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Solutions 13 46060
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131.
Determine the critical buckling load for the column. The material can be assumed rigid.
P
L 2 k
Equilibrium: The disturbing force F can be determined by summing moments about point A. a + MA = 0; P(Lu)  F a L b=0 2
L 2 A
F = 2Pu Spring Formula: The restoring spring force F1 can be determine using spring formula Fs = kx. Fs = k a L kLu ub = 2 2
Critical Buckling Load: For the mechanism to be on the verge of buckling, the disturbing force F must be equal to the restoring spring force F1. 2Pcr u = kLu 2 kL 4 Ans.
Pcr =
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132. Determine the critical load Pcr for the rigid bar and spring system. Each spring has a stiffness k. Equilibrium: The disturbing forces F1 and F2 can be related to P by writing the moment equation of equlibrium about point A. Using small angle ananlysis, where cos u 1 and sin u = u, + MA = 0; F2 a L 2 b + F1 a L b  PLu = 01 3 3 (1)
k
P
L 3 k L 3
F2 + 2F1 = 3Pu
Spring Force. The restoring spring force A Fsp B 1 and A Fsp B 2 can be determined using the spring formula, Fsp = kx, where x1 = 2 1 Lu and x2 = Lu, Fig. b. Thus, 3 3 2 2 = kx1 = k a Lu b = kLu 3 3
A
L 3
A Fsp B 1
A Fsp B 2 = kx2 = k a Lu b =
1 3
1 kLu 3
Critical Buckling Load. When the mechanism is on the verge of buckling the disturbing force F must be equal to the restoring force of the spring Fsp. Thus, F1 = A Fsp B 1 = 2 kLu 3 F2 = A Fsp B 2 = 1 kLu 3
Substituting this result into Eq. (1), 2 1 kLu + 2 a kLu b = 3Pcr u 3 3 Pcr = 5 kL 9 Ans.
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133. The leg in (a) acts as a column and can be modeled (b) by the two pinconnected members that are attached to a torsional spring having a stiffness k (torque rad). Determine the critical buckling load. Assume the bone material is rigid.  P(u) a L b + 2ku = 0 2
k
P
a + MA = 0; Require:
L 2
Pcr =
4k L
Ans.
L 2
(a)
(b)
*134. Rigid bars AB and BC are pin connected at B. If the spring at D has a stiffness k, determine the critical load Pcr for the system. Equilibrium. The disturbing force F can be related P by considering the equilibrium of joint A and then the equilibrium of member BC, Joint A (Fig. b) + c Fy = 0; FAB cos f  P = 0 FAB = P cos f
a a
P A
B
k D
Member BC (Fig. c) MC = 0; F(a cos u) P P cos f (2a sin u) sin f(2a cos u) = 0 cos f cos f
a C
F = 2P(tan u + tan f) Since u and f are small, tan u u and tan f f. Thus, (1)
F = 2P(u + f) Also, from the geometry shown in Fig. a, 2au = af Thus Eq. (1) becomes F = 2P(u + 2u) = 6Pu f = 2u
Spring Force. The restoring spring force Fsp can be determined using the spring formula, Fsp = kx, where x = au, Fig. a. Thus, Fsp = kx = kau
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134.
Continued
Critical Buckling Load. When the mechanism is on the verge of buckling the disturbing force F must be equal to the restoring spring force Fsp. F = Fsp
6Pcru = kau ka 6
Pcr =
Ans.
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135.
An A36 steel column has a length of 4 m and is pinned at both ends. If the cross sectional area has the dimensions shown, determine the critical load.
25 mm
Section Properties: A = 0.01(0.06) + 0..05(0.01) = 1.10 A 10  3 B m2 Ix = Iy = 1 1 (0.01) A 0.063 B + (0.05) A 0.013 B = 0.184167 A 10  6 B m4 12 12
25 mm 10 mm 25 mm 10 mm
25 mm
Critical Buckling Load: K = 1 for pin supported ends column. Applying Eulers formula, Pcr = p2EI (KL)2 p2 (200)(109)(0.184167)(10  6) = [1(4)]2 Ans.
= 22720.65 N = 22.7 kN Critical Stress: Eulers formula is only valid if scr 6 sg. scr = Pcr 22720.65 = 20.66 MPa 6 sg = 250 MPa = A 1.10(10  3)
O.K.
136. Solve Prob. 135 if the column is fixed at its bottom and pinned at its top. Section Properties: A = 0.01(0.06) + 0.05(0.01) = 1.10 A 10  3 B m2 1 1 Ix = Iy = (0.01) A 0.063 B + (0.05) A 0.013 B = 0.184167 A 10  6 B m4 12 12 Critical Buckling Load: K = 0.7 for one end fixed and the other end pinned column. Applying Eulers formula, Pcr = p EI (EL)2 p2 (200)(109)(0.184167)(10  6) = [0.7(4)]2 Ans.
2
25 mm
10 mm
25 mm
25 mm 10 mm
25 mm
= 46368.68 N = 46.4 kN Critical Stress: Eulers formula is only valid if scr 6 sg. scr = Pcr 46368.68 = 42.15 MPa 6 sg = 250 MPa = A 1.10(10  3)
O.K.
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137. A column is made of A36 steel, has a length of 20 ft, and is pinned at both ends. If the crosssectional area has the dimensions shown, determine the critical load. The cross sectional area and moment of inertia of the square tube is A = 6(6)  5.5(5.5) = 5.75 in2 I= 1 1 (6)(63) (5.5)(5.53) = 31.74 in4 12 12
6 in. 0.25 in.
5.5 in.
0.25 in. 0.25 in. 0.25 in.
The column is pinned at both of its end, k = 1. For A36 steel, E = 29.0(103) ksi and sg = 36 ksi (table in appendix). Applying Eulers formula, Pcr = p2 C 29.0(103) D (31.74) p2EI = (KL)2 C 1(20)(12) D 2 = 157.74 kip = 158 Critical Stress. Eulers formula is valid only if scr 6 sg. scr = Pcr 157.74 = = 27.4 ksi 6 sg = 36 ksi A 5.75 O.K.
Ans.
*138. A column is made of 2014T6 aluminum, has a length of 30 ft, and is fixed at its bottom and pinned at its top. If the crosssectional area has the dimensions shown, determine the critical load. The crosssectional area and moment of inertia of the square tube is A = 6(6)  5.5(5.5) = 5.75 in2 1 1 I= (6)(63) (5.5)(5.53) = 31.74 in4 12 12 The column is fixed at one end, K = 0.7. For 201476 aluminium, E = 10.6(103) ksi and sg = 60 ksi (table in appendix). Applying Eulers formula, Pcr = p2 C 10.6(103) D (31.74) p2EI = (KL)2 C 0.7(30)(12) D 2 = 52.29 kip = 52.3 kip Critical Stress. Eulers formula is valid only if scr 6 sg. scr = Pcr 52.3 = = 9.10 ksi 6 sg = 60 ksi A 5.75 O.K.
6 in. 0.25 in.
5.5 in.
0.25 in. 0.25 in. 0.25 in.
Ans.
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139.
The W14 * 38 column is made of A36 steel and is fixed supported at its base. If it is subjected to an axial load of P = 15 kip, determine the factor of safety with respect to buckling.
P
From the table in appendix, the crosssectional area and moment of inertia about weak axis (yaxis) for W14 * 38 are
20 ft
A = 11.2 in2
Iy = 26.7 in4
The column is fixed at its base and free at top, k = 2. Here, the column will buckle about the weak axis (y axis). For A36 steel, E = 29.0(103) ksi and sy = 36 ksi. Applying Eulers formula, Pcr = p2EIy (KL)2 = p2 C 29.0(103) D (26.7)
C 2 (20)(12) D 2
= 33.17 kip
Thus, the factor of safety with respect to buckling is F.S = Pcr 33.17 = = 2.21 P 15 Ans.
