Pb-9.1 in sec.9.
Indented line 1)Solve for the reactions at B and at A in the Example 2.04 (see textbook page 72) with the stress strain relation
σ
=
E
ϵ
1
/
2
{\displaystyle \sigma =E\ \epsilon ^{1/2}}
problem 2.04, Beer 2012 p.72 Given:
Length
L
A
D
=
L
D
C
=
L
=
150
m
m
{\displaystyle L_{AD}=L_{DC}=L=150\,mm}
L
C
K
=
L
K
B
=
L
=
150
m
m
{\displaystyle L_{CK}=L_{KB}=L=150\,mm}
Area
A
A
D
=
A
D
C
=
A
A
C
=
250
m
m
2
{\displaystyle A_{AD}=A_{DC}=A_{AC}=250\,mm^{2}}
A
C
K
=
A
K
B
=
A
C
B
=
400
m
m
2
{\displaystyle A_{CK}=A_{KB}=A_{CB}=400\,mm^{2}}
Young's modulus
E
A
D
=
E
D
C
=
E
C
K
=
E
K
B
=
E
{\displaystyle E_{AD}=E_{DC}=E_{CK}=E_{KB}=E}
Applied forces
F
D
=
300
k
N
{\displaystyle F_{D}=300\,kN}
F
K
=
600
k
N
{\displaystyle F_{K}=600\,kN}
Draw a Free Body Diagram for each section
2.04 problem, Vu-Quoc sec.9 p.9-6 Use
∑
F
=
0
{\displaystyle \sum \mathbf {F=0} }
P
C
K
=
R
B
−
F
K
{\displaystyle P_{CK}=R_{B}-F_{K}}
Knowing that the strain relation is found by:
δ
=
ϵ
L
=
σ
2
E
2
L
=
P
2
L
A
2
E
2
{\displaystyle \delta =\epsilon \,L={\frac {\sigma ^{2}}{E^{2}}}\,L={\frac {P^{2}\,L}{A^{2}\,E^{2}}}}
δ
C
K
=
−
P
C
K
2
L
C
K
(
A
C
K
)
2
(
E
C
K
)
2
=
−
(
R
B
−
F
K
)
2
L
(
A
C
B
)
2
E
2
{\displaystyle \delta _{CK}=-{\frac {P_{CK}^{2}L_{CK}}{(A_{CK})^{2}(E_{CK})^{2}}}=-{\frac {(R_{B}-F_{K})^{2}L}{(A_{CB})^{2}E^{2}}}}
R
B
{\displaystyle R_{B}}
δ
A
D
=
(
R
B
−
F
K
−
F
D
)
2
L
(
250
)
2
E
2
{\displaystyle \displaystyle \delta _{AD}={\frac {(R_{B}-F_{K}-F_{D})^{2}\,L}{(250)^{2}\,E^{2}}}}
δ
D
C
=
(
R
B
−
F
K
)
2
L
(
250
)
2
E
2
{\displaystyle \displaystyle \delta _{DC}={\frac {(R_{B}-F_{K})^{2}\,L}{(250)^{2}\,E^{2}}}}
δ
C
K
=
(
R
B
−
F
K
)
2
L
(
400
)
2
E
2
{\displaystyle \displaystyle \delta _{CK}={\frac {(R_{B}-F_{K})^{2}\,L}{(400)^{2}\,E^{2}}}}
δ
K
B
=
(
R
B
)
2
L
(
400
)
2
E
2
{\displaystyle \displaystyle \delta _{KB}={\frac {(R_{B})^{2}\,L}{(400)^{2}\,E^{2}}}}
δ
T
=
∑
δ
=
0
{\displaystyle \delta _{T}=\sum \delta =0}
δ
T
=
δ
A
D
+
δ
D
C
+
δ
C
K
+
δ
K
B
=
0
{\displaystyle \delta _{T}=\delta _{AD}+\delta _{DC}+\delta _{CK}+\delta _{KB}=0}
since all quantities are squared, therefore
δ
A
D
=
0
{\displaystyle \delta _{AD}=0}
δ
D
C
=
0
{\displaystyle \delta _{DC}=0}
δ
C
K
=
0
{\displaystyle \delta _{CK}=0}
δ
K
B
=
0
{\displaystyle \delta _{KB}=0}
0
=
(
R
B
−
F
K
−
F
D
)
2
L
(
250
)
2
E
2
