:=
ao 150mm:= ai 100mm:=
bo 2 1 2+ 3+( )⋅ 2+[ ] mm⋅:=
bi 2 1 3+( )⋅[ ] mm⋅:=
Solution:
A ao bo⋅ ai bi⋅−:= A 1300 mm2=
Pallow σallow A( )⋅:=
Pallow 1.092 kN= Ans
Problem 1-41
The cinder block has the dimensions shown. If it is subjected to a centrally applied force of P = 4 kN,
determine the average normal stress in the material. Show the result acting on a differential volume
element of the material.
Given: P 4kN:=
ao 150mm:= ai 100mm:=
bo 2 1 2+ 3+( )⋅ 2+[ ] mm⋅:=
bi 2 1 3+( )⋅[ ] mm⋅:=
Solution:
A ao bo⋅ ai bi⋅−:= A 1300 mm2=
σ P
A
:=
σ 3.08 MPa= Ans
Problem 1-42
The 250-N lamp is supported by three steel rods connected by a ring at A. Determine which rod is
subjected to the greater average normal stress and compute its value. Take θ = 30°. The diameter of
each rod is given in the figure.
Given: W 250N:= θ 30deg:= φ 45deg:=
dB 9mm:= dC 6mm:= dD 7.5mm:=
Solution: Initial guess: FAC 1N:= FAD 1N:=
Given
+ ΣFx=0; FAC cos θ( )⋅ FAD cos φ( )⋅− 0= [1]
+ ΣFy=0; FAC sin θ( )⋅ FAD sin φ( )⋅+ W− 0= [2]
Solving [1] and [2]:
FAC
FAD
⎛⎜⎜⎝
⎞
⎠
Find FAC FAD,( ):=
FAC
FAD
⎛⎜⎜⎝
⎞
⎠
183.01
224.14
⎛⎜⎝
⎞
⎠ N=
Rod AB:
AAB
π dB2⋅
4
:= AAB 63.61725 mm2=
σAB
W
AAB
:= σAB 3.93 MPa=
Rod AD :
AAD
π dD2⋅
4
:= AAD 44.17865 mm2=
σAD
FAD
AAD
:= σAD 5.074 MPa=
Rod AC:
AAC
π dC2⋅
4
:= AAC 28.27433 mm2=
σAC
FAC
AAC
:= σAC 6.473 MPa= Ans
Problem 1-43
Solve Prob. 1-42 for θ = 45°.
Given: W 250N:= θ 45deg:= φ 45deg:=
dB 9mm:= dC 6mm:= dD 7.5mm:=
Solution: Initial guess: FAC 1N:= FAD 1N:=
Given
+ ΣFx=0; FAC cos θ( )⋅ FAD cos φ( )⋅− 0= [1]
+ ΣFy=0; FAC sin θ( )⋅ FAD sin φ( )⋅+ W− 0= [2]
Solving [1] and [2]:
FAC
FAD
⎛⎜⎜⎝
⎞
⎠
Find FAC FAD,( ):=
FAC
FAD
⎛⎜⎜⎝
⎞
⎠
176.78
176.78
⎛⎜⎝
⎞
⎠ N=
Rod AB:
AAB
π dB2⋅
4
:= AAB 63.61725 mm2=
σAB
W
AAB
:= σAB 3.93 MPa=
Rod AD :
AAD
π dD2⋅
4
:= AAD 44.17865 mm2=
σAD
FAD
AAD
:= σAD 4.001 MPa=
Rod AC:
AAC
π dC2⋅
4
:= AAC 28.27433 mm2=
σAC
FAC
AAC
:= σAC 6.252 MPa= Ans
Problem 1-44
The 250-N lamp is supported by three steel rods connected by a ring at A. Determine the angle of
orientation θ of AC such that the average normal stress in rod AC is twice the average normal stress in
rod AD. What is the magnitude of stress in each rod? The diameter of each rod is given in the figure.
Given: W 250N:= φ 45deg:=
dB 9mm:= dC 6mm:= dD 7.5mm:=
Solution:
Rod AB: AAB
π dB2⋅
4
:= AAB 63.61725 mm2=
Rod AD : AAD
π dD2⋅
4
:= AAD 44.17865 mm2=
Rod AC: AAC
π dC2⋅
4
:= AAC 28.27433 mm2=
Since σAC 2σAD= Therefore
FAC
AAC
2
FAD
AAD
=
Initial guess: FAC 1N:= FAD 2N:= θ 30deg:=
Given FAC
AAC
2
FAD
AAD
= [1]
+ ΣFx=0; FAC cos θ( )⋅ FAD cos φ( )⋅− 0= [2]
+ ΣFy=0; FAC sin θ( )⋅ FAD sin φ( )⋅+ W− 0= [3]
Solving [1], [2] and [3]:
FAC
FAD
θ
⎛⎜⎜⎜⎝
⎞
⎟
⎠
Find FAC FAD, θ,( ):= FACFAD
⎛⎜⎜⎝
⎞
⎠
180.38
140.92
⎛⎜⎝
⎞
⎠ N=
θ 56.47 deg=
σAB
W
AAB
:= σAB 3.93 MPa= Ans
σAD
FAD
AAD
:= σAD 3.19 MPa= Ans
σAC
FAC
AAC
:= σAC 6.38 MPa= Ans
Problem 1-45
The shaft is subjected to the axial force of 30 kN. If the shaft passes through the 53-mm diameter hole
in the fixed support A, determine the bearing stress acting on the collar C. Also, what is the average
shear stress acting along the inside surface of the collar where it is fixed connected to
the 52-mm diameter shaft?
Given: P 30kN:=
dhole 53mm:= dshaft 52mm:=
dcollar 60mm:= hcollar 10mm:=
Solution:
Bearing Stress:
Ab
π
4
dcollar
2 dhole
2−⎛⎝ ⎞⎠⋅:=
σb
P
Ab
:= σb 48.3 MPa= Ans
Average Shear Stress:
As π dshaft( )⋅ hcollar( )⋅:=
τavg
P
As
:= τavg 18.4 MPa= Ans
Problem 1-46
The two steel members are joined together using a 60° scarf weld. Determine the average normal and
average shear stress resisted in the plane of the weld.
Given:
P 8kN:= θ 60deg:=
b 25mm:= h 30mm:=
Solution:
Equations of Equilibrium: 
 + ΣFx=0; N P sin θ( )⋅− 0=
N P sin θ( )⋅:= N 6.928 kN=
+ ΣFy=0; V P cos θ( )⋅− 0=
V P cos θ( )⋅:= V 4 kN=
A
h b⋅
sin θ( ):=
σ N
A
:= σ 8 MPa= Ans
τavg
V
A
:= τavg 4.62 MPa= Ans
Problem 1-47
The J hanger is used to support the pipe such that the force on the vertical bolt is 775 N. Determine the
average normal stress developed in the bolt BC if the bolt has a diameter of 8 mm. Assume A is a pin.
Given: P 775N:=
a 40mm:= b 30mm:=
d 8mm:= θ 20deg:=
Solution:
Support Reaction: 
 ΣFA=0; P a( )⋅ FBC cos θ( )⋅ a b+( )⋅− 0=
FBC
P a⋅
a b+( ) cos θ( )⋅:=
FBC 471.28 N=
Average Normal Stress: 
ABC
π d2⋅
4
:=
σ
FBC
ABC
:=
σ 9.38 MPa= Ans
Problem 1-48
The board is subjected to a tensile force of 425 N. Determine the average normal and average shear
stress developed in the wood fibers that are oriented along section a-a at 15° with the axis of the board.
Given: P 425N:= θ 15deg:=
b 25mm:= h 75mm:=
Solution:
Equations of Equilibrium: 
 + ΣFx=0; V P cos θ( )⋅− 0=
V P cos θ( )⋅:= V 410.518 N=
+ ΣFy=0; N P sin θ( )⋅− 0=
N P sin θ( )⋅:= N 1 N=
Average Normal Stress: 
A
h b⋅
sin θ( ):=
σ N
A
:= σ 0.0152 MPa= Ans
τavg
V
A
:= τavg 0.0567 MPa= Ans
Problem 1-49
The open square butt joint is used to transmit a force of 250 kN from one plate to the other. Determine
the average normal and average shear stress components that this loading creates on the face of the
weld, section AB.
Given: P 250kN:= θ 30deg:=
b 150mm:= h 50mm:=
Solution:
Equations of Equilibrium: 
 + ΣFx=0; V− P sin θ( )⋅+ 0=
V P sin θ( )⋅:= V 125 kN=
+ ΣFy=0; N P cos θ( )⋅− 0=
N P cos θ( )⋅:= N 216.506 kN=
Average Normal and Shear Stress: 
A
h b⋅
sin 2θ( ):=
σ N
A
:= σ 25 MPa= Ans
τavg
V
A
:= τavg 14.434 MPa= Ans
Problem 1-50
The specimen failed in a tension test at an angle of 52° when the axial load was 100 kN. If the diameter
of the specimen is 12 mm, determine the average normal and average shear stress acting on the area of
the inclined failure plane. Also, what is the average normal stress acting on the cross section when
failure occurs?