The Eulers formula is valid only if scr 6 sg. scr = Pcr 33.17 = = 2.96 ksi 6 sg = 36 ksi A 11.2 O.K.
1310. The W14 * 38 column is made of A36 steel. Determine the critical load if its bottom end is fixed supported and its top is free to move about the strong axis and is pinned about the weak axis. From the table in appendix, the crosssectional area and moment of inertia about weak axis (yaxis) for W14 * 38 are A = 11.2 in2 Ix = 385 in4 Iy = 26.7 in4
P
20 ft
The column is fixed at its base and free at top about strong axis. Thus, kx = 2. For A36 steel, E = 29.0(103) ksi and sg = 36 ksi. Pcr = p2EIx (KxLx)
2
=
p2 C 29.0(103) D (385)
C 2 (20)(12) D 2
= 478.28 kip
The column is fixed at its base and pinned at top about weak axis. Thus, ky = 0.7. Pcr = p2EIy (KyLy)
2
=
p2 C 29.0(103) D (26.7)
C 0.7(20)(12) D 2
= 270.76 kip = 271 kip (Control) The Eulers formula is valid only if scr 6 sg. scr = Pcr 270.76 = = 24.17 ksi 6 sg = 36 ksi A 11.2
Ans.
O.K.
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1311. The A36 steel angle has a crosssectional area of A = 2.48 in2 and a radius of gyration about the x axis of rx = 1.26 in. and about the y axis of ry = 0.879 in. The smallest radius of gyration occurs about the z axis and is rz = 0.644 in. If the angle is to be used as a pinconnected 10ftlong column, determine the largest axial load that can be applied through its centroid C without causing it to buckle.
y z x C x z
y
The least radius of gyration: r2 = 0.644 in. scr = controls. K = 1.0
p2 (29)(103) =
A KL B 2 r
p2E
;
C 1.0 (120) D 2 0.644
= 8.243 ksi 6 sg
O.K. Ans.
Pcr = scr A = 8.243 (2.48) = 20.4 kip
*1312. An A36 steel column has a length of 15 ft and is pinned at both ends. If the crosssectional area has the dimensions shown, determine the critical load.
8 in. 0.5 in.
0.5 in. 6 in. 0.5 in.
Ix =
1 1 (8)(73) (7.5)(63) = 93.67 in4 12 12 1 1 b (0.5)(83) + (6)(0.53) = 42.729 in4 (controls) 12 12
Iy = 2 a Pcr =
p2(29)(103)(42.729) p2EI = 2 (EL) [(1.0)(15)(12)]2 = 377 kip Ans.
Check: A = (2)(8)(0.5) + 6(0.5) = 11 in2 scr = Pcr 377 = = 34.3 ksi 6 sg A 11
Therefore, Eulers formula is valid
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1313.
An A36 steel column has a length of 5 m and is fixed at both ends. If the crosssectional area has the dimensions shown, determine the critical load. I= 1 1 (0.1)(0.053) (0.08)(0.033) = 0.86167 (10  6) m4 12 12 p2(200)(109)(0.86167)(10  6) p2EI = 2 (KL) [(0.5)(5)]2 = 272 138 N = 272 kN scr = Pcr ; A A = (0.1)(0.05)  (0.08)(0.03) = 2.6(10  3) m2 Ans.
10 mm 50 mm 100 mm 10 mm
Pcr =
=
272 138 = 105 MPa 6 sg 2.6 (10  3)
Therefore, Eulers formula is valid.
1314. The two steel channels are to be laced together to form a 30ftlong bridge column assumed to be pin connected at its ends. Each channel has a crosssectional area of A = 3.10 in2 and moments of inertia Ix = 55.4 in4, Iy = 0.382 in4. The centroid C of its area is located in the figure. Determine the proper distance d between the centroids of the channels so that buckling occurs about the xx and y y axes due to the same load. What is the value of this critical load? Neglect the effect of the lacing. Est = 2911032 ksi, sY = 50 ksi. Ix = 2(55.4) = 110.8 in.4 d2 Iy = 2(0.382) + 2 (3.10) a b = 0.764 + 1.55 d2 2 In order for the column to buckle about x  x and y  y at the same time, Iy must be equal to Ix Iy = Ix 0.764 + 1.55 d2 = 110.8 d = 8.43 in. Check: d 7 2(1.231) = 2.462 in. Pcr = p (29)(10 )(110.8) p2 EI = 2 (KL) [1.0(360)]2 Ans.
2 3
y 0.269 in.
y 1.231 in.
x C d y y C
x
Ans.
O.K.
= 245 kip Check stress: scr = Pcr 245 = = 39.5 ksi 6 sg A 2(3.10)
Therefore, Eulers formula is valid. 1 046
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1315. An A36steel W8 * 24 column is fixed at one end and free at its other end. If it is subjected to an axial load of 20 kip, determine the maximum allowable length of the column if F.S. = 2 against buckling is desired. Section Properties. From the table listed in the appendix, the crosssectional area and moment of inertia about the y axis for a W8 * 24 are A = 7.08 in2 Iy = 18.3 in4
Critical Buckling Load. The critical buckling load is Pcr = Pallow (F.S) = 20(2) = 40 kip Applying Eulers formula, Pcr = p2 EIy p2 C 29 A 103 B D (18.3) (2L)2 (KL)2
40 =
L = 180.93 in = 15.08 ft = 15.1 ft Critical Stress. Eulers formula is valid only if scr 6 sY. scr = Pcr 40 = = 5.65 ksi 6 sY = 36 ksi A 7.08
Ans.
O.K.
*1316. An A36steel W8 * 24 column is fixed at one end and pinned at the other end. If it is subjected to an axial load of 60 kip, determine the maximum allowable length of the column if F.S. = 2 against buckling is desired. Section Properties. From the table listed in the appendix, the crosssectional area and moment of inertia about the y axis for a W8 * 24 are A = 7.08 in2 Iy = 18.3 in4
Critical Buckling Load. The critical buckling load is Pcr = Pallow (F.S.) = 60(2) = 120 kip Applying Eulers formula, Pcr = p2EIy (KL)2 p2 C 24 A 103 B D (18.3) (0.7L)2 Ans.
120 =
L = 298.46 in = 24.87 ft = 24.9 ft Critical Stress. Eulers formula is valid only if scr 6 sY. scr = Pcr 120 = = 16.95 ksi 6 sY = 36 ksi A 7.08
O.K.
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1317.
The 10ft wooden rectangular column has the dimensions shown. Determine the critical load if the ends are assumed to be pin connected. Ew = 1.611032 ksi, sY = 5 ksi.
Section Properties:
10 ft
A = 4(2) = 8.00 in2 1 (2) A 43 B = 10.667 in4 12 1 (4) A 23 B = 2.6667 in4 (Controls !) 12
4 in. 2 in.
Ix =
Iy =
Critical Buckling Load: K = 1 for pin supported ends column. Applying Eulers formula,. Pcr = p2EI (KL)2 p2(1.6)(103)(2.6667) = [1(10)(12)]2 Ans.
= 2.924 kip = 2.92 kip Critical Stress: Eulers formula is only valid if scr 6 sg. scr = Pcr 2.924 = = 0.3655 ksi 6 sg = 5 ksi A 8.00
O.K.
