ENGR0135 - Statics and Mechanics of Materials 1 (2151)
Homework #9
Solution Set
1. There are different approaches one can take (e.g., one could begin with a free-body
diagram of the entire truss to determine the support reactions at A and F ). In this
case, let’s start at pin D, since it involves only two unknown forces, and then work our
way out from there.
• Pin D
D
TCD
60
45◦
◦
• Pin C
x
◦
C 60
y
60
x
TCD
◦
TCE
TBC
• Pin E
y
TCE
TDE
60◦
60◦
E
60◦
6 kN
TDE
TBE
y
x
Fy = −TDE sin 60◦ − 6 sin 45◦ = 0
=⇒
TDE = −4.89898 kN
Fx = 6 cos 45◦ − TCD − TDE cos 60◦ = 0
=⇒
TCD = 6.69213 kN
Fy = −TCD sin 60◦ − TCE sin 60◦ = 0
=⇒
TCE = −6.69213 kN
Fx = TCD cos 60◦ − TCE cos 60◦ − TBC = 0
=⇒
TBC = 6.69213 kN
Fy = TCE sin 60◦ + TDE sin 60◦ − TEF sin 60◦ = 0
=⇒
TEF = −11.5911 kN
Fx = TDE cos 60◦ − TCE cos 60◦ − TEF cos 60◦ − TBE = 0
=⇒
TBE = 6.69213 kN
TEF
• Pin B
x
TBC
y
60
TAB
60◦
B
TBE
◦
TBF
Fy = −TBE sin 60◦ − TBF sin 60◦ = 0
=⇒
TBF = −6.69213 kN
Fx = TBC + TBE cos 60◦ − TBF cos 60◦ − TAB = 0
=⇒
TAB = 13.38426 kN
• Pin F
y
TBF
TAF
TEF
60◦
60◦
F
x
Fx = TEF cos 60◦ − TBF cos 60◦ − TAF = 0
=⇒
TAF = −2.44949 kN
Fy
Thus, the forces in each member of the truss are
TAB = 13.38 kN (T) TBE = 6.69 kN (T) TCE = 6.69 kN (C)
TAF = 2.45 kN (C)
TBF = 6.69 kN (C) TDE = 4.90 kN (C)
TBC = 6.69 kN (T) TCD = 6.69 kN (T) TEF = 11.59 kN (C)
2. Consider the free-body diagram of the portion of the truss shown below:
1500 lb
1000 lb
6 ft
B TBC
θ = tan−1
6
8
TCH
cos θ = 0.8
H
6 ft
C
2000 lb
6 ft
D
θ
TGH
E
8 ft
G
F
The equilibrium equations can be used to determine the forces in members BC and
CH (we do not need the force in member GH for this problem).
X
Fy = −TCH cos θ − 1000 − 1500 − 2000 = 0
=⇒
TCH = −5625 lb
X
MH = 8TBC − 6(1000) − 12(1500) − 18(2000) = 0 =⇒
TBC = 7500 lb
(a) The normal stress in member BC is
σBC =
TBC
7500 lb
=
= 30, 000 psi = 30.0 ksi (T)
ABC
0.25 in2
(b) Noting that the unloaded length of member CH is
√
LCH = 62 + 82 = 10 ft = 120 in ,
the change in length of member CH is
δCH =
TCH LCH
(−5625 lb)(120 in)
=
= −9.31 × 10−2 in
ACH ECH
(0.25 in2 )(29 × 106 lb/in2 )
3. Consider the free-body diagram of the portion of the truss shown below:
1500 lb
c
D
30◦
TBD
B
b
TCD
C
12 ft
◦
A
30 a
θ
16 ft
Begin by determining the needed geometric parameters:
12
= 36.8699◦
θ = tan−1
16
√
a = 62 + 82 = 10 ft
b = a tan 30◦ = 5.7735 ft
√
c = a2 + b2 cos(θ + 30◦ ) = 4.5359 ft
Then, by summing moments about A,
X
MA = −(c)(1500) − (a)(TBD sin 30◦ ) − (b)(TBD cos 30◦ ) = 0 ,
it is seen that the force in member BD is
TBD =
−1500c
= −680.39 lb = 680 lb (C)
a sin 30◦ + b cos 30◦
4. The force at the end of the pipe system is
P = −75k (lb)
and the position vector from the support A to the point of application of the force is
r = −10i + 20j + 12k (in) .
Let RA and CA be the reaction force and couple, respectively, at the support A. It
follows from the equilibrium equations for the pipe system,
X
F = RA + P = 0
=⇒
RA = −P
X
MA = CA + r × P = 0 =⇒
CA = −r × P
Thus, the reaction at support A is
RA = 75k (lb) ,
CA = 1500i + 750j (in · lb)
5. The free-body diagram for the door is shown below, where TC is the tension in cable
CD.
z
59 in
D
42 in
Az
Bz
10 in
64 in
x
Ax
240 lb
A
Ay
50 in
B
25 in
E
42 in
By
TC
C
21 in
The force exerted on the door at C by cable CD is
38i − 50j + 42k
TC = TC √
= TC (0.5030i − 0.6618j + 0.5559k)
382 + 502 + 422
Summing moments about A,
X
MA = 0 = rD/A × TC + rE/A × W + rB/A × B
= (−5i + 42k) × TC (0.5030i − 0.6618j + 0.5559k)
+ (−32i + 25j) × (−240k) + (−64i) × (By j + Bz k)
= (27.7956TC − 6000)i + (23.9055TC − 7680 + 64Bz )j
+ (3.3090TC − 64By )k
it follows that
TC = 215.86 lb ,
By = 11.16 lb ,
Bz = 39.37 lb
Summing forces,
X
F = 0 = A + B + W + TC
= (Ax i + Ay j + Az k) + (11.16j + 39.37k) + (−240k)
+ 215.86(0.5030i − 0.6618j + 0.5559k)
it follows that
Ax = −108.58 lb ,
Ay = 131.70 lb ,
Az = 80.63 lb
y

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