Given: P 100kN:=
d 12mm:= θ 52deg:=
Solution:
Equations of Equilibrium: 
+ ΣFx=0; V P cos θ( )⋅− 0=
V P cos θ( )⋅:= V 61.566 kN=
+ ΣFy=0; N P sin θ( )⋅− 0=
N P sin θ( )⋅:= N 78.801 kN=
Inclined plane:
A
π
4
d2
sin θ( )
⎛⎜⎝
⎞
⎠:=
σ N
A
:= σ 549.05 MPa= Ans
τavg
V
A
:= τavg 428.96 MPa= Ans
Cross section: 
A
πd2
4
:=
σ P
A
:= σ 884.19 MPa= Ans
τavg 0:= τavg 0 MPa= Ans
Problem 1-51
A tension specimen having a cross-sectional area A is subjected to an axial force P. Determine the
maximum average shear stress in the specimen and indicate the orientation θ of a section on which it
occurs. 
Solution:
Equations of Equilibrium: 
+ ΣFy=0; V P cos θ( )⋅− 0=
V P cos θ( )⋅=
Inclined plane:
Aincl
A
sin θ( )=
τ V
Aincl
= τ P cos θ( )⋅ sin θ( )⋅
A
= τ P sin 2θ( )⋅
2A
=
dτ
dθ
P cos 2θ( )⋅
A
=
dτ
dθ 0=
cos 2θ( ) 0=
2θ 90deg=
θ 45deg:= Ans
τmax
P sin 90°( )⋅
2A
= τmax
P
2A
= Ans
Problem 1-52
The joint is subjected to the axial member force of 5 kN. Determine the average normal stress acting
on sections AB and BC. Assume the member is smooth and is 50-mm thick.
Given: P 5kN:= θ 45deg:= φ 60deg:=
dAB 40mm:= dBC 50mm:= t 50mm:=
Solution: α 90deg φ−:= α 30.00 deg=
AAB t dAB⋅:=
ABC t dBC⋅:=
+ ΣFx=0; NAB cos α( )⋅ P cos θ( )⋅− 0=
NAB
P cos θ( )⋅
cos α( ):=
NAB 4.082 kN=
+ ΣFy=0; NAB− sin α( )⋅ P sin θ( )⋅+ NBC− 0=
NBC NAB− sin α( )⋅ P sin θ( )⋅+:=
NBC 1.494 kN=
σAB
NAB
AAB
:= σAB 2.041 MPa= Ans
σBC
NBC
ABC
:= σBC 0.598 MPa= Ans
Problem 1-53
The yoke is subjected to the force and couple moment. Determine the average shear stress in the bolt
acting on the cross sections through A and B. The bolt has a diameter of 6 mm. Hint: The couple
moment is resisted by a set of couple forces developed in the shank of the bolt.
Given: P 2.5kN:= M 120N m⋅:=
ho 62mm:= hi 50mm:=
d 6mm:= θ 60deg:=
Solution:
As a force on bolt shank is zero, then
τA 0:= Ans
Equations od Equilibrium:
ΣFz=0; P 2Fz− 0=
Fz 0.5P:= Fz 1.25 kN=
ΣMz=0; M Fx hi( )⋅− 0=
Fx
M
hi
:= Fx 2.4 kN=
Average Shear Stress: 
A
πd2
4
:=
The bolt shank subjected to a shear force of VB Fx
2 Fz
2+:=
τB
VB
A
:= τB 95.71 MPa= Ans
Problem 1-54
The two members used in the construction of an aircraft fuselage are joined together using a 30°
fish-mouth weld. Determine the average normal and average shear stress on the plane of each weld.
Assume each inclined plane supports a horizontal force of 2 kN.
Given:
P 4.0kN:=
b 37.5mm:= hhalf 25m:=
θ 30deg:=
Solution:
Equations of Equilibrium: 
 + ΣFx=0; V− 0.5P cos θ( )⋅+ 0=
V 0.5P cos θ( )⋅:= V 1.732 kN=
+ ΣFy=0; N 0.5P sin θ( )⋅− 0=
N 0.5P sin θ( )⋅:= N 1 kN=
Average Normal and Shear Stress: 
A
hhalf( ) b⋅
sin θ( ):=
σ N
A
:= σ 533.33 Pa= Ans
τavg
V
A
:= τavg 923.76 Pa= Ans
Problem 1-55
The row of staples AB contained in the stapler is glued together so that the maximum shear stress the
glue can withstand is τ max = 84 kPa. Determine the minimum force F that must be placed on the
plunger in order to shear off a staple from its row and allow it to exit undeformed through the groove
at C. The outer dimensions of the staple are shown in the figure. It has a thickness of 1.25 mm
Assume all the other parts are rigid and neglect friction.
Given: τmax 0.084MPa:=
a 12.5mm:= b 7.5mm:=
t 1.25mm:=
Solution:
Average Shear Stress:
A a b⋅ a 2t−( ) b t−( )⋅[ ]−:=
τmax
V
A
= V τmax( ) A⋅:=
V 2.63 N=
Fmin V:=
Fmin 2.63 N= Ans
Problem 1-56
Rods AB and BC have diameters of 4mm and 6 mm, respectively. If the load of 8 kN is applied to the
ring at B, determine the average normal stress in each rod if θ = 60°.
Given: W 8kN:= θ 60deg:=
dA 4mm:= dC 6mm:=
Solution:
Rod AB: AAB
π dA2⋅
4
:=
Rod BC : ABC
π dC2⋅
4
:=
+ ΣFy=0; FBC sin θ( )⋅ W− 0=
FBC
W
sin θ( ):=
FBC 9.238 kN=
+ ΣFx=0; FBC cos θ( )⋅ FAB− 0=
FAB FBC cos θ( )⋅:=
FAB 4.619 kN=
σAB
FAB
AAB
:= σAB 367.6 MPa= Ans
σBC
FBC
ABC
:= σBC 326.7 MPa= Ans
Problem 1-57
Rods AB and BC have diameters of 4 mm and 6 mm, respectively. If the vertical load of 8 kN is
applied to the ring at B, determine the angle θ of rod BC so that the average normal stress in each rod
is equivalent. What is this stress?
Given: W 8kN:=
dA 4mm:= dC 6mm:=
Solution:
Rod AB: AAB
π dA2⋅
4
:=
Rod BC : ABC
π dC2⋅
4
:=
+ ΣFy=0; FBC sin θ( )⋅ W− 0=
+ ΣFx=0; FBC cos θ( )⋅ FAB− 0=
Since FAB σ AAB⋅=
FBC σ ABC⋅=
Initial guess: σ 100MPa:= θ 50deg:=
Given
σ ABC⋅ sin θ( )⋅ W− 0= [1]
σ ABC⋅ cos θ( )⋅ σ AAB⋅− 0= [2]
Solving [1] and [2]:
σ
θ
⎛⎜⎝
⎞
⎠ Find σ θ,( ):=
θ 63.61 deg= Ans
σ 315.85 MPa= Ans
Problem 1-58
The bars of the truss each have a cross-sectional area of 780 mm2. Determine the average normal
stress in each member due to the loading P = 40 kN. State whether the stress is tensile or compressive.
Given: P 40kN:=
a 0.9m:= b 1.2m:= A 780mm2:=
Solution: c a2 b2+:= c 1.5 m=
h
b
c
:= v a
c
:=
Joint A:
+ ΣFy=0; v( ) FAB⋅ P− 0= FAB
P
v
:= FAB 66.667 kN=
+ ΣFx=0; h( ) FAB⋅ FAE− 0= FAE h( ) FAB⋅:= FAE 53.333 kN=
σAB
FAB
A
:= σAB 85.47 MPa= (T) Ans
σAE
FAE
A
:= σAE 68.376 MPa= (C) Ans
Joint E:
+ ΣFy=0; FEB 0.75P− 0= FEB 0.75P:= FEB 30 kN=
+ ΣFx=0; FED FAE− 0= FED FAE:= FED 53.333 kN=
σEB
FEB
A
:= σEB 38.462 MPa= (T) Ans
σED
FED
A
:= σED 68.376 MPa= (C) Ans
Joint B:
+ ΣFy=0; v( ) FBD⋅ v( ) FAB⋅− FEB− 0= FBD FAB
FEB
v
⎛⎜⎝
⎞
⎠+:= FBD 116.667 kN=
+ ΣFx=0; FBC h( )FAB− h( )FBD− 0= FBC h( )FAB h( )FBD+:= FBC 146.667 kN=
σBC
FBC
A
:= σBC 188.034 MPa= (T) Ans
σBD
FBD
A
:= σBD 149.573 MPa= (C) Ans
Problem 1-59
The bars of the truss each have a cross-sectional area of 780 mm2. If the maximum average normal
stress in any bar is not to exceed 140 MPa, determine the maximum magnitude P of the loads that can
be applied to the truss.
σallow 140MPa:=Given:
a 0.9m:= b 1.2m:= A 780mm2:=
Solution: c a2 b2+:= c 1.5 m=
h
b
c
:= v a
c
:=
For comparison purpose, set P 1kN:=
Joint A:
+ ΣFy=0; v( ) FAB⋅ P− 0= FAB
P
v
:= FAB 1.667 kN=
+ ΣFx=0; h( ) FAB⋅ FAE− 0= FAE h( ) FAB⋅:= FAE 1.333 kN=
σAB
FAB
A
:= σAB 2.137 MPa= (T)
σAE
FAE
A
:= σAE 1.709 MPa= (C)
Joint E:
+ ΣFy=0; FEB 0.75P− 0= FEB 0.75P:= FEB 0.75 kN=
+ ΣFx=0; FED FAE− 0= FED FAE:= FED 1.333 kN=
σEB
FEB
A
:= σEB 0.962 MPa= (T)
σED
FED
A
:= σED 1.709 MPa= (C)
Joint B:
+ ΣFy=0; v( ) FBD⋅ v( ) FAB⋅− FEB− 0= FBD FAB
FEB
v
⎛⎜⎝
⎞
⎠+:= FBD 2.917 kN=
+ ΣFx=0; FBC h( )FAB− h( )FBD− 0= FBC h( )FAB h( )FBD+:= FBC 3.667 kN=
σBC
FBC
A
:= σBC 4.701 MPa= (T)
σBD
FBD
A
:= σBD 3.739 MPa= (C)
Since the cross-sectional areas are the same, the highest stress occurs in the member BC,
which has the greatest force
Fmax max FAB FAE, FEB, FED, FBD, FBC,( ):= Fmax 3.667 kN=
Pallow
P
Fmax
⎛⎜⎝
⎞
⎠
σallow A⋅( )⋅:= Pallow 29.78 kN= Ans
Problem 1-60
The plug is used to close the end of the cylindrical tube that is subjected to an internal pressure of p =
650 Pa. Determine the average shear stress which the glue exerts on the sides of the tube needed to
hold the cap in place.