1318. The 10ft column has the dimensions shown. Determine the critical load if the bottom is fixed and the top is pinned. Ew = 1.611032 ksi, sY = 5 ksi. Section Properties: A = 4(2) = 8.00 in2 1 (2) A 43 B = 10.667 in4 Ix = 12 1 Iy = (4) A 23 B = 2.6667 in4 (Controls!) 12 Critical Buckling Load: K = 0.7 for column with one end fixed and the other end pinned. Applying Eulers formula. Pcr = p2EI (KL)2 p2 (1.6)(103)(2.6667) = [0.7(10)(12)]2 Ans.
10 ft 4 in. 2 in.
= 5.968 kip = 5.97 kip Critical Stress: Eulers formula is only valid if scr 6 sg. scr = Pcr 5.968 = = 0.7460 ksi 6 sg = 5 ksi A 8.00
O.K.
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1319. Determine the maximum force P that can be applied to the handle so that the A36 steel control rod BC does not buckle. The rod has a diameter of 25 mm.
P 350 mm A 250 mm B 45 C
800 mm
Support Reactions: a + MA = 0; P(0.35)  FBC sin 45(0.25) = 0 FBC = 1.9799P Section Properties: A= p A 0.0252 B = 0.15625 A 10  3 B p m2 4 p A 0.01254 B = 19.17476 A 10  9 B m4 4
I=
Critical Buckling Load: K = 1 for a column with both ends pinned. Appyling Eulers formula, Pcr = FBC = p2EI (KLBC)2
1.9799P =
p2(200)(109) C 19.17476(10  9) D [1(0.8)]2 Ans.
P = 29 870 N = 29.9 kN Critical Stress: Eulers formula is only valid if scr 6 sg. scr = 1.9799(29 870) Pcr = 120.5 MPa 6 sg = 250 MPa = A 0.15625(10  3)p
O.K.
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*1320. The W10 * 45 is made of A36 steel and is used as a column that has a length of 15 ft. If its ends are assumed pin supported, and it is subjected to an axial load of 100 kip, determine the factor of safety with respect to buckling. Critical Buckling Load: Iy = 53.4 in4 for a W10 * 45 wide flange section and K = 1 for pin supported ends column. Applying Eulers formula, Pcr = p2EI (KL)2 p2 (29)(103)(53.4) = [1(15)(12)]2
P
15 ft
= 471.73 kip Critical Stress: Eulers formula is only valid if scr 6 sg. A = 13.3 in2 for the W10 * 45 wideflange section. scr = Factor of Safety: F.S = Pcr 471.73 = = 4.72 P 100 Ans. Pcr 471.73 = = 35.47 ksi 6 sg = 36 ksi A 13.3 O.K.
P
1321.
The W10 * 45 is made of A36 steel and is used as a column that has a length of 15 ft. If the ends of the column are fixed supported, can the column support the critical load without yielding?
P
Critical Buckling Load: Iy = 53.4 in4 for W10 * 45 wide flange section and K = 0.5 for fixed ends support column. Applying Eulers formula, Pcr = p2EI (KL)2 p2 (29)(103)(53.4) = [0.5(15)(12)]2
P
15 ft
= 1886.92 kip Critical Stress: Eulers formula is only valid if scr 6 sg. A = 13.3 in2 for W10 * 45 wide flange section. scr = Pcr 1886.92 = = 141.87 ksi 7 sg = 36 ksi (No!) A 13.3 Ans.
The column will yield before the axial force achieves the critical load Pcr and so Eulers formula is not valid.
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1322. The W12 * 87 structural A36 steel column has a length of 12 ft. If its bottom end is fixed supported while its top is free, and it is subjected to an axial load of P = 380 kip, determine the factor of safety with respect to buckling. A = 25.6 in2 Ix = 740 in4 Iy = 241 in4 (controls)
P
W 12 * 87 K = 2.0 Pcr =
12 ft
p2(29)(103)(241) p2EI = = 831.63 kip 2 (KL) [(2.0)(12)(12)]2 Pcr 831.63 = = 2.19 P 380 Ans.
F.S. = Check: scr =
Pcr A 831.63 = 32.5 ksi 6 sg 25.6 O.K.
=
1323. The W12 * 87 structural A36 steel column has a length of 12 ft. If its bottom end is fixed supported while its top is free, determine the largest axial load it can support. Use a factor of safety with respect to buckling of 1.75. A = 25.6 in2 Ix = 740 in4 Iy = 241 in4
P
W 12 * 87 K = 2.0
(controls)
12 ft
Pcr
p2(29)(103)(241) p2EI = = = 831.63 kip 2 (KL) (2.0(12)(12))2 Pcr 831.63 = = 475 ksi F.S 1.75 Ans.
P=
Check: scr = P 831.63 = = 32.5 ksi 6 sg A 25.6 O.K.
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*1324. An L2 tool steel link in a forging machine is pin connected to the forks at its ends as shown. Determine the maximum load P it can carry without buckling. Use a factor of safety with respect to buckling of F.S. = 1.75. Note from the figure on the left that the ends are pinned for buckling, whereas from the figure on the right the ends are fixed.
P
P
1.5 in. 24 in.
0.5 in.
Section Properties: A = 1.5(0.5) = 0.750 in2 Ix = 1 (0.5) A 1.53 B = 0.140625 in4 12 1 (1.5) A 0.53 B = 0.015625 in4 12
P
P
Iy =
Critical Buckling Load: With respect to the x  x axis, K = 1 (column with both ends pinned). Applying Eulers formula, Pcr = p2EI (KL)2 p2(29.0)(103)(0.140625) = [1(24)]2
= 69.88 kip With respect to the y  y axis, K = 0.5 (column with both ends fixed). Pcr = p2EI (KL)2 p2(29.0)(103)(0.015625) = [0.5(24)]2 (Controls!)
= 31.06 kip
Critical Stress: Eulers formula is only valid if scr 6 sg. scr = Factor of Safety: F.S = Pcr P 31.06 P Ans. Pcr 31.06 = = 41.41 ksi 6 sg = 102 ksi A 0.75 O.K.
1.75 =
P = 17.7 kip
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The W14 * 30 is used as a structural A36 steel column that can be assumed pinned at both of its ends. Determine the largest axial force P that can be applied without causing it to buckle.
1325.
P
From the table in appendix, the crosssectional area and the moment of inertia about weak axis (yaxis) for W14 * 30 are A = 8.85 in2 Iy = 19.6 in4
25 ft
Critical Buckling Load: Since the column is pinned at its base and top, K = 1. For A36 steel, E = 29.0(103) ksi and sg = 36 ksi. Here, the buckling occurs about the weak axis (yaxis). P = Pcr = p2EIy (KL)2 = p2 C 29.0(103) D (19.6)
C 1(25)(12) D 2
= 62.33 kip = 62.3 kip Eulers formula is valid only if scr 6 sg. scr = Pcr 62.33 = = 7.04 ksi 6 sg = 36 ksi A 8.85
Ans.
O.K.
1326. The A36 steel bar AB has a square cross section. If it is pin connected at its ends, determine the maximum allowable load P that can be applied to the frame. Use a factor of safety with respect to buckling of 2. a + MA = 0; FBC sin 30(10)  P(10) = 0 FBC = 2 P + : Fx = 0; FA  2P cos 30 = 0 FA = 1.732 P Buckling load: Pcr = FA(F.S.) = 1.732 P(2) = 3.464 P L = 10(12) = 120 in. I= 1 (1.5)(1.5)3 = 0.421875 in4 12 p2 EI (KL)2 p2 (29)(103)(0.421875) [(1.0)(120)]2
C
A
1.5 in.
30 B 10 ft P 1.5 in. 1.5 in.
Pcr =
3.464 P =
P = 2.42 kip Pcr = FA(F.S.) = 1.732(2.42)(2) = 8.38 kip Check: scr = Pcr 8.38 = = 3.72 ksi 6 sg A 1.5 (1.5)
Ans.
O.K.
