110
Prob. 11-1
(a) 0040.0
)12(10
48.0
axial L
—————————————————-
Prob. 11-2
(O.K.) psi 34,000 psi 000,25
12
000,300
A
P
t
—————————————————-
Prob. 11-3
250.01
GPa (2)44
GPa 110
1
2G
E
—————————————————
Prob. 11-4
)000,800,5(2
—————————————————
Prob. 11-5
E = 207×103 MPa; = 0.25
Stresses
MPa 3.53
10600 3
P
y (Comp.)
= 106 MPa < 230 MPa (O.K.)
Strains: transverse = axial
)(
1
yxz E
(increase) 10576
10207
)3.53(25.0106
)(
6
3
yxx E
Dimensional changes:
z = 63.6×10-6(25)
= 1.590×10-3 mm (decrease)
= L( T)
= 0.0000065(1000)(70-32) = 0.247 ft
(Distance measured 0.247 ft too long.)
—————————————————
Prob. 11.7
Initial temp. = 68°F
(a) At 32°: wire contracts, stress increases.
T = 6832 = 36°
111
11-2
Prob. 11.8 E = 103×103 MPa
= 16.7×10-6 1/C°
T to close gap:
————————————————
Prob. 11.9
L = 60 ft
@ 30°: T = 40°
= L T
——————————————————
Prob. 11.11
Wood: S.Pine
c(all) = 1250 psi
E = 1700 ksi
Steel: A36
c(all) = 20 ksi
Es = 30,000 ksi
—————————————————–
Prob. 11.12
Post: (Doug. fir)
150 mm x 150 mm
n = Est/Ew = 17.25
Assume wood at allowable:
st =n w = 17.25(7.24)
= 124.9 MPa (<147 MPa (O.K.))
P = Ast st + Aw w
cu Acu + st Ast = 8000 lb
cu(0.1104) + 2 st(0.1964) = 8000 lb
cu = 15,900 psi
st = 31,800 psi
—————————————————–
Prob. 11.14
ADF = AHF = 11.5(5.5) = 63.25 in.2
EDF = 1700 ksi; EHF = 1400 ksi
n =EDF/EHF = 1.2143
DF = n HF = 1.2143 HF
112
11-3
Prob. 11.15
AS = 4.0 in.2, LS = 12 in.
ACI = 9.0 in.2 , LCI = 14 in.
ES = 30,000 ksi, ECI = 15,000 ksi
—————————————————-
Prob. 11.16
—————————————————–
Prob. 11.17
(a)
r = 8 mm; r/d = 8/104 = 0.077
(c) r = 38 mm, r/d = 38/44 = 0.86
Fig. 11.14: k = 1.25
Anet = 13(120-2(38)) = 572 mm2
)00050(25.1
kP
)375.0(4
A
t
(b) psi 700,10
)375.0(3
000,12
net
A
P
t
(c) r/d = 0.5/3 = 0.167: Fig 11.14: k = 1.7
psi 190,18)700,10(7.1
kP
Fig. 11.14: k = 2.22
lbP
k
A
P
A
kP tt
t
150,11
22.2
)125.1(000,22
ne(all)
net
(max)
—————————————————-
Prob. 11.20
113
11-4
Prob. 11.20 contd
2sin
2
(a) ‘
A
P
—————————————————–
Prob. 11.21
A = 50 mm × 75 mm = 3.75×103 mm2
(a) max. on 45° plane:
—————————————————–
Prob. 11.22
2sin
2
‘
A
P
Prob. 11.23
P = 67 k
A = 0.7854(6)2
= 28.3 in.2
P
psi 940070sin000,10
2sin(b) n
—————————————————–
Prob. 11.25
(a) See element of Prob. 11.24:
2cos‘
Use element of unit thickness and use diagonal d =
1. Therefore, h = sin and
114
11-5
Prob. 11.30
(a) Pa 105.137
101963
10279 6
6
3
A
P
t
(b)
= axial L
= 664×10-6(1 m)
=0.664 mm
—————————————————-
Prob. 11.31
psi 000,16
000,64
P
t
—————————————————–
Prob. 11.32
d = 150 mm
A = 0.7854d2
= 17.67×103 mm2
300
275.0
z
Prob. 11.33
d = 1.5 in.
(O.K.) psi 34,000
2496.0
00115.0
000287.0
000287.0
5.1
00043.0
psi 000,530,29
00115.0
960,33
x
z
x
x
x
z
z
L
E
(more)
115
Prob. 11.35 contd
Strains:
(decrease) 0009.0
E
zyx
x
—————————————————–
Prob. 11.36
Stresses:
(O.K.) ksi 34 ksi 8
)5.0(10
40
(O.K.) ksi 34 ksi 20
)5.0(2
20
x
A
P
y
Prob. 11.37
(a) = L ( T)
6.46
)33(0000065.0
01.0
L
T
BLB( T ) + SLS( T ) = 0.030 in.
F 6.35
in./ft) 12)](6(105.6)3(104.10[
in. 03.0
030.0
66
SSBB LL
T
T to cause thermal stress:
T = 27 21.65 = 5.35 C°
= E T
= 207×103(11.7×10-6)(5.35) = 12.96 MPa
—————————————————–
Prob. 11.40
116
Prob. 11.41
Cut section a-a
as shown.
CD = 72 kN (Comp.)