Given: p 650Pa:= a 25mm:=
di 35mm:= do 40mm:=
Solution:
Ap
π di2⋅
4
:= As π do⋅( ) a( ):=
P a( )⋅ FBC cos θ( )⋅ a b+( )⋅− 0=
P p Ap( )⋅:= P 0.625 N=
Average Shear Stress: 
τavg
P
As
:=
τavg 199.1 Pa= Ans
Problem 1-61
The crimping tool is used to crimp the end of the wire E. If a force of 100 N is applied to the handles,
determine the average shear stress in the pin at A. The pin is subjected to double shear and has a
diameter of 5 mm. Only a vertical force is exerted on the wire.
Given: P 100N:=
a 37.5mm:= b 50mm:= c 25mm:=
d 125mm:= dpin 5mm:=
Solution:
From FBD (a):
+ ΣFx=0; Bx 0:=
Bx 0 N=
+ ΣΜD=0; P d( )⋅ By c( )⋅− 0=
By P
d
c
⋅:=
By 500 N=
From FBD (b):
+ ΣFx=0; Ax 0:=
Ax 0 N=
+ ΣΜE=0; Ay a( )⋅ By a b+( )⋅− 0=
Ay By
a b+
a
⋅:=
Ay 1166.67 N=
Average Shear Stress: 
Apin
π dpin2⋅
4
:=
VA 0.5 Ay( )⋅:= VA 583.333 N=
τavg
VA
Apin
:= τavg 29.709 MPa= Ans
Problem 1-62
Solve Prob. 1-61 for pin B. The pin is subjected to double shear and has a diameter of 5 mm.
Given: a 37.5mm:= b 50mm:= c 25mm:=
d 125mm:= dpin 5mm:= P 100N:=
Solution:
From FBD (a):
+ ΣFx=0; Bx 0:=
Bx 0 N=
+ ΣΜD=0; P d( )⋅ By c( )⋅− 0=
By P
d
c
⋅:=
By 500 N=
Average Shear Stress: Pin B is subjected to doule shear
Apin
π dpin2⋅
4
:=
VB 0.5 By( )⋅:= VB 250 N=
τavg
VB
Apin
:= τavg 12.732 MPa= Ans
Problem 1-63
The railcar docklight is supported by the 3-mm-diameter pin at A. If the lamp weighs 20 N, and the
extension arm AB has a weight of 8 N/m, determine the average shear stress in the pin needed to
support the lamp. Hint: The shear force in the pin is caused by the couple moment required for
equilibrium at A.
Given: w 8
N
m
:= P 20N:=
a 900mm:= h 32mm:=
dpin 3mm:=
Solution:
From FBD (a):
+ ΣFx=0; Bx 0:=
+ ΣΜA=0; V h( )⋅ w a⋅( ) 0.5a( )⋅− P a( )⋅− 0=
V w a⋅( ) 0.5 a
h
⎛⎜⎝
⎞
⎠⋅ P
a
h
⋅+:= V 663.75 N=
Average Shear Stress: 
Apin
π dpin2⋅
4
:=
τavg
V
Apin
:= τavg 93.901 MPa= Ans
Problem 1-64
The two-member frame is subjected to the distributed loading shown. Determine the average normal
stress and average shear stress acting at sections a-a and b-b. Member CB has a square cross section
of 35 mm on each side. Take w = 8 kN/m.
Given: w 8
kN
m
:=
a 3m:= b 4m:= A 0.0352( )m2:=
Solution: c a2 b2+:= c 5 m=
h
a
c
:= v b
c
:=
Member AB:
 ΣMA=0; By a( )⋅ w a⋅( ) 0.5a( )⋅− 0=
By 0.5w a⋅:= By 12 kN=
+ ΣFy=0; v( ) FAB⋅ By− 0=
FAB
By
v
:= FAB 15 kN=
Section a-a: 
σa_a
FAB
A
:= σa_a 12.24 MPa= Ans
τa_a 0:= τa_a 0 MPa= Ans
Section b-b: 
+ ΣFx=0; N FAB h( )⋅− 0= N FAB h( )⋅:= N 9 kN=
+ ΣFy=0; V FAB v( )⋅− 0= V FAB v( )⋅:= V 12 kN=
Ab_b
A
h
:=
σb_b
N
Ab_b
:= σb_b 4.41 MPa= Ans
τb_b
V
Ab_b
:= τb_b 5.88 MPa= Ans
Problem 1-65
Member A of the timber step joint for a truss is subjected to a compressive force of 5 kN. Determine
the average normal stress acting in the hanger rod C which has a diameter of 10 mm and in member B
which has a thickness of 30 mm.
Given: P 5kN:= θ 60deg:= φ 30deg:=
drod 10mm:= h 40mm:= t 30mm:=
Solution:
AB t h⋅:=
Arod
π
4
drod
2⋅:=
+ ΣFx=0; P cos θ( )⋅ FB− 0=
FB P cos θ( )⋅:= FB 2.5 kN=
+ ΣFy=0; Fc P sin θ( )⋅− 0=
FC P sin θ( )⋅:= FC 4.33 kN=
Average Normal Stress:
σB
FB
AB
:= σB 2.083 MPa= Ans
σC
FC
Arod
:= σC 55.133 MPa= Ans
Problem 1-66
Consider the general problem of a bar made from m segments, each having a constant cross-sectional
area Am and length Lm. If there are n loads on the bar as shown, write a computer program that can be
used to determine the average normal stress at any specified location x. Show an application of the
program using the values L1 = 1.2 m, d1 = 0.6 m, P1 = 2 kN, A1 = 1875 mm2, L2 = 0.6 m, d2 = 1.8 m,
P2 = -1.5 kN, A2 = 625 mm2.
Problem 1-67
The beam is supported by a pin at A and a short link BC. If P = 15 kN, determine the average shear
stress developed in the pins at A, B, and C. All pins are in double shear as shown, and each has a
diameter of 18 mm.
Given: P 15kN:=
a 0.5m:= b 1m:= c 1.5m:=
d 1.5m:= e 0.5m:=
θ 30deg:= dpin 18mm:=
Solution: L a b+ c+ d+ e+:=
Support Reactions:
 ΣΜA=0; By− L( )⋅ P L a−( )⋅+ 4P c d+ e+( )⋅+ 4P d e+( )⋅+ 2P e( )⋅+ 0=
By P
L a−
L
⋅ 4 P⋅ c d+ e+
L
⋅+ 4 P⋅ d e+
L
⋅+ 2 P⋅ e
L
⋅+:=
By 82.5 kN=
 + ΣFy=0; By− P+ 4 P⋅+ 4 P⋅+ 2 P⋅+ Ay− 0=
Ay By− P+ 4 P⋅+ 4P+ 2 P⋅+:=
Ay 82.5 kN=
FBC
By
sin θ( ):= FBC 165 kN=
Ax FBC cos θ( )⋅:= Ax 142.89 kN=
Average Shear Stress: 
Apin
π dpin2⋅
4
:=
For Pins B and C: 
τB_and_C
0.5FBC
Apin
:= τB_and_C 324.2 MPa= Ans
For Pin A: 
FA Ax
2 Ay
2+:= FA 165 kN=
τA
0.5FA
Apin
:= τA 324.2 MPa= Ans
Problem 1-68
The beam is supported by a pin at A and a short link BC. Determine the maximum magnitude P of the
loads the beam will support if the average shear stress in each pin is not to exceed 80 MPa. All pins are
in double shear as shown, and each has a diameter of 18 mm.
Given: τallow 80MPa:=
a 0.5m:= b 1m:= c 1.5m:=
d 1.5m:= e 0.5m:=
θ 30deg:= dpin 18mm:=
Solution: L a b+ c+ d+ e+:=
For comparison purpose, set P 1kN:=
Support Reactions:
 ΣΜA=0; By− L( )⋅ P L a−( )⋅+ 4P c d+ e+( )⋅+ 4P d e+( )⋅+ 2P e( )⋅+ 0=
By P
L a−
L
⋅ 4 P⋅ c d+ e+
L
⋅+ 4 P⋅ d e+
L
⋅+ 2 P⋅ e
L
⋅+:= By 5.5 kN=
 + ΣFy=0; By− P+ 4 P⋅+ 4 P⋅+ 2 P⋅+ Ay− 0=
Ay By− P+ 4 P⋅+ 4P+ 2 P⋅+:= Ay 5.5 kN=
FBC
By
sin θ( ):= FBC 11 kN=
Ax FBC cos θ( )⋅:= Ax 9.53 kN=
FA Ax
2 Ay
2+:= FA 11 kN=
Require: 
Fmax max FBC FA,( ):= Fmax 11 kN=
Apin
π dpin2⋅
4
:=
Pallow
P
Fmax
⎛⎜⎝
⎞
⎠
τallow 2Apin( )⋅⎡⎣ ⎤⎦⋅:= Pallow 3.70 kN= Ans
Problem 1-69
The frame is subjected to the load of 1 kN. Determine the average shear stress in the bolt at A as a
function of the bar angle θ. Plot this function, 0 θ≤ 90o≤ , and indicate the values of θ for which this
stress is a minimum. The bolt has a diameter of 6 mm and is subjected to single shear.