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1327. Determine the maximum allowable intensity w of the distributed load that can be applied to member BC without causing member AB to buckle. Assume that AB is made of steel and is pinned at its ends for xx axis buckling and fixed at its ends for yy axis buckling. Use a factor of safety with respect to buckling of 3. Est = 200 GPa, sY = 360 MPa.
w
C
1.5 m
B 0.5 m
2m 30 mm x 20 mm y x 30 mm A
Moment of inertia: Ix = 1 (0.02)(0.033) = 45.0(10  9)m4 12 1 (0.03)(0.023) = 20(10  9) m4 12
y
Iy = x  x axis:
Pcr = FAB (F.S.) = 1.333w(3) = 4.0 w K = 1.0, Pcr = p2EI (KL)2 p2(200)(109)(45.0)(10  9) [(1.0)(2)]2 (controls) Ans. L = 2m
4.0w =
w = 5552 N> m = 5.55 kN> m y  y axis K = 0.5, 4.0w = L = 2m
p2 (200)(109)(20)(10  9) [(0.5)(2)]2
w = 9870 N> m = 9.87 kN> m Check: scr = 4(5552) Pcr = = 37.0 MPa 6 sg A (0.02)(0.03) O.K.
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*1328. Determine if the frame can support a load of w = 6 kN> m if the factor of safety with respect to buckling of member AB is 3. Assume that AB is made of steel and is pinned at its ends for xx axis buckling and fixed at its ends for yy axis buckling. Est = 200 GPa, sY = 360 MPa. Check x  x axis buckling: Ix = 1 (0.02)(0.03)3 = 45.0(10  9) m4 12 L = 2m
y x
w
C
1.5 m
B 0.5 m
2m 30 mm x 20 mm y 30 mm A
K = 1.0 Pcr
p2(200)(109)(45.0)(10  9) p2EI = = 2 (KL) ((1.0)(2))2
Pcr = 22.2 kN a + MC = 0; FAB(1.5)  6(2)(1) = 0 FAB = 8 kN Preqd = 8(3) = 24 kN 7 22.2 kN No, AB will fail. Ans.
The beam supports the load of P = 6 kip. As a result, the A36 steel member BC is subjected to a compressive load. Due to the forked ends on the member, consider the supports at B and C to act as pins for xx axis buckling and as fixed supports for yy axis buckling. Determine the factor of safety with respect to buckling about each of these axes.
1329.
P 4 ft 4 ft
A 3 ft 3 in. Cx
B
y
a + MA = 0;
3 FBC a b (4)  6000(8) = 0 5 FBC = 20 kip
1 in.
y
x
x  x axis buckling: Pcr =
1 p2(29)(103)(12)(1)(3)3 p2EI = = 178.9 kip (KL)2 (1.0(5)(12))2
F.S. =
178.9 = 8.94 20
Ans.
y  y axis buckling: Pcr =
1 p2 (29)(103)(12)(3)(1)3 p2EI = = 79.51 2 (KL) (0.5(5)(12))2
F.S. =
79.51 = 3.98 20
Ans.
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1330. Determine the greatest load P the frame will support without causing the A36 steel member BC to buckle. Due to the forked ends on the member, consider the supports at B and C to act as pins for xx axis buckling and as fixed supports for yy axis buckling.
3 ft
P 4 ft 4 ft
A 3 in. Cx
B
y
a + MA = 0;
3 FBC a b (4)  P(8) = 0 5 FBC = 3.33 P
1 in.
y
x  x axis buckling: Pcr =
1 p2(29)(103)(12)(1)(3)3 p2EI = = 178.9 kip (KL)2 (1.0(5)(12))2
x
y  y axis buckling: Pcr = Thus, 3.33 P = 79.51 P = 23.9 kip Ans.
1 p2(29)(103)(12)(3)(1)3 p2EI = = 79.51 kip (KL)2 (0.5(5)(12))2
1331. Determine the maximum distributed load that can be applied to the bar so that the A36 steel strut AB does not buckle. The strut has a diameter of 2 in. It is pin connected at its ends. The compressive force developed in member AB can be determined by writing the moment equation of equilibrium about C. a + MC = 0; FAB(2)  w(2)(3) = 0 I= FAB = 3w
w
C
2 ft
A
2 ft
4 ft
A = p(12) = p in2
p4 p (1 ) = in4 4 4
Since member AB is pinned at both ends, K = 1. For A36 steel, E = 29.0(103) ksi and sg = 36 ksi. Pcr = p EI ; (KL)2
2
3w =
p C 29.0(10 ) D (p> 4)
2 3
B
w = 32.52 kip> ft = 32.5 kip> ft The Eulers formula is valid only if scr 6 sg. scr = 3(32.52) Pcr = = 31.06 ksi 6 sg = 36 ksi p A
C 1(4)(12) D 2
Ans.
O.K.
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*1332. The members of the truss are assumed to be pin connected. If member AC is an A36 steel rod of 2 in. diameter, determine the maximum load P that can be supported by the truss without causing the member to buckle.
P B
C
4 ft
A 3 ft
D
Section the truss through a  a, the FBD of the top cut segment is shown in Fig. a. The compressive force developed in member AC can be determined directly by writing the force equation of equilibrium along x axis. + : Fx = 0; 3 FAC a b  P = 0 5 I= FAC = 5 P (C) 3
A = p(12) = p in2
and sg = 36 ksi. The length of member AC is LAC = 232 + 42 = 5 ft. Pcr = p2EI ; (KL)2 p2 C 29.0(103) D (p> 4) 5 P= 3 C 1(5)(12) D 2
p4 p (1 ) = in4 4 4
Since both ends of member AC are pinned, K = 1. For A36 steel, E = 29.0(103) ksi
P = 37.47 kip = 37.5 kip Eulers formula is valid only if scr 6 sg. 5 (37.47) Pcr 3 = = = 19.88 ksi 6 sg = 36 ksi p A
Ans.
scr
O.K.
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1333.
The steel bar AB of the frame is assumed to be pin connected at its ends for yy axis buckling. If w = 3 kN> m, determine the factor of safety with respect to buckling about the yy axis due to the applied loading. Est = 200 GPa, sY = 360 MPa.
6m w B C 3m 40 mm 40 mm y A 4m x 40 mm
The force with reference to the FBD shown in Fig. a. a + MC = 0; 3 3(6)(3)  FAB a b (6) = 0 5 Iy = FAB = 15 kN
The length of member AB is L = 232 + 42 = 5m. Here, buckling will occur about the weak axis, (yaxis). Since both ends of the member are pinned, Ky = 1. A = 0.04(0.08) = 3.2(10  3) m2 1 (0.08)(0.043) = 0.4267(10  6)m4 12 Pcr = p2EIy (KyLy)2 = p2 C 200(109) D C 0.4267(10  6) D
C 1.0(5) D 2
= 33.69 kN
Eulers formula is valid only if scr 6 sg. scr = 33.69(103) Pcr = 10.53(106)Pa = 10.53 MPa 6 sg = 360 MPa = A 3.2(10  3) O.K.
Thus, the factor of safety against buckling is F.S = Pcr 33.69 = = 2.25 FAB 15 Ans.
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1334. The members of the truss are assumed to be pin connected. If member AB is an A36 steel rod of 40 mm diameter, determine the maximum force P that can be supported by the truss without causing the member to buckle.