{\displaystyle \displaystyle 0={\frac {(R_{B}-F_{K}-F_{D})^{2}\,L}{(250)^{2}\,E^{2}}}}
0
=
(
R
B
−
F
K
)
2
L
(
250
)
2
E
2
{\displaystyle \displaystyle 0={\frac {(R_{B}-F_{K})^{2}\,L}{(250)^{2}\,E^{2}}}}
0
=
(
R
B
−
F
K
)
2
L
(
400
)
2
E
2
{\displaystyle \displaystyle 0={\frac {(R_{B}-F_{K})^{2}\,L}{(400)^{2}\,E^{2}}}}
0
=
(
R
B
)
2
L
(
400
)
2
E
2
{\displaystyle \displaystyle 0={\frac {(R_{B})^{2}\,L}{(400)^{2}\,E^{2}}}}
R
B
=
0
{\displaystyle R_{B}=0}
R
B
−
F
K
=
0
{\displaystyle R_{B}-F_{K}=0}
F
K
=
0
{\displaystyle F_{K}=0}
R
B
−
F
K
−
F
D
=
0
{\displaystyle R_{B}-F_{K}-F_{D}=0}
F
D
=
0
{\displaystyle F_{D}=0}
Solution 2.04
P2.12, Beer 2012
A nylon thread is to be subjected to a 10-N tension. Knowing that
E = 3.2 GPa, that the maximum allowable normal stress is 40 MPa,
and that the length of the thread must not increase by more than
1%, determine the required diameter of the thread
GIVEN:
P
=
10
N
{\displaystyle P=10N}
E
=
3.3
G
P
a
{\displaystyle E=3.3GPa}
σ
=
40
M
P
a
{\displaystyle \sigma =40MPa}
δ
=
P
L
A
E
{\displaystyle \delta ={\frac {PL}{AE}}}
AND IT IS REQUIRED THAT
δ
<
0.001
L
{\displaystyle \delta <0.001L}
⇒
P
L
A
E
<
0.001
L
{\displaystyle \Rightarrow {\frac {PL}{AE}}<0.001L}
⇒
A
E
P
<
100
{\displaystyle \Rightarrow {\frac {AE}{P}}<100}
⇒
A
>
100
P
E
{\displaystyle \Rightarrow A>{\frac {100P}{E}}}
AND WE ALSO KNOW THAT
A
=
π
4
×
d
2
{\displaystyle A={\frac {\pi }{4}}\times d^{2}}
π
4
×
d
2
<
100
P
E
{\displaystyle {\frac {\pi }{4}}\times d^{2}<{\frac {100P}{E}}}
⇒
d
2
<
400
P
π
E
{\displaystyle \Rightarrow d^{2}<{\frac {400P}{\pi E}}}
⇒
d
>
20
p
π
E
{\displaystyle \Rightarrow d>20{\sqrt {\frac {p}{\pi E}}}}
⇒
d
>
20
10
π
×
3.2
×
10
9
{\displaystyle \Rightarrow d>20{\sqrt {\frac {10}{\pi \times 3.2\times 10^{9}}}}}
d
>
0.63
m
m
{\displaystyle d>0.63mm}
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
P2.16, Beer 2012
The brass tube AB (E = 105 GPa) has a cross-sectional area of
140 mm^2 and is fitted with a plug at A. The tube is attached at B
to a rigid plate that is itself attached at C to the bottom of an
aluminum cylinder (E = 72 GPa) with a cross-sectional area of
250 mm^2. The cylinder is then hung from a support at D. In order
to close the cylinder, the plug must move down through 1 mm.
Determine the force P that must be applied to the cylinder.