Given: P 1kN:= dbolt 6mm:=
a 0.6m:= b 0.45m:= c 0.15m:=
Solution:
Support Reactions:
 ΣΜC=0; FAB cos θ( )⋅ c( )⋅ FAB sin θ( )⋅ a( )⋅+ P a b+( )⋅− 0=
FAB
P a b+( )⋅
cos θ( ) c( )⋅ sin θ( ) a( )⋅+=
Average Shear Stress: Pin B is subjected to doule shear
τ
FAB
Abolt
= Abolt
π dbolt2⋅
4
:=
τ 4P a b+( )⋅
cos θ( ) c( )⋅ sin θ( ) a( )⋅+⎡⎣ ⎤⎦ π dbolt2⋅⎛⎝ ⎞⎠⋅
=
dτ
dθ
4P a b+( )⋅
π dbolt2⋅
sin θ( ) c( )⋅ cos θ( ) a( )⋅−
cos θ( ) c( )⋅ sin θ( ) a( )⋅+⎡⎣ ⎤⎦2
⋅=
dτ
dθ 0= sin θ( ) c( )⋅ cos θ( ) a( )⋅− 0= tan θ( )
a
c
=
θ atan a
c
⎛⎜⎝
⎞
⎠:=
θ 75.96 deg= Ans
Problem 1-70
The jib crane is pinned at A and supports a chain hoist that can travel along the bottom flange of the
beam, 1ft x≤ 12ft≤ . If the hoist is rated to support a maximum of 7.5 kN, determine the maximum
average normal stress in the 18-mm-diameter tie rod BC and the maximum average shear stress in the
16-mm-diameter pin at B.
Given: P 7.5kN:= xmax 3.6m:=
a 3m:= θ 30deg:=
drod 18mm:= dpin 16mm:=
Solution:
Support Reactions:
 
ΣΜC=0; FBC sin θ( ) a( )⋅⋅ P x( )⋅− 0=
FBC
P x( )⋅
sin θ( ) a( )⋅=
Maximum FBC occurs when x= xmax . Therefore, 
FBC
P xmax( )⋅
sin θ( ) a( )⋅:= FBC 18.00 kN=
Arod
π drod2⋅
4
:= Apin
π dpin2⋅
4
:=
τpin
0.5 FBC⋅
Apin
:= τpin 44.762 MPa= Ans
σrod
FBC
Arod
:= σrod 70.736 MPa= Ans
Problem 1-71
The bar has a cross-sectional area A and is subjected to the axial load P. Determine the average normal
and average shear stresses acting over the shaded section, which is oriented at θ from the horizontal.
Plot the variation of these stresses as a function of θ (0o θ≤ 90o≤ ).
Solution:
Equations of Equilibrium: 
 + ΣFx=0; V P cos θ( )⋅− 0=
V P cos θ( )⋅=
 + ΣFy=0; N P sin θ( )⋅− 0=
N P sin θ( )⋅=
Inclined plane:
Aθ
A
sin θ( )=
σ N
A
= σ P
A
sin θ( )2⋅= Ans
τavg
V
A
= τavg
P
2A
sin 2θ( )⋅= Ans
Problem 1-72
The boom has a uniform weight of 3 kN and is hoisted into position using the cable BC. If the cable has
a diameter of 15 mm, plot the average normal stress in the cable as a function of the boom position θ for
0o θ≤ 90o≤ .
Given: W 3kN:=
a 1m:=
do 15mm:=
Solution: Angle B: φB 0.5 90deg θ+( )=
φB 45deg 0.5θ+=
Support Reactions:
 ΣΜA=0; FBC sin φB( )⋅ a( )⋅ W 0.5a( )⋅ cos θ( )− 0=
FBC
0.5W cos θ( )⋅
sin 45deg 0.5θ+( )=
Average Normal Stress: 
σBC
FAB
ABC
= ABC
π do2⋅
4
:=
σBC
2W
π do2⋅
⎛⎜⎝
⎞
⎠
cos θ( )
sin 45deg 0.5θ+( )⋅= Ans
Problem 1-73
The bar has a cross-sectional area of 400 (10-6) m2. If it is subjected to a uniform axial distributed
loading along its length and to two concentrated loads as shown, determine the average normal stress in
the bar as a function of for 0 x< 0.5m≤ .
Given: P1 3kN:= P2 6kN:=
w 8
kN
m
:= A 400 10 6−( )⋅ m2:=
a 0.5m:= b 0.75m:=
Solution: L a b+:=
+ ΣFx=0; N− P1+ P2+ w L x−( )⋅+ 0=
N P1 P2+ w L x−( )⋅+=
Average Normal Stress:
σ N
A
=
σ
P1 P2+ w L x−( )⋅+
A
= Ans
Problem 1-74
The bar has a cross-sectional area of 400 (10-6) m2. If it is subjected to a uniform axial distributed
loading along its length and to two concentrated loads as shown, determine the average normal stress in
the bar as a function of for 0.5m x< 1.25m≤ .
Given: P1 3kN:= P2 6kN:=
w 8
kN
m
:= A 400 10 6−( )⋅ m2:=
a 0.5m:= b 0.75m:=
Solution: L a b+:=
+ ΣFx=0; N− P1+ w L x−( )⋅+ 0=
N P1 w L x−( )⋅+=
Average Normal Stress:
σ N
A
=
σ
P1 w L x−( )⋅+
A
= Ans
Problem 1-75
The column is made of concrete having a density of 2.30 Mg/m3. At its top B it is subjected to an axial
compressive force of 15 kN. Determine the average normal stress in the column as a function of the
distance z measured from its base. Note: The result will be useful only for finding the average normal
stress at a section removed from the ends of the column, because of localized deformation at the ends.
Given: P 3kN:= ρ 2.3 103( ) kg
m3
:= g 9.81 m
s2
:=
r 180mm:= h 0.75m:=
Solution:
A π r2⋅:= w ρ g⋅ A⋅:=
+ ΣFz=0; N P− w h z−( )⋅− 0=
N P w h z−( )⋅+=
Average Normal Stress:
σ N
A
=
σ P w h z−( )⋅+
A
= Ans
Problem 1-76
The two-member frame is subjected to the distributed loading shown. Determine the largest
intensity of the uniform loading that can be applied to the frame without causing either the average
normal stress or the average shear stress at section b-b to exceed σ = 15 MPa, and τ = 16 MPa
respectively. Member CB has a square cross-section of 30 mm on each side.
Given: σallow 15MPa:= τallow 16MPa:=
a 4m:= b 3m:= A 0.0302( )m2:=
Solution: c a2 b2+:= c 5 m=
v
b
c
:=h a
c
:=
Set wo 1
kN
m
:=
Member AB:
 ΣMA=0; By− b( )⋅ wo b⋅( ) 0.5b( )⋅− 0=
By 0.5wo b⋅:=
By 1.5 kN=
Section b-b: 
By FBC h( )⋅= FBC
By
h
:=
FBC 1.88 kN=
+ ΣFx=0; FBC h( )⋅ Vb_b− 0= Vb_b FBC h( )⋅:=
Vb_b 1.5 kN=
+ ΣFy=0; Nb_b− FBC v( )⋅+ 0= Nb_b FBC v( )⋅:=
Nb_b 1.125 kN=
Ab_b
A
v
:=
σb_b
Nb_b
Ab_b
:= σb_b 0.75 MPa=
τb_b
Vb_b
Ab_b
:= τb_b 1 MPa=
Assume failure due to normal stress: wallow wo
σallow
σb_b
⎛⎜⎝
⎞
⎠
⋅:= wallow 20.00
kN
m
=
Assume failure due to shear stress: wallow wo
τallow
τb_b
⎛⎜⎝
⎞
⎠
⋅:= wallow 16.00
kN
m
= Ans
Controls ! 
Problem 1-77
The pedestal supports a load P at its center. If the material has a mass density ρ, determine the radial
dimension r as a function of z so that the average normal stress in the pedestal remains constant. The
cross section is circular.
Solution:
Require: σ P w1+
A
= σ P w1+ dw+
A dA+=
P dA⋅ w1 dA⋅+ A dw⋅=
dw
dA
P w1+
A
=
dw
dA
σ= [1]
dA π r dr+( )2 πr2−=
dA 2πr dr⋅=
dw πr2 ρ g⋅( )⋅ dz⋅=
From Eq.[1],
πr2 ρ g⋅( )⋅ dz⋅
2πr dr⋅ σ=
r ρ g⋅( )⋅ dz⋅
2dr
σ=
ρ g⋅
2σ 0
z
z1( )
⌠⎮⌡ d
r1
r
r
1
r
⌠⎮
⎮⌡
d=
ρ g⋅ z⋅
2σ ln
r
r1
⎛⎜⎝
⎞
⎠= r r1 e
ρ g⋅
2σ
⎛⎜⎝
⎞
⎠ z⋅⋅=
However, 
σ P
π r12⋅
=
Ans
r r1 e
π r12⋅ ρ⋅ g⋅
2P
⎛⎜⎜⎝
⎞
⎠ z⋅⋅=
Problem 1-78
The radius of the pedestal is defined by r = (0.5e-0.08y2) m, where y is given in meters. If the material
has a density of 2.5 Mg/m3, determine the average normal stress at the support.