E
C
2m D
1.5 m B A 2m P
By inspecting the equilibrium of joint E, FAB = 0. Then, the compressive force developed in member AB can be determined by analysing the equilibrium of joint A, Fig. a. + c Fy = 0; + : Fx = 0; 3 FAC a b  P = 0 5 5 4 P a b  FAB = 0 3 5 I= FAC = 5 P (T) 3 4 P(c) 3
FAB =
A = p(0.022) = 0.4(10  3)p m2
p (0.024) = 40(10  9) p m4 4
Since both ends of member AB are pinned, K = 1. For A36 steel, E = 200 GPa and sg = 250 MPa. Pcr = p2EI ; (KL)2 p2 C 200(109) D C 40(10  9)p D 4 P= 3 C 1(2) D 2 P = 46.51(103) N = 46.5 kN
Ans.
The Eulers formula is valid only if scr 6 sg. 4 (46.51)(103) Pcr 3 = 49.35(106) Pa = 49.35 MPa 6 sg = 250 MPa O.K. = = A 0.4(10  3)p
scr
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1335. The members of the truss are assumed to be pin connected. If member CB is an A36 steel rod of 40 mm diameter, determine the maximum load P that can be supported by the truss without causing the member to buckle.
E
C
2m D
1.5 m B A 2m P
Section the truss through aa, the FBD of the left cut segment is shown in Fig. a. The compressive force developed in member CB can be obtained directly by writing the force equation of equilibrium along y axis. + c Fy = 0; FCB  P = 0 FCB = P (C) I= p (0.024) = 40(10  9)p m4 4
A = p(0.022) = 0.4(10  3)p m2
Since both ends of member CB are pinned, K = 1. For A36 steel, E = 200 GPa and sg = 250 MPa. Pcr = p2EI ; (KL)2 P= p2 C 200(109) D C 40(10  9)p D
= 110.24(103) N = 110 kN The Eulers formula is valid only if scr 6 sg. scr = 110.24(103) Pcr = 87.73(106) Pa = 87.73 MPa 6 sg = 250 MPa = A 0.4(10  3)p
C 1(1.5) D 2
Ans.
O.K.
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*1336. If load C has a mass of 500 kg, determine the required minimum diameter of the solid L2steel rod AB to the nearest mm so that it will not buckle. Use F.S. = 2 against buckling.
45
A
D
4m
Equilibriun. The compressive force developed in rod AB can be determined by analyzing the equilibrium of joint A, Fig. a. Fy = 0; FAB sin 15  500(9.81) cos 45 = 0 FAB = 13 400.71 N
B
60
C
Section Properties. The crosssectional area and moment of inertia of the solid rod are A= p2 d 4 I= p d4 p4 ab= d 42 64
Critical Buckling Load. Since the rod is pinned at both of its ends, K = 1. Here, Pcr = FAB (F.S.) = 13400.71(2) = 26801.42 N. Applying Eulers formula, p2EIy (KL)2 p2 C 200 A 109 B D c [1(4)]2 p4 dd 64
Pcr =
26801.42 =
d = 0.04587 m = 45.87 mm Use d = 46 mm Critical Stress. Eulers formula is valid only if scr 6 sY. scr = Pcr 26801.42 = = 16.13 MPa 6 sY = 703 MPa p A 2 A 0.046 B 4 O.K. Ans.
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1337. If the diameter of the solid L2steel rod AB is 50 mm, determine the maximum mass C that the rod can support without buckling. Use F.S. = 2 against buckling.
A
45
D
4m
Equilibrium. The compressive force developed in rod AB can be determined by analyzing the equilibrium of joint A, Fig. a. Fy = 0; FAB sin 15  m(9.81) cos 45 = 0 FAB = 26.8014m
B
60
C
Section Properties. The crosssectional area and moment of inertia of the rod are A= p A 0.052 B = 0.625 A 10  3 B pm2 4 p A 0.0254 B = 97.65625 A 10  9 B pm4 4
I=
Critical Buckling Load. Since the rod is pinned at both of its ends, K = 1. Here, Pcr = FAB (F.S.) = 26.8014m(2) = 53.6028m. Applying Eulers formula, Pcr = p2EIy (KL)2 p2 c 200 A 109 B d c 97.65625 A 10  9 B p d [1(4)]2 Ans.
53.6028m =
m = 706.11 kg = 7.06 kg Critical Stress. Eulers formula is valid only if scr 6 sY. scr = 53.6028(706.11) Pcr = = 19.28 MPa 6 sY = 703 MPa A p 0.625 A 10  3 B
O.K.
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1338. The members of the truss are assumed to be pin connected. If member GF is an A36 steel rod having a diameter of 2 in., determine the greatest magnitude of load P that can be supported by the truss without causing this member to buckle.
H
G
F
E
12 ft
A 16 ft
B 16 ft P P
C 16 ft
D
Support Reactions: As shown on FBD(a). Member Forces: Use the method of sections [FBD(b)]. + MB = 0; Section Properties: A= p2 A 2 B = p in2 4 p4 A 1 B = 0.250p in4 4 FGF (12)  P(16) = 0 FGF = 1.3333P (C)
I=
Critical Buckling Load: K = 1 for a column with both ends pinned. Applying Eulers formula, Pcr = FGF = p2EI (KLGF)2 p2 (29)(103)(0.250p) [1(16)(12)]2 Ans.
1.3333P =
P = 4.573 kip = 4.57 kip Critical Stress: Eulers formula is only valid if scr 6 sg. scr = 1.3333(4.573) Pcr = = 1.94 ksi 6 sg = 36 ksi p A
O.K.
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1339. The members of the truss are assumed to be pin connected. If member AG is an A36 steel rod having a diameter of 2 in., determine the greatest magnitude of load P that can be supported by the truss without causing this member to buckle.
H
G
F
E
12 ft
A 16 ft
B 16 ft P P
C 16 ft
D
Support Reactions: As shown on FBD(a). Member Forces: Use the method of joints [FBD(b)]. + c Fy = 0; Section Properties: P3 =0 F 5 AG FAG = 1.6667P (C)
LAG = 2162 + 122 = 20.0 ft p2 A 2 B = p in2 4 p4 A 1 B = 0.250p in4 4
A=
I=
Critical Buckling Load: K = 1 for a column with both ends pinned. Applying Eulers formula, Pcr = FGF = p2EI (KLGF)2 p2 (29)(103)(0.250p) [1(20)(12)]2 Ans.
1.6667P =
P = 2.342 kip = 2.34 kip Critical Stress: Eulers formula is only valid if scr = sg. scr = 1.6667(2.342) Pcr = = 1.24 ksi 6 sg = 36 ksi p A
O.K.
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*1340. The column is supported at B by a support that does not permit rotation but allows vertical deflection. Determine the critical load Pcr . EI is constant.
Pcr A
L
B
Elastic curve: EI d2y = M = P y dx2 P P x d + C2 cos c xd A EI A EI
P d2y + y=0 EI dx2 y = C1 sin c
Boundry conditions: At x = 0; 0 = 0 + C2; At x = L; y=0 C2 = 0 dv =0 dx P Z0 A EI n = 1, 3, 5
P P dv = C1 cos c L] d = 0; dx A EI A EI P L d = 0; A EI
cos c
P p L = na b A EI 2
C1
For n = 1 ; p2EI 4L2
p2 P = EI 4L2 Ans.
Pcr =
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The ideal column has a weight w (force length) and rests in the horizontal position when it is subjected to the axial load P. Determine the maximum moment in the column at midspan. EI is constant. Hint: Establish the differential equation for deflection, Eq. 131, with the origin at the mid span. The general solution is v = C1 sin kx + C2 cos kx + 1w>12P22x2  1wL>12P22x  1wEI> P22 where k2 = P> EI.