Given:
Brass Tube AB:
E
A
B
=
105
∗
10
9
P
a
{\displaystyle E_{AB}=105*10^{9}Pa}
A
A
B
=
140
m
m
2
{\displaystyle A_{AB}=140mm^{2}}
L
A
B
=
1
+
375
=
376
m
m
{\displaystyle L_{AB}=1+375=376mm}
Aluminum Cylinder DC:
E
D
C
=
72
∗
10
9
P
a
{\displaystyle E_{DC}=72*10^{9}Pa}
A
D
C
=
250
m
m
2
{\displaystyle A_{DC}=250mm^{2}}
L
D
C
=
375
m
m
{\displaystyle L_{DC}=375mm}
Equations:
Compression of Brass Tube AB:
δ
A
B
=
P
∗
L
A
B
A
A
B
∗
E
A
B
=
(
376
)
P
(
105
∗
10
9
)
(
140
)
=
(
2.56
∗
10
−
11
)
∗
P
{\displaystyle \delta _{AB}={\frac {P*L_{AB}}{A_{AB}*E_{AB}}}={\frac {(376)P}{(105*10^{9})(140)}}=(2.56*10^{-}11)*P}
Tension of Aluminum Cylinder DC:
δ
D
C
=
P
∗
L
D
C
A
D
C
∗
E
D
C
=
(
375
)
P
(
72
∗
10
9
)
(
250
)
=
(
2.1
∗
10
−
11
)
∗
P
{\displaystyle \delta _{DC}={\frac {P*L_{DC}}{A_{DC}*E_{DC}}}={\frac {(375)P}{(72*10^{9})(250)}}=(2.1*10^{-}11)*P}
We know the total deflection:
δ
T
=
1
m
m
{\displaystyle \delta _{T}=1mm}
δ
T
=
δ
A
B
+
δ
D
C
⇒
1
=
(
2.56
∗
10
−
11
)
P
+
(
2.1
∗
10
−
11
)
P
=
(
4.66
∗
10
−
11
)
P
{\displaystyle \delta _{T}=\delta _{AB}+\delta _{DC}\Rightarrow 1=(2.56*10^{-}11)P+(2.1*10^{-}11)P=(4.66*10^{-}11)P}
⇒
P
=
2.15
∗
10
1
0
m
N
,
o
r
21.5
k
N
{\displaystyle \Rightarrow P=2.15*10^{1}0mN,or21.5kN}
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
P2.24, Beer 2012
For the steel truss (E = 29 x 10^6 psi) and loading shown, determine
the deformations of members BD and DE, knowing that their
cross-sectional areas are 2 in^2 and 3 in^2, respectively.
∑
F
x
=
0
=
−
F
x
+
90
k
i
p
s
⇒
F
x
=
90
k
i
p
s
{\displaystyle \sum F_{x}=0=-F_{x}+90kips\Rightarrow F_{x}=90kips}
∑
M
F
=
0
=
−
30
k
i
p
s
∗
16
f
t
−
30
k
i
p
s
∗
8
f
t
−
24
f
t
∗
30
k
i
p
s
+
G
y
∗
15
f
t
⇒
G
y
=
96
k
i
p
s
(
t
e
n
s
i
o
n
)
{\displaystyle \sum M_{F}=0=-30kips*16ft-30kips*8ft-24ft*30kips+G_{y}*15ft\Rightarrow G_{y}=96kips(tension)}
∑
F
y
=
0
=
F
y
+
G
y
⇒
F
y
=
−
G
y
=
−
96
k
i
p
s
{\displaystyle \sum F_{y}=0=F_{y}+G_{y}\Rightarrow F_{y}=-G_{y}=-96kips}
∑
F
x
=
0
=
−
F
x
+
30
k
i
p
s
+
F
D
E
⇒
F
D
E
=
90
k
i
p
s
−
30
k
i
p
s
=
60
k
i
p
s
{\displaystyle \sum F_{x}=0=-F_{x}+30kips+F_{DE}\Rightarrow F_{DE}=90kips-30kips=60kips}
∑
M
G
=
0
=
−
30
k
i
p
s
∗
8
f
t
+
F
y
∗
15
f
t
−
F
B
D
∗
15
f
t
−
F
D
E
∗
8
f
t
⇒
F
B
D
=
60
k
i
p
s
∗
8
f
t
+
30
k
i
p
s
∗
8
f
t
−