Given:
ro 0.5m:= h 3m:= g 9.81
m
s2
=
r ro e
0.08− y2m⋅= ρ 2.5 103( )⋅ kg
m3
:=
yunit 1m:=
Solution:
dr π e 0.08− y
2⎛⎝ ⎞⎠ dy⋅=
Ao πro2:= Ao
0.7854 m2=
dV π r2( ) dy⋅=
dV πro2 e 0.08− y
2⎛⎝ ⎞⎠
2
dy⋅=
V
0
3
yπro2 e 0.08− y
2⎛⎝ ⎞⎠
2
yunit( )⋅⎡⎢⎣
⎤⎥⎦
⌠⎮
⎮⌡ d:=
V 1.584 m3=
W ρ g⋅ V⋅:=
W 38.835 kN=
σ W
Ao
:=
σ 0.04945 MPa= Ans
Problem 1-79
The uniform bar, having a cross-sectional area of A and mass per unit length of m, is pinned at its
center. If it is rotating in the horizontal plane at a constant angular rate of ω, determine the average
normal stress in the bar as a function of x.
Solution:
Equation of Motion :
+ ΣFx=MaN=Mω r2 ;
N m
L
2
x−⎛⎜⎝
⎞
⎠⋅ ω x
1
2
L
2
x−⎛⎜⎝
⎞
⎠+
⎡⎢⎣
⎤⎥⎦⋅=
N
m ω⋅
8
L2 4 x2⋅−( )⋅=
Average Normal Stress:
σ N
A
=
σ m ω⋅
8A
L2 4 x2⋅−( )⋅= Ans
Problem 1-80
Member B is subjected to a compressive force of 4 kN. If A and B are both made of wood and are
10mm. thick, determine to the nearest multiples of 5mm the smallest dimension h of the support so that
the average shear stress does not exceed τallow = 2.1 MPa.
Given: P 4kN:=
t 10mm:= τallow 2.1MPa:=
a 300mm:= b 125mm:=
Solution: c a2 b2+:= c 325 mm=
h
a
c
:= v b
c
:=
V P v( )⋅:= V 1.54 kN=
τallow
V
t h⋅=
h
V
t τallow⋅
:= h 73.26 mm=
Use h 75mm:= h 75 mm= Ans
Problem 1-81
The joint is fastened together using two bolts. Determine the required diameter of the bolts if the failure
shear stress for the bolts is τfail = 350 MPa. Use a factor of safety for shear of F.S. = 2.5.
Given: P 80kN:= τfail 350MPa:= γ 2.5:=
Solution:
τallow
τfail
γ:= τallow 140 MPa=
Vbolt 0.5
P
2
⎛⎜⎝
⎞
⎠⋅:= Vbolt 20 kN=
Abolt
Vbolt
τallow
=
π
4
⎛⎜⎝
⎞
⎠ d
2⋅
Vbolt
τallow
=
d
4
π
Vbolt
τallow
⎛⎜⎝
⎞
⎠
⋅:=
d 13.49 mm= Ans
Problem 1-82
The rods AB and CD are made of steel having a failure tensile stress of σfail = 510 MPa. Using a factor
of safety of F.S. = 1.75 for tension, determine their smallest diameter so that they can support the load
shown. The beam is assumed to be pin connected at A and C.
Given: P1 4kN:= P2 6kN:= P3 5kN:=
a 2m:= b 2m:= c 3m:= d 3m:=
γ 1.75:= σfail 510MPa:=
Solution: L a b+ c+ d+:=
Support Reactions:
 ΣΜA=0; FCD L( )⋅ P1 a( )⋅− P2 a b+( )⋅− P3 a b+ c+( )⋅− 0=
FCD P1
a
L
⎛⎜⎝
⎞
⎠⋅ P2
a b+
L
⋅+ P3
a b+ c+
L
⋅+:= FCD 6.70 kN=
 ΣΜC=0; FAB− L( )⋅ P1 b c+ d+( )⋅+ P2 c d+( )⋅+ P3 d( )⋅+ 0=
FAB P1
b c+ d+
L
⋅ P2
c d+
L
⋅+ P3
d
L
⎛⎜⎝
⎞
⎠⋅+:= FAB 8.30 kN=
Average Normal Stress: Design of rod sizes
σallow
σfail
γ:= σallow 291.43 MPa=
For Rod AB
Abolt
FAB
σallow
=
π
4
⎛⎜⎝
⎞
⎠ dAB
2⋅
FAB
σallow
=
dAB
4
π
FAB
σallow
⎛⎜⎝
⎞
⎠
⋅:= dAB 6.02 mm= Ans
For Rod CD
Abolt
FCD
σallow
=
π
4
⎛⎜⎝
⎞
⎠ dCD
2⋅
FCD
σallow
=
dCD
4
π
FCD
σallow
⎛⎜⎝
⎞
⎠
⋅:= dCD 5.41 mm= Ans
Problem 1-83
The lever is attached to the shaft A using a key that has a width d and length of 25 mm. If the shaft is
fixed and a vertical force of 200 N is applied perpendicular to the handle, determine the dimension d if
the allowable shear stress for the key is τallow = 35 MPa.
Given: P 200N:= τallow 35MPa:=
L 500mm:= a 20mm:=
b 25mm:=
Solution:
 ΣΜA=0; Fa_a a( )⋅ P L( )⋅− 0=
Fa_a P
L
a
⋅:=
Fa_a 5000 N=
For the key
Aa_a
Fa_a
τallow
= b d⋅
Fa_a
τallow
=
d
1
b
Fa_a
τallow
⎛⎜⎝
⎞
⎠
⋅:=
d 5.71 mm= Ans
Problem 1-84
The fillet weld size a is determined by computing the average shear stress along the shaded plane,
which has the smallest cross section. Determine the smallest size a of the two welds if the force
applied to the plate is P = 100 kN. The allowable shear stress for the weld material is τallow = 100 MPa.
Given: P 100kN:= τallow 100MPa:=
L 100mm:= θ 45deg:=
Solution:
Shear Plane in the Weld: Aweld L a⋅ sin θ( )⋅=
Aweld
0.5P
τallow
=
L a⋅ sin θ( )⋅ 0.5Pτallow=
a
1
L sin θ( )⋅
0.5P
τallow
⎛⎜⎝
⎞
⎠
⋅:=
a 7.071 mm= Ans
Problem 1-85
The fillet weld size a = 8 mm. If the joint is assumed to fail by shear on both sides of the block along
the shaded plane, which is the smallest cross section, determine the largest force P that can be applied
to the plate. The allowable shear stress for the weld material is τallow = 100 MPa.
Given: a 8mm:= L 100mm:=
θ 45deg:= τallow 100MPa:=
Solution:
Shear Plane in the Weld: Aweld L a⋅ sin θ( )⋅=
P τallow 2Aweld( )⋅=
P τallow 2 L a⋅ sin θ( )⋅( )⎡⎣ ⎤⎦⋅:=
P 113.14 kN= Ans
Problem 1-86
The eye bolt is used to support the load of 25 kN. Determine its diameter d to the nearest multiples of
5mm and the required thickness h to the nearest multiples of 5mm of the support so that the washer
will not penetrate or shear through it. The allowable normal stress for the bolt is σallow = 150 MPa and
the allowable shear stress for the supporting material is τallow = 35 MPa.
Given: P 25kN:= dwasher 25mm:=
σallow 150MPa:= τallow 35MPa:=
Solution:
Allowable Normal Stress: Design of bolt size
Abolt
P
σallow
=
π
4
⎛⎜⎝
⎞
⎠ d
2⋅ Pσallow
=
d
4
π
P
σallow
⎛⎜⎝
⎞
⎠
⋅:=
d 14.567 mm=
Use d 15mm:= d 15 mm= Ans
Allowable Shear Stress: Design of support thickness
Asupport
P
τallow
= π dwasher( )⋅ h⋅ Pτallow=
h
1
π dwasher( )⋅
P
τallow
⎛⎜⎝
⎞
⎠
⋅:=
h 9.095 mm=
Use d 10mm:= d 10 mm= Ans
Problem 1-87
The frame is subjected to the load of 8 kN. Determine the required diameter of the pins at A and B if
the allowable shear stress for the material is τallow = 42 MPa. Pin A is subjected to double shear,
whereas pin B is subjected to single shear.
Given: P 8kN:= τallow 42MPa:=
a 1.5m:= b 1.5m:= c 1.5m:= d 0.6m:=
Solution: θBC 45deg:=
Support Reactions: From FBD (a),
 ΣΜD=0; FBC sin θBC( )⋅ c( )⋅ P c d+( )⋅− 0=
FBC P
c d+
sin θBC( ) c( )⋅⋅:=
FBC 15.839 kN=
From FBD (b),
 ΣΜA=0; Dy a b+( )⋅ P c d+( )⋅− 0=
Dy P
c d+
a b+⋅:= Dy 5.6 kN=
+ ΣFx=0; Ax P− 0= Ax P:= Ax 8 kN=
 + ΣFy=0; Dy Ay− 0= Ay Dy:=
FA Ax
2 Ay
2+:= FA 9.77 kN=
Apin
0.5FA
τallow
=
π
4
⎛⎜⎝
⎞
⎠ d
2⋅
0.5FA
τallow
=
d
4
π
0.5FA
τallow
⎛⎜⎝
⎞
⎠
⋅:= d 12.166 mm= Ans
For pin B: Pin A is subjected to single shear, and FB FBC:=
Apin
FB
τallow
=
π
4
⎛⎜⎝
⎞
⎠ d
2⋅
FB
τallow
=
d
4
π
FB
τallow
⎛⎜⎝
⎞
⎠
⋅:= d 21.913 mm= Ans
Problem 1-88
The two steel wires AB and AC are used to support the load. If both wires have an allowable tensile
stress of σallow = 200 MPa, determine the required diameter of each wire if the applied load is P = 5
kN.