1341.
w P
L
Moment Functions: FBD(b). a + Mo = 0; M(x) = wL x wx a b  M(x)  a b x  Pv = 0 2 2 [1]
w2 A x  Lx B  Pv 2
Differential Equation of The Elastic Curve: EI d2y = M(x) dx2 d2y w2 = A x  Lx B  Py 2 dx2
EI
w d2y P y= + A x2  Lx B EI 2EI dx2 The solution of the above differential equation is of the form v = C1 sin a and P P w2 wL wEI x b + C2 cos xb+ xxA EI A EI 2P 2P P2
[2]
The integration constants can be determined from the boundary conditions. Boundary Condition: At x = 0, y = 0. From Eq. [2], 0 = C2 wEI P2 C2 = wEI P2
dv P P P P w wL = C1 cos x  C2 sin x + xdx A EI A EI A EI A EI P 2P
[3]
At x =
0 = C1
P wEI P wL PL PL wL cos sin + a bA EI A EI 2 A EI 2 P2 2P P2 A EI C1 = wEI PL tan A EI 2 P2
L dy = 0. From Eq.[3], , 2 dx
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1341.
Continued
Elastic Curve: y=
However, y = ymax at x =
w EI PL P EI P x2 L EI x + x + tan cos  xB sin R PP A EI 2 A EI P A EI 2 2 P wEI PL PL2  1R B sec A EI 2 8EI P2 w L2 L  L a b R  Pymax B 24 2 PL PL PL EI w EI EI L2 tan cos B sin + R PP A EI 2 A EI 2 P A EI 2 8 P L . From, Eq.[1], 2 L . Then, 2
ymax =
=
Maximum Moment: The maximum moment occurs at x =
Mmax =
=
=
wL2 wEI PL PL2  P b 2 B sec  1R r 8 A EI 2 8EI P PL wEI B sec  1R P A EI 2
Ans.
1342. The ideal column is subjected to the force F at its midpoint and the axial load P. Determine the maximum moment in the column at midspan. EI is constant. Hint: Establish the differential equation for deflection, Eq. 131. The general solution is v = C1 sin kx + C2 cos kx  c2x> k2, where c2 = F> 2EI, k2 = P> EI. Moment Functions: FBD(b). a + Mo = 0; M(x) + F x + P(v) = 0 2 M(x) = F x  Pv 2 [1]
F P
L 2
L 2
Differential Equation of The Elastic Curve: EI d2y = M(x) dx2 F d2y =  x  Py 2 2 dx
EI
d2y F P y=x + EI 2EI dx2 The solution of the above differential equation is of the form, v = C1 sin a P P F x b + C2 cos xb x A EI A EI 2P 1 067
[2]
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1342. and
Continued dv P P P P F = C1 cos x  C2 sin x dx A EI A EI A EI A EI 2P
[3]
The integration constants can be determined from the boundary conditions. Boundary Conditions: At x = 0, y = 0. From Eq.[2], C2 = 0 At x = L dy = 0. From Eq.[3], , 2 dx 0 = C1 PL P F cos A EI A EI 2 2P F EI PL sec 2P A P A EI 2
C1 = Elastic Curve: y=
=
F EI PL P F sec x x sin 2P A P A EI 2 A EI 2P F EI PL P sec x  xR B sin 2P A P A EI 2 A EI L . Then, 2 F EI PL L tan B  R 2P A P A EI 2 2 FL a b  Pymax 22 F EI PL PL L sec B sin  R 2P A P A EI 2 A EI 2 2 L . From Eq.[1], 2
However, y = ymax at x =
ymax =
=
Maximum Moment: The maximum moment occurs at x =
Mmax = 
=
=
FL F EI PL L  Pb tan B  Rr 4 2P A P A EI 2 2 PL F EI tan 2AP A EI 2
Ans.
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1343. The column with constant EI has the end constraints shown. Determine the critical load for the column.
P
L
Moment Function. Referring to the freebody diagram of the upper part of the deflected column, Fig. a, a + MO = 0; M + Pv = 0 M =  Pv
Differential Equation of the Elastic Curve. EI d2v =M dx2 d2v =  Pv dx2
EI
d2v P + v=0 2 EI dx The solution is in the form of v = C1 sin a P P x b + C2 cos xb A EI A EI
Boundary Conditions. At x = 0, v = 0. Then Eq. (1) gives 0 = 0 + C2
dv P P P P = C1 cos x  C2 sin x dx A EI A EI A EI A EI C2 = 0 dv = 0. Then Eq. (2) gives dx 0 = C1
(1)
(2)
At x = L,
C1 = 0 is the trivial solution, where v = 0. This means that the column will remain straight and buckling will not occur regardless of the load P. Another possible solution is P np L= A EI 2 p Pcr L= A EI 2 p2EI 4L2 cos P L = 0 A EI
P P cos L A EI A EI
n = 1, 3, 5
The smallest critical load occurs when n = 1, then
Pcr =
Ans.
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*1344. Consider an ideal column as in Fig. 1310c, having both ends fixed. Show that the critical load on the column is given by Pcr = 4p2EI> L2. Hint: Due to the vertical deflection of the top of the column, a constant moment M will be developed at the supports. Show that d2v> dx2 + 1P> EI2v = M > EI. The solution is of the form v = C1 sin1 1P> EIx2 + C2 cos1 1P> EIx2 + M > P. Moment Functions: M(x) = M  Py Differential Equation of The Elastic Curve: EI d2y = M(x) dx2
EI
d2y = M  Py dx2 (Q.E.D.)
M d2y P + y= 2 EI EI dx The solution of the above differential equation is of the form v = C1 sin a and P P M x b + C2 cos xb + A EI P A EI
[1]
The integration constants can be determined from the boundary conditions. Boundary Conditions: At x = 0, y = 0. From Eq.[1], C2 = At x = 0, M P
dv P P P P = C1 cos x  C2 sin x dx A EI A EI A EI A EI
[2]
dy = 0. From Eq.[2], C1 = 0 dx M P x R B 1  cos P A EI
Elastic Curve: y= and
However, due to symmetry sin B
The smallest critical load occurs when n = 1. Pce = 4p2EI L2 (Q.E.D.)
PL a bR = 0 A EI 2
M P P dy = sin x dx P A EI A EI or
L dy = 0 at x = . Then, dx 2
PL a b = np A EI 2
where n = 1, 2, 3,...
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1345. Consider an ideal column as in Fig. 1310d, having one end fixed and the other pinned. Show that the critical load on the column is given by Pcr = 20.19EI> L2. Hint: Due to the vertical deflection at the top of the column,a constant moment M will be developed at the fixed support and horizontal reactive forces R will be developed at both supports. Show that d2v> dx2 + 1P> EI2v = 1R > EI21L  x2. The solution is of the form v = C1 sin 1 1P> EIx2 + C2 cos 1 1P> EIx2 + 1R > P21L  x2. After application of the boundary conditions show that tan 1 1P> EIL2 = 1P> EI L. Solve by trial and error for the smallest nonzero root.
Equilibrium. FBD(a). Moment Functions: FBD(b). M(x) = R (L  x)  Py Differential Equation of The Elastic Curve: EI d2y = M(x) dx2 d2y = R (L  x)  Py dx2 (Q.E.D.)
EI
d2y P R + y= (L  x) 2 EI EI dx The solution of the above differential equation is of the form v = C1 sin a and P P R (L  x) x b + C2 cos xb + A EI P A EI
[1]
The integration constants can be determined from the boundary conditions. Boundary Conditions: At x = 0, y = 0. From Eq.[1], C2 = RL P
dv P P P P R = C1 cos x  C2 sin x dx A EI A EI A EI A EI P
[2]
At x = 0,
dy R EI = 0. From Eq.[2], C1 = dx PAP
Elastic Curve: y=
=
EI P P R sin x  L cos x + (Lx) R B P AP A EI A EI
R EI P RL P R sin x cos x + (L  x) PAP A EI P A EI P
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1345.