Force :
AE
PL
——————————————————
Prob. 11.42 Refer to Prob. 5-25
T = +40F°
L
(ft)
A
(in.2 )
P
(k)
AE
PL
(in.)
L T
(in.)
CD 15 1.00 13.33
0.80
0.0468
—————————————————–
Prob. 11.43
= L ( T)
= 0.0000065(100)(70-25) = 0.02925 ft
Prob. 11.44
1-in. diam. rod: A = 0.7854 in.2
Due to tensioning:
psi 730,12
7854.0
000,10
A
P (tension)
Due to temp. rise of 30F°:
= E ( T)
= 0.0000093, E = 15,000,000 psi
(a) = E ( T) T = s/(E )
F 5.107
)0000093.0(000,000,15
000,15
T
(b) In addition to (a):
= L ( T) T = /( L)
)107.11(10207
63
E
Initial temp to which member must be heated:
20°C + 28.9 C° = 48.9°C
117
11-8
Prob. 11.47 contd
t1 = L1 ( T)
= 0.0000065(60)(100) = 0.0390 in.
—————————————————–
Prob. 11.48
——————————————————
Prob. 11.49
(a) Unrestrained contraction:
(b) Force P applied to restore member to its original
length:
PL
PL
—————————————————–
Prob. 11.50
Copper: Acu = 2(100)(3.2) = 640 mm2
Ecu = 103×103 MPa
L5x5x ½ : A = 4.79 in.2
Redwoood: E = 1300 ksi, c(all) = 1050 psi
n = Est/Ew = 30,000/1300 = 23.1
Assume steel at allowable:
118
Prob. 11.52
stAst + cAc + CI ACI = 400 k (Eqn. I)
st = c = CI
CI
CI
c
c
st
st
EEE
000,153120000,30
CIcst
CI = 6.50 ksi
st = 2(6.50) = 13.0 ksi
c = 0.208(6.50) = 1.352 ksi
——————————————————
Prob. 11.53
n = Est/Ec = 8.322
st = n c = 8.322 c
P = Ast st + Ac c
100 = 2.41(8.322 c) + 193.6 c
c = 0.468 ksi
100
——————————————————
Prob. 11.54
Est = 207×103 MPa, Ast = 1250 mm2
Ecu = 103×103 MPa, Acu = 1800 mm2
The steel rod will initially carry the load until the
0.06 mm gap closes:
119
11-10
Prob. 11.55
Subs. into Eqn. I:
Pst = Pal = 12,000 lb
Subs. into Eqn. II:
a = 5.00 ft
Stresses:
——————————————————
Prob. 11.56
Steel: Ast = 5162×10-6 m2
Prob. 11.57
Each wire: 1/4 diam., A = 0.49 in.2, L=20
Fy = 2Pst + Pbr 2000 = 0 —– Eqn. I
psi 6800
049.0
:Bronze st
psi 010,17
049.0
3.833
:Steel st
)12)(20(101,17
L
PL
Each wire: 1/4 diam., A = 0.49 in.2, L=20
Fy = 2Pal + Pst 1500 = 0 —– Eqn. I
Box remains horizontal:
al = st
stal AE
PL
AE
PL
120
(a) Stresses:
psi 6120
049.0
300
:Aluminum al
——————————————————
Prob. 11.59
Steel: Diam. = 25 mm
Ast = 490.9×10-6 m2
Fy = 2Pst + Pbr (180×103) = 0 —–Eqn. I
—————————————————–
Prob. 11.60
Prob. 11.61
Ag = 75(10) = 750 mm2 = 750×10-6 m2
Ah = 19(10) = 190 mm2 = 190×10-6 m2
Anet = 560×10-6 m2
r/d = 9.5/56 = 0.17; Fig. 11.14: k = 2.35
Pa 1014.32
10560
10180 6
6
3
avg
avg A
P
(b) W = 2.5 in., D = 0.50 in.
Ag = 2.5(0.375) = 0.938 in.2
Ah = 0.50(0.375) = 0.1875 in.2
Anet = 0.938 0.1875 = 0.75 in.2
d = W D = 2.5 0.50 = 2.00 in.
121
Prob. 11.62 contd
(c) W = 3.0 in., D = 1.00 in.
—————————————————-
Prob. 11.63
d = 125 38 = 87 mm
r = 38/2 = 19 mm
——————————————————
Prob. 11.64
6 diam. compression member:
A = 28.27 in.2
Based on shear stress (max. when = 45°):
Prob. 11.65
A = 150(200) = 30×103 mm2
(Controls) N 100.96
90sin
)1030)(2(106.1
2sin
)2(
3
36
(all)
max
A
P
—————————————————–
Prob. 11.66
A = 4.0 in.2
(ult) = 1200 psi
A = 490.9×10-6 m2
122
11-13
Prob. 11.67 contd
60sin
)109.490(2
1080
2sin
2
6
3
‘
A
P
(b) Max. normal stress ( = 0°)
0cos
1080
cos
2
6
3
2
(max) A
P
n
—————————————————–
Prob. 11.68
A = 1.0 in.2
—————————————————–
Prob. 11.69
On plane CD:
‘ = 35 MPa
—————————————————————
Prob. 11.70
Tens. and/or comp. stress (max. at 45°)
(a) n = sin2
= 3750×10-6 m2
On 50° plane:
‘ = 138 MPa
( = 40°)
MPa 1382sin
(a)
‘
P
123