229
18-1
Prob. 18.1 L = 12 ft = 144 in.
2
2
L
E
fe
—————————————————–
Prob. 18.2
(a) L = 50 ft
(O.K.) ksi 34 ksi 109.7
94.2
)12(50
)000,30(
2
2
e
f
e
(c) L = 20 ft
.applicablenot Euler
ksi 34 ksi 4.44
94.2
)12(20
)000,30(
2
2
e
f
(d) L = 15 ft
15 ft < 20 ft (part c) therefore, Euler not applicable.
—————————————————–
—————————————————–
Prob. 18.4
Steel bar – 1×2, A = 2.00 in.2
in. 2887.0
1
axisweak r
000,30
For 5-ft long bar:
2887.0
)12(5
)000,000,30(
2
2
e
f
L/r fe
(MPa)
0
50 817
100 204
150 90.8
200 51.1
——————————————————
230
Prob. 18.6
W12x22: A = 6.48 in.2, ry = 0.848
—————————————————–
Prob. 18.7 P.L. = 40 ksi
r
KL
E
fe
ksi 40
2
2
Prob. 18.8 P.L.= 30,000 psi
A = 2(1)  0.8(1.8)
= 0.560 in.2
Prob. 18.9 L = 7 m, K = 1.0
P.L. = 331 MPa, E = 207×103 MPa
111.9 MPa < 331 MPa (O.K.)
Fa = 111.9/2.5 = 44.8MPa
Pa = FaA = 45.1(7550) = 337×103 N
Pa = 337 kN
—————————————————–
76.3
Check:
(O.K.)k 350 k 389
5.2
)8.38(1.25
a
P
Use W14x132
——————————————————
k 656)5.26(
67.1
)50(658.0
9.109
50
a
P
——————————————————
231
18-3
Prob. 18.12
K KL/r fe (ksi) Pa (k)
——————————————————
Prob. 18.13
W250×115: A = 14 600 mm2, ry = 66.0 mm,
Fy = 345 MPa, E = 207×103 MPa
P = 1300 kN
kN 1680
)mm 600,14(
1.67
N/mm 345658.0 2
2
247
345
a
P
1300 kN < 1680 kN (O.K.)
(b) K = 1.0, L = 9 m
Prob. 18.14
Steel pipe: A = 11.9 in.2, r = 3.67 in
35
y
F
r
.)(short/int 9.371 4.65
67.3
)12)(20(0.1
r
KL
Use Eqn. 18.8:
)000,30(
2
2
E
4
3
in. 3.200
12
)12(5.0
23.56
y
I
——————————————————
232
18-4
Prob. 18.16
Select standard wt. steel pipe.
Fy = 240 MPa, P = 100 kN, L = 5.5 m,
K = 1.0
Try Fa = Fy /3 = 80 MPa
(O.K.) kN 100 kN 106
)2040)(1.99(525.0
MPa1.99
6.143
)10207(
2
32
2
2
a
e
P
r
KL
E
f
Use 102 mm diam. Pipe
—————————————————–
Prob. 18.17
Select W14, Fy = 50 ksi, E = 30,000 ksi,
P = 360 k, L = 24 ft, K = 1.0
(N.G.)k 360 k 252
)8.21)(0.22(525.0
ksi0.22
1.116
)000,30(
2
2
a
e
P
f
(O.K.)k 360 k 517
)5.26(
67.1
)50(658.0
9.48
8.77
)000,30(
9.48
50
2
2
2
2
a
e
P
rKL
E
f
Use W14x94
—————————————————–
233
18-5
Prob. 18.19
A = 0.7854d2, r = 0.25d
Assume rod will be intermediate and J. B.
Johnson formula applies.
Check :
4.91
)875.0(25.0
)20(0.1
r
KL
7.118
42
)000,30(22
trans.
22
y
F
E
r
KL
91.4 < 118.7 (O.K.)
Use 7/8 diam. rod
—————————————————–
Prob. 18.20
Assume rod to be slender.
7.118
42
)000,30(22
trans.
22
y
F
E
r
KL
192 > 118.7 (O.K.)
6.153
)50.2(25.0
)96(0.1
r
KL
5.83
85
)000,30(22
trans.
22
y
F
E
r
KL
153.6 > 83.5 (O.K.)
Use 2½ diam. rod
—————————————————–
Prob. 18.22
8×8 (S4S) Doug. fir, E = 1700 ksi, K = 1.0,
A = 56.3 in.2, Fc = 1050 psi, d = 7.5 in.
234
18-6
Prob. 18.22 (cont.)
(b) ) Le = KL = 1.0(17) = 17.0 ft
——————————————————
Prob. 18.23 (Rework Prob. 18.22)
250 mm x 250 mm, Eastern white pine,
E = 7.6×103 MPa, K = 0.8,
949.0
8.0
45.4
6.1
45.41
6.1
45.41 2
P
C
Fa = FcCP = 5.00(0.949) = 4.75 MPa
Pa = FaA = 4.75(58.1×103) = 276 kN
(b) ) Le = KL = 0.80(5180) = 4144 mm
Prob. 18.24
Doug. fir: E = 1700 ksi, Fc = 1050 psi
Fa = FcCP = 1050(0.670) = 704 psi
(N.G.) in. 3.90in. 8.127
704
000,90
dreq’
22
A
Fa = FcCP = 1050(0.789) = 829 psi
Pa = FaA = 829(132) = 109,400 lb
109.4 k > 90 k (O.K.)
Use 12×12 (S4S)
—————————————————–
Prob. 18.25
Southern pine: E = 1700 ksi
235
Prob. 18.25 Contd
Fa = FcCP = 1250(0.514) = 643 psi
(b) P = 60,000 lb
From part (a), 8×12 too small.