Continued P P EI sin L  L cos L A EI A EI AP tan P P L = L A EI A EI P L = 4.49341 A EI
However, y = 0 at x = L. Then, 0=
(Q.E.D.)
By trial and error and choosing the smallest root, we have
Then, Pcr = 20.19EI L2 (Q.E.D.)
1346. Determine the load P required to cause the A36 steel W8 * 15 column to fail either by buckling or by yielding. The column is fixed at its base and free at its top. Section properties for W8 * 15: A = 4.44 in2 rx = 3.29 in. Ix = 48.0 in4 d = 8.11 in. Iy = 3.41 in4
1 in.
P
Buckling about y  y axis: K = 2.0 P = Pcr = L = 8(12) = 96 in. p2EIy (KL)2 = p2(29)(103)(3.41) [(2.0)(96)]2 = 26.5 kip (controls) Ans.
8 ft
Check: scr =
Pcr 26.5 = = 5.96 ksi 6 sg A 4.44
O.K.
Check yielding about x  x axis: smax =
P ec KL P c 1 + 2 sec a bd A 2r A EA r
26.5 P = = 5.963 ksi A 4.44
2.0(96) P 26.5 KL = = 0.4184 2r A EA 2(3.29) A 29(103)(4.44) smax = 5.963[1 + 0.37463 sec (0.4184)] = 8.41 ksi 6 sg = 36 ksi O.K.
(1) A 8.11 B ec 2 = = 0.37463 r2 (3.29)2
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1347. The hollow red brass C83400 copper alloy shaft is fixed at one end but free at the other end. Determine the maximum eccentric force P the shaft can support without causing it to buckle or yield. Also, find the corresponding maximum deflection of the shaft.
2m a
P 150 mm 30 mm
a
Section Properties. A = p A 0.032  0.022 B = 0.5 A 10  3 B p m2 I= 0.1625 A 10 B p I = = 0.01803 m C 0.5 A 10  3 B p AA
6
20 mm Section a a
p A 0.034  0.024 B = 0.1625 A 10  6 B p m4 4
r=
e = 0.15 m
c = 0.03 m
For a column that is fixed at one end and free at the other, K = 2. Thus, KL = 2(2) = 4 m Yielding. In this case, yielding will occur before buckling. Applying the secant formula, smax = P ec KL P B 1 + 2 sec R A 2rx A EA rx 0.5 A 10 P
3
70.0 A 106 B = 70.0 A 106 B =
Bp
0.5 A 10  3 B p
P
D1 +
0.15(0.03) 0.018032
a 1 + 13.846 sec 8.8078 A 10  3 B 2P b
sec C
P 4 ST 2(0.01803) A 101 A 109 B C 0.5 A 10  3 B p D
Solving by trial and error, P = 5.8697 kN = 5.87 kN Maximum Deflection. vmax = e B sec P KL  1R A EI 2 Ans.
= 0.04210 m = 42.1 mm
= 0.15 D sec C
C 101 A 109 B C 0.1625 A 10  6 B p D 2
5.8697 A 103 B
4 a b S  1T
Ans.
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*1348. The hollow red brass C83400 copper alloy shaft is fixed at one end but free at the other end. If the eccentric force P = 5 kN is applied to the shaft as shown, determine the maximum normal stress and the maximum deflection.
2m a
P 150 mm 30 mm
a
Section Properties. A = p A 0.032  0.022 B = 0.5 A 10  3 B p m2 I= 0.1625 A 10  6 B p I = 0.01803 m = r= C 0.5 A 10  3 B p AA e = 0.15 m p A 0.034  0.024 B = 0.1625 A 10  6 B p m4 4
20 mm Section a a
c = 0.03 m
For a column that is fixed at one end and free at the other, K = 2. Thus, KL = 2(2) = 4 m Yielding. Applying the secant formula, smax = P ec KL P B 1 + 2 sec R A 2r A EA r 5 A 103 B 0.15(0.03)
5 A 103 B 4 ST sec C D1 + = 2(0.01803) C 101 A 109 B C 0.5 A 10  3 B p D 0.018032 0.5 A 10  3 B p = 57.44 MPa = 57.4 MPa Since smax 6 sY = 70 MPa, the shaft does not yield. Maximum Deflection. vmax = e B sec P KL  1R A EI 2
Ans.
= 0.03467 m = 34.7 mm
= 0.15 D sec C
C 101 A 109 B C 0.1625 A 10  6 B p D 2
5 A 103 B
4 a b S  1T
Ans.
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1349.
The tube is made of copper and has an outer diameter of 35 mm and a wall thickness of 7 mm. Using a factor of safety with respect to buckling and yielding of F.S. = 2.5, determine the allowable eccentric load P. The tube is pin supported at its ends. Ecu = 120 GPa, sY = 750 MPa.
P 14 mm
2m
P
Section Properties: A= p (0.0352  0.0212) = 0.61575(10  3) m2 4 I 64.1152(10  9) = = 0.010204 m AA A 0.61575(10  3) p (0.01754  0.01054) = 64.1152(10  9) m4 4
I=
r=
For a column pinned at both ends, K = 1. Then KL = 1(2) = 2 m. Buckling: Applying Eulers formula, Pmax = Pcr = p2 (120)(109) C 64.1152(10  9) D p2EI = = 18983.7 N = 18.98 kN (KL)2 22
Critical Stress: Eulers formula is only valid if scr 6 sg. scr = Pcr 18983.7 = 30.83 MPa 6 sg = 750 MPa = A 0.61575(10  3) (KL) Pmax Pmax ec B 1 + 2 sec R A 2r A EA r Pmax 0.61575(10  3) Pmax 0.61575(10  3) O.K.
Yielding: Applying the secant formula, smax =
750 A 106 B = 750 A 106 B =
B1 +
A 1 + 2.35294 sec 0.0114006 2Pmax B
0.0102042
0.014(0.0175)
sec
Pmax 2 R 2(0.010204) A 120(109)[0.61575(10  3)]
Solving by trial and error, Pmax = 16 885 N = 16.885 kN (Controls!) Factor of Safety: P= Pmax 16.885 = = 6.75 kN F.S. 2.5 Ans.
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1350. The tube is made of copper and has an outer diameter of 35 mm and a wall thickness of 7 mm. Using a factor of safety with respect to buckling and yielding of F.S. = 2.5, determine the allowable eccentric load P that it can support without failure. The tube is fixed supported at its ends. Ecu = 120 GPa, sY = 750 MPa.
P 14 mm
2m
P
Section Properties: A= p A 0.0352  0.0212 B = 0.61575 A 10  3 B m2 4 I 64.1152(10  9) = = 0.010204 ms AA A 0.61575(10  3) p A 0.01754  0.01054 B = 64.1152 A 10  9 B m4 4
I=
r=
For a column fixed at both ends, K = 0.5. Then KL = 0.5(2) = 1 m. Buckling: Applying Eulers formula, Pmax = Pcr = p2(120)(109) C 64.1152(10  9) D p2EI = = 75 935.0 N = 75.93 kN (KL)2 12
Critical Stress: Eulers formula is only valid if scr 6 sg. scr = Pcr 75 935.0 = 123.3 MPa 6 sg = 750 MPa = A 0.61575(10  3) (KL) Pmax Pmax ec B 1 + 2 sec R A 2r A EA r Pmax 0.61575(10  3) Pmax 0.61575(10  3) O. K.
Yielding: Applying the secant formula, smax =
750 A 106 B = 750 A 106 B =
B1 +
A 1 + 2.35294 sec 5.70032 A 10  3 B 2P B
0.0102042
0.014(0.0175)
sec
2 Pmax R 2(0.010204) A 120(109)[0.61575(10  3)]
Solving by trial and error, Pmax = 50 325 N = 50.325 kN (Controls!) Factor of Safety: P= Pmax 50.325 = = 20.1 kN F.S. 2.5 Ans.