96
k
i
p
s
∗
15
f
t
15
f
t
=
−
48
k
i
p
s
(
t
e
n
s
i
o
n
)
{\displaystyle \sum M_{G}=0=-30kips*8ft+F_{y}*15ft-F_{BD}*15ft-F_{DE}*8ft\Rightarrow F_{BD}={\frac {60kips*8ft+30kips*8ft-96kips*15ft}{15ft}}=-48kips(tension)}
δ
B
D
=
P
L
A
E
=
48
k
i
p
s
∗
8
f
t
∗
12
i
n
f
t
2
i
n
2
∗
29
∗
10
3
k
i
p
s
i
n
2
=
0.0794
i
n
{\displaystyle \delta _{BD}={\frac {PL}{AE}}={\frac {48kips*8ft*12{\frac {in}{ft}}}{2in^{2}*29*10^{3}{\frac {kips}{in^{2}}}}}=0.0794in}
δ
D
E
=
P
L
A
E
=
60
k
i
p
s
∗
15
f
t
∗
12
i
n
f
t
3
i
n
2
∗
29
∗
10
3
k
i
p
s
i
n
2
=
0.124
i
n
{\displaystyle \delta _{DE}={\frac {PL}{AE}}={\frac {60kips*15ft*12{\frac {in}{ft}}}{3in^{2}*29*10^{3}{\frac {kips}{in^{2}}}}}=0.124in}
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
P2.40, Beer 2012
A polystyrene rod consisting of two cylindrical portions AB and BC
is restrained at both ends and supports two 6-kip loads as shown.
Knowing that E = 0.45 X 10^6 psi, determine (a) the reactions at
A and C, (b) the normal stress in each portion of the rod.
Splitting the diagram into two parts we get part 1 and part 2 So we are given:
Part 1:
d
1
=
1.25
i
n
{\displaystyle d_{1}=1.25in}
L
1
=
25
i
n
{\displaystyle L_{1}=25in}
Part 2:
d
2
=
2
i
n
{\displaystyle d_{2}=2in}
L
2
=
15
i
n
{\displaystyle L_{2}=15in}
E
=
0.45
×
10
6
p
s
i
{\displaystyle E=0.45\times 10^{6}psi}
A
1
=
π
d
1
2
4
=
1.227
i
n
2
{\displaystyle A_{1}={\frac {\pi d_{1}^{2}}{4}}=1.227in^{2}}
A
2
=
π
d
2
2
4
=
3.14
i
n
2
{\displaystyle A_{2}={\frac {\pi d_{2}^{2}}{4}}=3.14in^{2}}
R
1
+
R
2
=
12
k
i
p
{\displaystyle R_{1}+R_{2}=12kip}
δ
t
o
t
a
l
{\displaystyle \delta _{total}}
δ
t
o
t
a
l
=
0
=
δ
1
+
δ
2
{\displaystyle \delta _{total}=0=\delta _{1}+\delta _{2}}
δ
=
P
L
A
E
{\displaystyle \delta ={\frac {PL}{AE}}}
P
1
L
1
A
1
E
1
+
P
2
L
2
A
2
E
2
=
0
{\displaystyle {\frac {P_{1}L_{1}}{A_{1}E_{1}}}+{\frac {P_{2}L_{2}}{A_{2}E_{2}}}=0}
Substituting in the reaction forces
R
1
{\displaystyle R_{1}}
R
2
{\displaystyle R_{2}}
(
R
2
{\displaystyle (R_{2}}
R
1
)
{\displaystyle R_{1})}
R
1
L
1
A
1
E
1
−
R
2
L
2
A
2
E
2
=
0
{\displaystyle {\frac {R_{1}L_{1}}{A_{1}E_{1}}}-{\frac {R_{2}L_{2}}{A_{2}E_{2}}}=0}
R
1
{\displaystyle R_{1}}
R
1
=
R
2
A
1
L
2
L
1
A
2
{\displaystyle R_{1}={\frac {R_{2}A_{1}L_{2}}{L_{1}A_{2}}}}
So
R
1
=
.2344
R
2
.
.
.
.