Given: P 5kN:= σallow 200MPa:=
a 4m:= b 3m:= θ 60deg:=
Solution:
c a2 b2+:= h a
c
:= v b
c
:=
At joint A:
Initial guess: FAB 1kN:= FAC 2kN:=
Given
+ ΣFx=0; FAC h( )⋅ FAB sin θ( )⋅− 0= [1]
 + ΣFy=0; FAC v( )⋅ FAB cos θ( )⋅+ P− 0= [2]
Solving [1] and [2]:
FAB
FAC
⎛⎜⎜⎝
⎞
⎠
Find FAB FAC,( ):=
FAB
FAC
⎛⎜⎜⎝
⎞
⎠
4.3496
4.7086
⎛⎜⎝
⎞
⎠ kN=
For wire AB
AAB
FAB
σallow
=
π
4
⎛⎜⎝
⎞
⎠ dAB
2⋅
FAB
σallow
=
dAB
4
π
FAB
σallow
⎛⎜⎝
⎞
⎠
⋅:= dAB 5.26 mm= Ans
For wire AC
AAC
FAC
σallow
=
π
4
⎛⎜⎝
⎞
⎠ dAC
2⋅
FAC
σallow
=
dAC
4
π
FAC
σallow
⎛⎜⎝
⎞
⎠
⋅:= dAC 5.48 mm= Ans
Problem 1-89
The two steel wires AB and AC are used to support the load. If both wires have an allowable tensile
stress of σallow = 180 MPa, and wire AB has a diameter of 6 mm and AC has a diameter of 4 mm,
determine the greatest force P that can be applied to the chain before one of the wires fails.
Given: σallow 180MPa:=
a 4m:= b 3m:= θ 60deg:=
dAB 6mm:= dAC 4mm:=
Solution:
c a2 b2+:= h a
c
:= v b
c
:=
Assume failure of AB: FAB AAB( ) σallow⋅=
FAB
π
4
⎛⎜⎝
⎞
⎠ dAB
2⋅ σallow⋅:= FAB 5.09 kN=
At joint A:
Initial guess: P1 1kN:= FAC 2kN:=
Given
+ ΣFx=0; FAC h( )⋅ FAB sin θ( )⋅− 0= [1]
 + ΣFy=0; FAC v( )⋅ FAB cos θ( )⋅+ P1− 0= [2]
Solving [1] and [2]:
P1
FAC
⎛⎜⎜⎝
⎞
⎠
Find P1 FAC,( ):= P1FAC
⎛⎜⎜⎝
⎞
⎠
5.8503
5.5094
⎛⎜⎝
⎞
⎠ kN=
Assume failure of AC: FAC AAC( ) σallow⋅=
FAC
π
4
⎛⎜⎝
⎞
⎠ dAC
2⋅ σallow⋅:= FAC 2.26 kN=
At joint A:
Initial guess: P2 1kN:= FAB 2kN:=
Given
+ ΣFx=0; FAC h( )⋅ FAB sin θ( )⋅− 0= [1]
 + ΣFy=0; FAC v( )⋅ FAB cos θ( )⋅+ P2− 0= [2]
Solving [1] and [2]:
FAB
P2
⎛⎜⎜⎝
⎞
⎠
Find FAB P2,( ):= FABP2
⎛⎜⎜⎝
⎞
⎠
2.0895
2.4019
⎛⎜⎝
⎞
⎠ kN=
Chosoe the smallest value: P min P1 P2,( ):= P 2.40 kN= Ans
Problem 1-90
The boom is supported by the winch cable that has a diameter of 6 mm and an allowable normal stress
of σallow = 168 MPa. Determine the greatest load that can be supported without causing the cable to fail
when θ = 30° and φ = 45°. Neglect the size of the winch.
Given: σallow 168MPa:= do 6mm:=
θ 30deg:= φ 45deg:=
Solution:
For the cable:
Tcable Acable( ) σallow⋅=
Tcable
π
4
⎛⎜⎝
⎞
⎠ do
2⋅ σallow⋅:=
Tcable 4.7501 kN=
At joint B:
Initial guess: FAB 1 kN:= W 2 kN:=
Given
+ ΣFx=0; Tcable− cos θ( ) FAB cos φ( )⋅+ 0= [1]
 + ΣFy=0; W− FAB sin φ( )⋅+ Tcable sin θ( )⋅− 0= [2]
Solving [1] and [2]:
FAB
W
⎛⎜⎝
⎞
⎠ Find FAB W,( ):=
FAB
W
⎛⎜⎝
⎞
⎠
5.818
1.739
⎛⎜⎝
⎞
⎠ kN= Ans
Problem 1-91
The boom is supported by the winch cable that has an allowable normal stress of σallow = 168 MPa. If
it is required that it be able to slowly lift 25 kN, from θ = 20° to θ = 50°, determine the smallest
diameter of the cable to the nearest multiples of 5mm. The boom AB has a length of 6 m. Neglect the
size of the winch. Set d = 3.6 m.
Given: σallow 168MPa:= W 25 kN:=
d 3.6m:= a 6m:=
Solution:
Maximum tension in canle occurs when θ 20deg:=
sin θ( )
a
sin ψ( )
d
= ψ asin d
a
⎛⎜⎝
⎞
⎠ sin θ( )⋅
⎡⎢⎣
⎤⎥⎦:= ψ 11.842 deg=
At joint B:
Initial guess: FAB 1 kN:= Tcable 2 kN:=
Given φ θ ψ+:=
+ ΣFx=0; Tcable− cos θ( ) FAB cos φ( )⋅+ 0= [1]
 + ΣFy=0; W− FAB sin φ( )⋅+ Tcable sin θ( )⋅− 0= [2]
Solving [1] and [2]:
FAB
Tcable
⎛⎜⎜⎝
⎞
⎠
Find FAB Tcable,( ):=
FAB
Tcable
⎛⎜⎜⎝
⎞
⎠
114.478
103.491
⎛⎜⎝
⎞
⎠ kN=
For the cable:
Acable
P
σallow
=
π
4
⎛⎜⎝
⎞
⎠ do
2⋅
Tcable
σallow
=
do
4
π
Tcable
σallow
⎛⎜⎝
⎞
⎠
⋅:=
do 28.006 mm=
Use do 30mm:= do 30 mm= Ans
Problem 1-92
The frame is subjected to the distributed loading of 2 kN/m. Determine the required diameter of the
pins at A and B if the allowable shear stress for the material is τallow = 100 MPa. Both pins are
subjected to double shear.
Given: w 2
kN
m
:= τallow 100MPa:=
r 3m:=
Solution: Member AB is atwo-force member 
θ 45deg:=
Support Reactions:
 ΣΜA=0; FBC sin θ( )⋅ r( )⋅ w r⋅( ) 0.5r( )⋅− 0=
FBC
0.5w r⋅
sin θ( ):= FBC 4.243 kN=
 + ΣFy=0; Ay FBC sin θ( )⋅+ w r⋅− 0=
Ay FBC− sin θ( )⋅ w r⋅+:= Ay 3 kN=
+ ΣFx=0; Ax FBC cos θ( )⋅− 0=
Ax FBC cos θ( )⋅:= Ax 3 kN=
Average Shear Stress: Pin A and pin B are subjected to double shear
FA Ax
2 Ay
2+:= FA 4.243 kN=
FB FBC:= FB 4.243 kN=
Since both subjected to the same shear force V = 0.5 FA and V 0.5FB:=
Apin
V
τallow
=
π
4
⎛⎜⎝
⎞
⎠ dpin
2⋅ Vτallow
=
dpin
4
π
V
τallow
⎛⎜⎝
⎞
⎠
⋅:=
dpin 5.20 mm= Ans
Problem 1-93
Determine the smallest dimensions of the circular shaft and circular end cap if the load it is required to
support is P = 150 kN. The allowable tensile stress, bearing stress, and shear stress is (σt)allow = 175
MPa, (σb)allow = 275 MPa, and τallow = 115 MPa.
Given: P 150kN:= σt_allow 175MPa:=
τallow 115MPa:= σb_allow 275MPa:=
d2 30mm:=
Solution:
Allowable Normal Stress: Design of end cap outer diameter
A
P
σt_allow
=
π
4
⎛⎜⎝
⎞
⎠ d1
2 d2
2−⎛⎝ ⎞⎠⋅ Pσt_allow=
d1
4
π
P
σt_allow
⎛⎜⎝
⎞
⎠
⋅ d22+:= d1 44.62 mm= Ans
Allowable Bearing Stress: Design of circular shaft diameter
A
P
σb_allow
=
π
4
⎛⎜⎝
⎞
⎠ d3
2⎛⎝ ⎞⎠⋅ Pσb_allow=
d3
4
π
P
σb_allow
⎛⎜⎝
⎞
⎠
⋅:= d3 26.35 mm= Ans
Allowable Shear Stress: Design of end cap thickness
A
P
τallow
= π d3⋅( ) t⋅ Pτallow=
t
1
π d3⋅
P
τallow
⎛⎜⎝
⎞
⎠
⋅:= t 15.75 mm= Ans
Problem 1-94
If the allowable bearing stress for the material under the supports at A and B is (σb)allow = 2.8 MPa,
determine the size of square bearing plates A' and B' required to support the loading. Take P = 7.5 kN.