Try 10×10: A = 90.3 in.2, d = 9.5 in.
—————————————————–
Prob. 18.26 (See Prob. 18.25)
Southern pine: E = 1700 ksi
Fc = 1250 psi
K = 0.8, L = 16 ft
(O.K.) 50 48.20
5.7
)12)(16(8.0
6
d
Le
O.K.)(Say in. 3.56in. 4.56
Use 8×8 (S4S)
(b) P = 60,000 lb
Assume an 8 nominal thickness from part (a) and
k 2.64)767.1(1.24
(O.K.) ksi 32 ksi 1.24
375.0
e
P
r
(b) L = 60 in.
236
18-8
Prob. 18.31 (cont.)
)000,10(
2
2
E
—————————————————–
Prob. 18.32
A = 0.7854(0.5)2 = 0.1964 in.2
——————————————————
Prob. 18.33
For minimum length for Euler, se = P.L.
—————————————————–
Prob. 18.34
d = 25 mm, A = 490.9 mm2,
r = 0.25(25) = 6.25 mm – steel
(a) L = 1 m
25.6
1000
)10207(
2
32
2
2
e
r
L
E
f
(c) L = 4 m
)10207(
32
2
E
—————————————————–
Prob. 18.35 (Rework Prob. 18.34)
d = 25 mm, A = 490.9 mm2,
kN 3.251N 1025.13)9.490(0.27
3
e
P
(b) L = 2 m
(c) L = 4 m
N 828)9.490(687.1
(O.K.) MPa 220 MPa 687.1
25.6
4000
)1070(
2
32
2
2
e
e
P
r
L
E
f
237
18-9
Prob. 18.36
A = 14.1 + 2(16) = 46.1 in.2
—————————————————–
Prob. 18.37
2 diam. standard weight pipe, fixed-pinned, L = 84
in., P.L. = 34 ksi
——————————————————
Prob. 18.38
4
2
2
2
2
in. 51.17
)000,30(
)0.2()12300.1(20
)F.S()(
dreq‘ E
.KLP
I
Prob. 18.39
W250x73: A = 9290 mm2, ry = 64.6 mm
A992: Fy = 345 MPa, E = 207×103 MPa
4.115
345
10207
71.4 trans.
3
r
KL
K L
(m)
KL/r Eqn fe
(MPa)
Pa
(kN)
a 1.0 5 77.5 18-8 340 1255
ksi 7.72
2
2
e
r
KL
E
f
238
18-10
Prob. 18.41 (cont.)
0.136
36
000,30
71.471.4 Trans.
y
F
E
r
KL
—————————————————–
Prob. 18.42 Select a W10 shape.
A992: Fy = 50 ksi, E = 30,000 ksi
P = 280 k, K = 1.0, L = 18 ft
4.115
50
)000,30(
71.471.4 trans.
Y
s
E
r
KL
k 365)9.19(
67.1
)50(658.0
6.42
50
a
P
365 k > 280 k (O.K.  over by 31%)
Try a smaller shape; try W10x54:
A = 15.8 in.2, ry = 2.56 in.
—————————————————–
Prob. 18.43 Select Pipe X-strong
Fy = 35 ksi, E = 30,000 ksi
P = 90 k, K = 1.0, L = 16 ft
k 112)83.7(
67.1
)35(658.0
5.38
35
a
P
r
112 k > 90 k (O.K.)
0.3
7854.0k 10
2
d
Solve for d:
Reqd d = 1.074 in., try 1 1/8 diam. rod.
Check validity of J.B.Johnson formula:
239
18-11
Prob. 18.45
19 mm diam. steel rod, K = 1.0,
L = 350 mm, F.S. = 2.5
r = 0.25(19) = 4.75 mm
—————————————————–
Prob. 18.46
Cross section is d×2d
K = 1.0, L = 16 in., P = 4.60 k, H.S. steel with
Fy = 110,000 psi, S.F. = 3.0
A = 2d2 in.2
107
)517.0(2887.0
)16(0.1
r
KL
107 > 73.4 (O.K.  bar is slender.)
Reqd dimensions: 0.517 × 1.035
—————————————————–
Prob. 18.47
Doug. fir: E = 1700 ksi, Fc = 1050 psi
10×12 (S4S): A = 109 in.2, d = 9.5 in.
K = 0.8, L = 16 ft
Southern pine: E = 1700 ksi, Fc = 1250 psi
Full nom. 12×12: A = 144 in.2, d = 12 in.
K = 1.0, L = 20 ft
(O.K.) 50 20
12
)12(20
d
Le
02.1
)101700(3.0
2
3
1050
Try 10×10 (S4S): A = 90.3 in.2, d = 9.5 in.
943.0
)1050(7.22
)107.1(3.0
(O.K.) 50 7.22
5.9
)12)(18(0.1
2
6
d
Le
240
Prob. 18.49 contd
—————————————————–
Prob. 18.50
Southern pine: E = 1700 ksi
Assume Fa = Fc:
2
in. 40
1250
000,50
dreq’ A
Try 8×8 (S4S): A = 56.3 in.2, d = 7.5 in.
(O.K.) 50 32
5.7
)12)(20(0.1
d
Le
Try 10×10 (S4S): A = 90.3 in.2, d = 9.5 in.
(O.K.) 50 3.25
5.9
)12)(20(0.1
d
Le
——————————————————
Prob. 18.51
62.8
Try 250×250 (S4S)
250×250 300×300
CP0.544 0.689
Pa (kN) 272 507
Decision N.G. O.K.
Use 300×300 (S4S)
——————————————————
241