(
1
)
{\displaystyle R_{1}=.2344R_{2}....(1)}
R
1
+
R
2
=
12
k
i
p
.
.
.
.
(
2
)
{\displaystyle R_{1}+R_{2}=12kip....(2)}
R
1
=
2.27
k
i
p
s
{\displaystyle R_{1}=2.27kips}
R
2
=
9.72
k
i
p
s
{\displaystyle R_{2}=9.72kips}
(b) Using the formula
σ
=
P
A
{\displaystyle \sigma ={\frac {P}{A}}}
So
σ
1
=
R
1
A
1
=
1.85
k
s
i
{\displaystyle \sigma _{1}={\frac {R_{1}}{A_{1}}}=1.85ksi}
σ
2
=
R
2
A
2
=
3.09
k
s
i
{\displaystyle \sigma _{2}={\frac {R_{2}}{A_{2}}}=3.09ksi}
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
P2.44, Beer 2012
The rigid bar AD is supported by two steel wires of 1
16-in. diameter
(E = 29 X 10^6 psi) and a pin and bracket at D. Knowing that the
wires were initially taut, determine (a) the additional tension in
each wire when a 120-lb load P is applied at B, (b) the corresponding
deflection of point B.
δ
A
E
=
P
A
E
L
A
E
A
E
{\displaystyle \delta _{AE}={\frac {P_{AE}L_{AE}}{AE}}}
⇒
P
A
E
=
δ
A
E
∗
A
E
L
A
E
=
24
i
n
∗
θ
∗
π
4
∗
(
1
16
i
n
)
2
∗
29
∗
10
6
p
s
i
15
i
n
=
142.4
k
i
p
s
∗
θ
{\displaystyle \Rightarrow P_{AE}={\frac {\delta _{AE}*AE}{L_{AE}}}={\frac {24in*\theta *{\frac {\pi }{4}}*({\frac {1}{16}}in)^{2}*29*10^{6}psi}{15in}}=142.4kips*\theta }
δ
C
F
=
P
C
F
∗
L
C
F
A
E
{\displaystyle \delta _{CF}={\frac {P_{CF}*L_{CF}}{AE}}}
⇒
P
C
F
=
δ
C
F
A
E
L
C
F
=
8
i
n
∗
π
4
∗
(
1
16
i
n
)
2
∗
29
∗
10
6
p
s
i
8
i
n
=
88.97
k
i
p
s
∗
θ
{\displaystyle \Rightarrow P_{CF}={\frac {\delta _{CF}AE}{L_{CF}}}={\frac {8in*{\frac {\pi }{4}}*({\frac {1}{16}}in)^{2}*29*10^{6}psi}{8in}}=88.97kips*\theta }
∑
M
D
=
0
=
−
24
i
n
∗
P
A
E
+
16
i
n
∗
P
−
8
i
n
∗
P
C
F
=
−
24
i
n
∗
142.4
∗
10
3
l
b
∗
θ
+
16
i
n
∗
120
l
b
−
8
i
n
∗
88.97
∗
10
3
l
b
∗
θ
{\displaystyle \sum M_{D}=0=-24in*P_{AE}+16in*P-8in*P_{CF}=-24in*142.4*10^{3}lb*\theta +16in*120lb-8in*88.97*10^{3}lb*\theta }
⇒
θ
=
465
∗
10
−
6
{\displaystyle \Rightarrow \theta =465*10^{-6}}
P
A
E
=
142.4
∗
10
3
l
b
∗
465
∗
10
−
6
=
66.21
l
b
{\displaystyle P_{AE}=142.4*10^{3}lb*465*10^{-6}=66.21lb}
P
C
F
=
88.97
∗
10
3
l
b
∗
465
∗
10
−
6
=
41.37
l
b
{\displaystyle P_{CF}=88.97*10^{3}lb*465*10^{-6}=41.37lb}
δ
B
=
16
i
n
∗
θ
=
16
i
n
∗
465
∗
10
−
6
=
7.44
∗
10
−
3
(
d
o
w
n
w
a
r
d
)
{\displaystyle \delta _{B}=16in*\theta =16in*465*10^{-6}=7.44*10^{-3}(downward)}
On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