Dimension the plates to the nearest multiples of 10mm. The reactions at the supports are vertical.
Given: σb_allow 2.8MPa:= P 7.5 kN:=
P1 10 kN:= P2 10 kN:=
P3 15 kN:= P4 10 kN:=
a 1.5m:= b 2.5m:=
Solution: L 3 a⋅ b+:=
Support Reactions:
 ΣΜA=0; By 3a( ) P2 a( )⋅− P3 2a( )⋅− P4 3a( )⋅− P L( )⋅− 0=
By P2
a
3a
⎛⎜⎝
⎞
⎠⋅ P3
2a
3a
⎛⎜⎝
⎞
⎠⋅+ P4
3a
3a
⎛⎜⎝
⎞
⎠⋅+ P
L
3a
⎛⎜⎝
⎞
⎠⋅+:= By 35 kN=
 ΣΜB=0; Ay− 3 a⋅( )⋅ P1 3 a⋅( )⋅+ P2 2 a⋅( )⋅+ P3 a( )⋅+ P b( )⋅− 0=
Ay P1
3a
3a
⎛⎜⎝
⎞
⎠⋅ P2
2a
3a
⎛⎜⎝
⎞
⎠⋅+ P3
a
3a
⎛⎜⎝
⎞
⎠⋅+ P
b
3a
⎛⎜⎝
⎞
⎠⋅−:= Ay 17.5 kN=
For Plate A:
Aplate_A
Ay
σb_allow
= aA
2 Ay
σb_allow
=
aA
Ay
σb_allow
:=
aA 79.057 mm=
Use aA x aA plate: aA 80mm= Ans
For Plate B
Aplate_B
By
σb_allow
= aB
2 By
σb_allow
=
aB
By
σb_allow
:=
aB 111.803 mm=
Use aB x aB plate: aB 120mm= Ans
ULoHS
Text Box
Rb = 35kNnullnull(b is in subscript)
Problem 1-95
If the allowable bearing stress for the material under the supports at A and B is (σb)allow = 2.8 MPa,
determine the maximum load P that can be applied to the beam. The bearing plates A' and B' have
square cross sections of 50mm x 50mm and 100mm x 100mm, respectively.
Given: σb_allow 2.8MPa:=
P1 10 kN:= P2 10 kN:=
P3 15 kN:= P4 10 kN:=
a 1.5m:= b 2.5m:=
aA 50mm:= aB 100mm:=
Solution: L 3 a⋅ b+:=
Support Reactions:
 ΣΜA=0; By 3a( ) P2 a( )⋅− P3 2a( )⋅− P4 3 a⋅( )⋅− P L( )⋅− 0=
By P2
a
3a
⎛⎜⎝
⎞
⎠⋅ P3 2
a
3a
⋅⎛⎜⎝
⎞
⎠⋅+ P4
3 a⋅
3a
⎛⎜⎝
⎞
⎠⋅+ P
L
3a
⎛⎜⎝
⎞
⎠⋅+=
 ΣΜB=0; Ay− 3 a⋅( )⋅ P1 3 a⋅( )⋅+ P2 2 a⋅( )⋅+ P3 a( )⋅+ P b( )⋅− 0=
Ay P1
3a
3a
⎛⎜⎝
⎞
⎠⋅ P2
2a
3a
⎛⎜⎝
⎞
⎠⋅+ P3
a
3a
⎛⎜⎝
⎞
⎠⋅+ P
b
3a
⎛⎜⎝
⎞
⎠⋅−=
For Plate A: Ay aA
2⎛⎝ ⎞⎠ σb_allow⋅:=
aA
2⎛⎝ ⎞⎠ σb_allow⋅ P1 3a3a
⎛⎜⎝
⎞
⎠⋅ P2
2a
3a
⎛⎜⎝
⎞
⎠⋅+ P3
a
3a
⎛⎜⎝
⎞
⎠⋅+ P
b
3a
⎛⎜⎝
⎞
⎠⋅−=
P P1
3a
b
⎛⎜⎝
⎞
⎠⋅ P2
2a
b
⎛⎜⎝
⎞
⎠⋅+ P3
a
b
⎛⎜⎝
⎞
⎠⋅+ aA
2⎛⎝ ⎞⎠ σb_allow⋅ 3ab
⎛⎜⎝
⎞
⎠⋅−:= P 26.400 kN=
Pcase_1 P:=
For Plate B: By aB
2⎛⎝ ⎞⎠ σb_allow⋅:=
aB
2⎛⎝ ⎞⎠ σb_allow⋅ P2 a3a
⎛⎜⎝
⎞
⎠⋅ P3 2
a
3a
⋅⎛⎜⎝
⎞
⎠⋅+ P4
3 a⋅
3a
⎛⎜⎝
⎞
⎠⋅+ P
L
3a
⎛⎜⎝
⎞
⎠⋅+=
P P2−
a
L
⎛⎜⎝
⎞
⎠⋅ P3 2
a
L
⋅⎛⎜⎝
⎞
⎠⋅− P4
3 a⋅
L
⎛⎜⎝
⎞
⎠⋅− aB
2⎛⎝ ⎞⎠ σb_allow⋅ 3aL⋅+:= P 3.000 kN=
Pcase_2 P:=
Pallow min Pcase_1 Pcase_2,( ):= Pallow 3 kN= Ans
Problem 1-96
Determine the required cross-sectional area of member BC and the diameter of the pins at A and B if
the allowable normal stress is σallow = 21 MPa and the allowable shear stress is τallow = 28 MPa.
Given: σallow 21MPa:= τallow 28MPa:=
P 7.5kip:= θ 60deg:=
a 0.6m:= b 1.2m:= c 0.6m:=
Solution: L a b+ c+:=
Support Reactions:
 ΣΜA=0; By L( )⋅ P a( )⋅− P a b+( )⋅− 0=
By P
a
L
⋅ P a b+
L
⋅+:=
FBC
By
sin θ( ):= FBC 38.523 kN=By 33.362 kN=
Bx FBC cos θ( )⋅:= Bx 19.261 kN=
 + ΣFy=0; By− P+ P+ Ay− 0= Ay By− P+ P+:= Ay 33.362 kN=
+ ΣFx=0; Bx Ax− 0= Ax Bx:= Ax 19.261 kN=
FA Ax
2 Ay
2+:= FA 38.523 kN=
Member BC: 
ABC
FBC
σallow
:= ABC 1834.416 mm2= Ans
Pin A:
AA
FA
τallow
=
π
4
⎛⎜⎝
⎞
⎠ dA
2⋅
FA
τallow
=
dA
4
π
FA
τallow
⎛⎜⎝
⎞
⎠
⋅:= dA 41.854 mm= Ans
Pin B:
AB
0.5FBC
τallow
=
π
4
⎛⎜⎝
⎞
⎠ dB
2⋅
0.5FBC
τallow
=
dB
4
π
0.5FBC
τallow
⎛⎜⎝
⎞
⎠
⋅:= dB 29.595 mm= Ans
Problem 1-97
The assembly consists of three disks A, B, and C that are used to support the load of 140 kN.
Determine the smallest diameter d1 of the top disk, the diameter d2 within the support space, and the
diameter d3 of the hole in the bottom disk. The allowable bearing stress for the material is (τallow)b =
350 MPa and allowable shear stress is τallow = 125 MPa.
Given: P 140kN:=
τallow 125MPa:= σb_allow 350MPa:=
hB 20mm:= hC 10mm:=
Solution:
Allowable Shear Stress: Assume shear failure dor disk C
A
P
τallow
= π d2⋅( ) hC⋅ Pτallow=
d2
1
π hC⋅
P
τallow
⎛⎜⎝
⎞
⎠
⋅:= d2 35.65 mm= Ans
Allowable Bearing Stress: Assume bearing failure dor disk C
A
P
σb_allow
=
π
4
⎛⎜⎝
⎞
⎠ d2
2 d3
2−⎛⎝ ⎞⎠⋅ Pσb_allow=
d3 d2
2 4
π
P
σb_allow
⎛⎜⎝
⎞
⎠
⋅−:= d3 27.60 mm= Ans
Allowable Bearing Stress: Assume bearing failure dor disk B
A
P
σb_allow
=
π
4
⎛⎜⎝
⎞
⎠ d1
2⎛⎝ ⎞⎠⋅ Pσb_allow=
d1
4
π
P
σb_allow
⎛⎜⎝
⎞
⎠
⋅:= d1 22.57 mm=
Since d3 > d1, disk B might fail due to shear.
τ P
A
= τ Pπ d1⋅ hB⋅
:= τ 98.73 MPa= < τallow (O.K.!)
Therefore d1 22.57 mm= Ans
Problem 1-98
Strips A and B are to be glued together using the two strips C and D. Determine the required thickness
t of C and D so that all strips will fail simultaneously. The width of strips A and B is 1.5 times that of
strips C and D.
Given: P 40N:= t 30mm:=
bA 1.5m:= bB 1.5m:=
bC 1m:= bD 1m:=
Solution:
Average Normal Stress: Requires,
σA σB= σB σC= σC σD=
N
bA( ) t⋅
0.5N
bC( ) tC( )⋅=
tC
0.5 bA( ) t⋅
bC
:= tC 22.5 mm= Ans
Problem 1-99
If the allowable bearing stress for the material under the supports at A and B is (σb)allow = 2.8 MPa,
determine the size of square bearing plates A' and B' required to support the loading. Dimension the
plates to the nearest multiples of 10mm. The reactions at the supports are vertical. Take P = 7.5 kN.
Given: σb_allow 2.8MPa:= P 7.5kN:=
w 10
kN
m
:= a 4.5m:= b 2.25m:=
Solution: L a b+:=
Support Reactions:
 ΣΜA=0; By a( ) w a( )⋅ 0.5 a⋅( )⋅− P L( )⋅− 0=
By w 0.5 a⋅( )⋅ P
L
a
⎛⎜⎝
⎞
⎠⋅+:= By 33.75 kN=
 ΣΜB=0; Ay− a( )⋅ w a( )⋅ 0.5 a⋅( )⋅+ P L( )⋅− 0=
Ay w 0.5 a⋅( )⋅ P
b
a
⎛⎜⎝
⎞
⎠⋅−:= Ay 18.75 kN=
Allowable Bearing Stress: Design of bearing plates
For Plate A:
Area
Ay
σb_allow
= aA
2 Ay
σb_allow
=
aA
Ay
σb_allow
:=
aA 81.832 mm=
Use aA x aA plate: aA 90mm:= aA 90 mm= Ans
For Plate B
Area
By
σb_allow
= aB
2 By
σb_allow
=
aB
By
σb_allow
:=
aB 109.789 mm=
Use aB x aB plate: aB 110mm:= aB 110.00 mm= Ans
Problem 1-100
If the allowable bearing stress for the material under the supports at A and B is (σb)allow = 2.8 MPa,
determine the maximum load P that can be applied to the beam. The bearing plates A' and B' have
square cross sections of 50mm x 50mm and 100mm x 100mm, respectively.
Given: σb_allow 2.8MPa:=
w 10
kN
m
:= a 4.5m:= b 2.25m:=
aA 50mm:= aB 100mm:=
Solution: L a b+:=
Support Reactions:
 ΣΜA=0;
By a( )⋅ w a( )⋅ 0.5 a⋅( )⋅− P L( )⋅− 0=
P By
a
L
⎛⎜⎝
⎞
⎠⋅ w
a
L
⎛⎜⎝
⎞
⎠⋅ 0.5 a⋅( )⋅−=
 ΣΜB=0; Ay− a( )⋅ w a( )⋅ 0.5 a⋅( )⋅+ P b( )⋅− 0=
P Ay−
a
b
⎛⎜⎝
⎞
⎠⋅ w
a
b
⎛⎜⎝
⎞
⎠⋅ 0.5 a⋅( )⋅+=
Allowable Bearing Stress:
Assume failure of material occurs under plate A. Ay aA
2⎛⎝ ⎞⎠ σb_allow⋅:=
P aA
2⎛⎝ ⎞⎠ σb_allow⋅⎡⎣ ⎤⎦− ab
⎛⎜⎝
⎞
⎠⋅ w a⋅( )
0.5a
b
⋅+:= P 31 kN=
Pcase_1 P:=
Assume failure of material occurs under plate B. By aB
2⎛⎝ ⎞⎠ σb_allow⋅:=
P By
a
L
⎛⎜⎝
⎞
⎠⋅ w
a
L
⎛⎜⎝
⎞
⎠⋅ 0.5 a⋅( )⋅−:= P 3.67 kN=
Pcase_2 P:=
Pallow min Pcase_1 Pcase_2,( ):= Pallow 3.67 kN= Ans
Problem 1-101
The hanger assembly is used to support a distributed loading of w = 12 kN/m. Determine the average
shear stress in the 10-mm-diameter bolt at A and the average tensile stress in rod AB, which has a
diameter of 12 mm. If the yield shear stress for the bolt is τy = 175 MPa, and the yield tensile stress for
the rod is σy = 266 MPa, determine the factor of safety with respect to yielding in each case.
Given: τy 175MPa:= w 12
kN
m
:=
σy 266MPa:=
a 1.2m:= b 0.6m:= e 0.9m:=
do 10mm:= drod 12mm:=
Solution:
c a2 e2+:= h a
c
:= v e
c
:=
Support Reactions: L a b+:=
 ΣΜC=0; FAB v( )⋅⎡⎣ ⎤⎦ a( )⋅ w L( )⋅ 0.5 L⋅( )⋅− 0=
FAB w
L
a v⋅
⎛⎜⎝
⎞
⎠⋅ 0.5 L⋅( )⋅:=
FAB 27 kN=
For bolt A: Bolt A is subjected to double shear, and V 0.5FAB:= V 13.5 kN=
A
π
4
⎛⎜⎝
⎞
⎠ do
2⋅:= τ V
A
:= τ 171.89 MPa= Ans
FS
τy
τ:= FS 1.02= Ans
For rod AB: N FAB:= N 27 kN=
A
π
4
⎛⎜⎝
⎞
⎠ drod
2⋅:= σ N
A
:= σ 238.73 MPa= Ans
FS
σy
σ:= FS 1.11= Ans
Problem 1-102
Determine the intensity w of the maximum distributed load that can be supported by the hanger
assembly so that an allowable shear stress of τallow = 95 MPa is not exceeded in the 10-mm-diameter
bolts at A and B, and an allowable tensile stress of σallow = 155 MPa is not exceeded in the
12-mm-diameter rod AB.
Given: τallow 95MPa:= σallow 155MPa:=
a 1.2m:= b 0.6m:= e 0.9m:=
do 10mm:= drod 12mm:=
Solution: c a2 e2+:= h a
c
:= v e
c
:=
Support Reactions: L a b+:=
 ΣΜC=0; FAB v( )⋅⎡⎣ ⎤⎦ a( )⋅ w L( )⋅ 0.5 L⋅( )⋅− 0=
FAB w
L
a v⋅
⎛⎜⎝
⎞
⎠⋅ 0.5 L⋅( )⋅=
Assume failure of pin A or B:
V 0.5FAB= V τallow A⋅= A
π
4
⎛⎜⎝
⎞
⎠ do
2⋅:=
0.5 w⋅ L
a v⋅
⎛⎜⎝
⎞
⎠⋅ 0.5 L⋅( )⋅ τallow
π
4
⎛⎜⎝
⎞
⎠ do
2⋅⎡⎢⎣
⎤⎥⎦⋅=
w
a v⋅
0.5L( )2
τallow⋅
π
4
⎛⎜⎝
⎞
⎠ do
2⋅⎡⎢⎣
⎤⎥⎦⋅:=
w 6.632
kN
m
= (controls!) Ans
Assuming failure of rod AB:
N FAB= N σallow A⋅= A
π
4
⎛⎜⎝
⎞
⎠ drod
2⋅:=
w
L
a v⋅
⎛⎜⎝
⎞
⎠⋅ 0.5 L⋅( )⋅ σallow
π
4
⎛⎜⎝
⎞
⎠ drod
2⋅⎡⎢⎣
⎤⎥⎦⋅=
w
a v⋅
0.5L2
σallow⋅
π
4
⎛⎜⎝
⎞
⎠ drod
2⋅⎡⎢⎣
⎤⎥⎦⋅:=
w 7.791
kN
m
=
Problem 1-103
The bar is supported by the pin. If the allowable tensile stress for the bar is (σt)allow = 150 MPa, and
the allowable shear stress for the pin is τallow = 85 MPa, determine the diameter of the pin for which
the load P will be a maximum. What is this maximum load? Assume the hole in the bar has the same
diameter d as the pin. Take t =6 mm and w = 50 mm. 
Given: τallow 85MPa:=
σt_allow 150MPa:=
t 6mm:= w 50mm:=
Solution: Given
Allowable Normal Stress: The effective cross-sectional
area Ae for the bar must be considered here by taking
into account the reduction in cross-sectional area
introduced by the hole. Here, effective area Ae is equal
to (w - d) t, and σallow equals to P /Ae .
σt_allow
P
w d−( ) t⋅= [1]
Allowable Shear Stress: The pin is subjected to double
shear and therefore the allowable τ equals to 0.5P /Apin,
and the area Apin is equal to ( π/4) d2. 
τallow
2
π
⎛⎜⎝
⎞
⎠
P
d2
⎛⎜⎝
⎞
⎠
⋅= [2]
Solving [1] and [2]: Initial guess: d 20mm:= P 10kN:=
P
d
⎛⎜⎝
⎞
⎠ Find P d,( ):= P 31.23 kN= Ans
d 15.29 mm= Ans
Problem 1-104
The bar is connected to the support using a pin having a diameter of d = 25 mm. If the allowable
tensile stress for the bar is (σt)allow = 140 MPa, and the allowable bearing stress between the pin and
the bar is (σb)allow =210 MPA, determine the dimensions w and t such that the gross area of the cross
section is wt = 1250 mm2 and the load P is a maximum. What is this maximum load? Assume the hole
in the bar has the same diameter as the pin.
Given: σt_allow 140 MPa:= σb_allow 210 MPa:=
A 1250mm2:= d 25mm:=
Solution: A w t⋅= Given
Allowable Normal Stress: The effective cross-sectional
area Ae for the bar must be considered here by taking
into account the reduction in cross-sectional area
introduced by the hole. Here, effective area Ae is equal