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1351. The wood column is fixed at its base and can be assumed pin connected at its top. Determine the maximum eccentric load P that can be applied without causing the column to buckle or yield. Ew = 1.811032 ksi, sY = 8 ksi.
P
y 4 in. x P y 10 in. x
10 ft
Section Properties: A = 10(4) = 40 in2 Ix 333.33 = = 2.8868 in. AA A 40 Iy = 1 (4)(103) = 333.33 in4 12 Ix = 1 (10)(43) = 53.33 in4 12
ry =
Buckling about x  x axis: P = Pcr = p2(1.8)(103)(53.33) p2EI = = 134 kip (KL)2 [(0.7)(10)(12)]2 Pcr 134 = = 3.36 ksi 6 sg A 40 O.K.
Check: scr =
Yielding about y  y axis: smax =
P ec KL P a 1 + 2 sec a b b A 2r A EA r
5(5) ec = = 3.0 r2 2.88682 a
8(40) = P[1 + 3.0 sec (0.054221 2P)] By trial and error: P = 73.5 kip (controls)
0.7(10)(12) P P KL b = = 0.054221 2P 2r A EA 2(2.8868) A 1.8(103)(40)
Ans.
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*1352. The wood column is fixed at its base and can be assumed fixed connected at its top. Determine the maximum eccentric load P that can be applied without causing the column to buckle or yield. Ew = 1.811032 ksi, sY = 8 ksi.
P
y 4 in. x P y 10 in. x
10 ft
Section Properties: A = 10(4) = 40 in2 Iy 333.33 = = 2.8868 in. AA A 40 Iy = 1 (4)(103) = 333.33 in4 12 Ix = 1 (10)(43) = 53.33 in4 12
ry =
Buckling about x  x axis: P = Pcr = p2(1.8)(103)(53.33) p2EI = = 263 kip 2 (KL) [(0.5)(10)(12)]2 Pcr 263 = = 6.58 ksi 6 sg A 40 O.K.
Check: scr =
Yielding about y  y axis: smax =
P ec KL P a 1 + 2 sec a bb A 2r A EA r
5(5) ec = = 3.0 r2 2.88682 a
8(40) = P[1 + 3.0 sec (0.038729 2P)] By trial and error: P = 76.5 kip (controls)
0.5(10)(12) P P KL b = = 0.038729 2P 2r A EA 2(2.8868) A 1.8(103)(40)
Ans.
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1353.
The W200 * 22 A36steel column is fixed at its base. Its top is constrained to rotate about the yy axis and free to move along the yy axis. Also, the column is braced along the xx axis at its midheight. Determine the allowable eccentric force P that can be applied without causing the column either to buckle or yield. Use F.S. = 2 against buckling and F.S. = 1.5 against yielding.
100 mm 5m y x
P y
x
Section Properties. From the table listed in the appendix, the necessary section properties for a W200 * 22 are A = 2860 mm2 = 2.86 A 10  3 B m2 Ix = 20.0 A 106 B mm4 = 20.0 A 10  6 B m4 e = 0.1m Buckling About the Strong Axis. Since the column is fixed at the base and free at the top, Kx = 2. Applying Eulers formula, p2EIx (KL)x 2 p2 c 200 A 109 B d c 20.0 A 10  6 B d [2(10)]2 ry = 22.3 mm = 0.0223 m c= bf 2 = 102 = 51 mm = 0.051 m 2
5m
Pcr =
=
= 98.70kN
Eulers formula is valid if scr 6 sY. scr = Then, Pallow = Pcr 98.70 = = 49.35 kN F.S. 2
98.70 A 103 B Pcr = = 34.51 MPa 6 sY = 250MPa A 2.86 A 10  3 B
O.K.
Yielding About Weak Axis. Since the support provided by the bracing can be considered a pin connection, the upper portion of the column is pinned at both of its ends. Then Ky = 1 and L = 5 m. Applying the secant formula, smax =
250 A 106 B =
A KL B y Pmax Pmax ec C 1 + 2 sec B RS A 2ry A EA ry
2.86 A 10 Pmax
3
B
250 A 106 B =
2.86 A 10  3 B
Pmax
D1 +
0.1(0.051) 0.02232
c 1 + 10.2556 sec 4.6875 A 10  3 B 2Pmax d
sec C
1(5) Pmax ST 2(0.0223) A 200 A 109 B C 2.86 A 10  3 B D
Solving by trial and error, Pmax = 39.376 kN Then, Pallow = Pmax 39.376 = = 26.3 kN (controls) 1.5 1.5 Ans.
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1354. The W200 * 22 A36steel column is fixed at its base. Its top is constrained to rotate about the yy axis and free to move along the yy axis. Also, the column is braced along the xx axis at its midheight. If P = 25 kN, determine the maximum normal stress developed in the column.
100 mm 5m y x
P y
x
5m
Section Properties. From the table listed in the appendix, necessary section properties for a W200 * 22 are A = 2860 mm2 = 2.86 A 10  3 B m2 Ix = 20.0 A 106 B mm4 = 20.0 A 10  6 B m4 e = 0.1m Buckling About the Strong Axis. Since the column is fixed at the base and free at the top, Kx = 2. Applying Eulers formula, p2EIx (KL)x 2 p2 c 200 A 109 B d c 20.0 A 10  6 B d [2(10)]2 ry = 22.3 mm = 0.0223 m c= bf 2 = 102 = 51 mm = 0.051 m 2
Pcr =
=
= 98.70kN
Eulers formula is valid only if scr 6 sY. scr =
98.70 A 103 B Pcr = = 34.51 MPa 6 sY = 250 MPa A 2.86 A 10  3 B
O.K.
Since P = 25 kN 6 Pcr, the column does not buckle. Yielding About Weak Axis. Since the support provided by the bracing can be considered a pin connection, the upper portion of the column is pinned at both of its ends. Then Ky = 1 and L = 5 m. Applying the secant formula, smax = (KL) P P ec C 1 + 2 sec B RS A 2ry A EA ry 2.86 A 10  3 B 2.5 A 103 B D1 + 0.1(0.051) 0.02232
=
= 130.26 MPa = 130 MPa
sec C
25 A 103 B 1(5) ST 2(0.0223) C 200 A 109 B C 2.86 A 10  3 B D
Ans.
Since smax 6 sY = 250 MPa, the column does not yield.
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1355. The wood column is fixed at its base, and its top can be considered pinned. If the eccentric force P = 10 kN is applied to the column, investigate whether the column is adequate to support this loading without buckling or yielding. Take E = 10 GPa and sY = 15 MPa.
P 150 mm x y
x 25 mm 25 mm 75 mm 75 mm
5m
Section Properties. A = 0.05(0.15) = 7.5 A 10  3 B m2 Ix = 14.0625 A 10  6 B Ix = 0.04330 m = AA C 7.5 A 10  3 B 1 (0.05) A 0.153 B = 14.0625 A 10  6 B m4 12
rx =
Iy =
1 (0.15) A 0.053 B = 1.5625 A 10  6 B m4 12 e = 0.15 m c = 0.075 m
For a column that is fixed at one end and pinned at the other K = 0.7. Then, (KL)x = (KL)y = 0.7(5) = 3.5 m Buckling About the Weak Axis. Applying Eulers formula, Pcr = p2EIy (KL)y 2 = p2 C 10 A 109 B D C 1.5625 A 10  6 B D 3.52 = 12.59 kN
Eulers formula is valid if scr 6 sY. scr =
12.59 A 103 B Pcr = = 1.68 MPa 6 sY = 15 MPa A 7.5 A 10  3 B (KL)x P ec P C 1 + 2 sec B RS A 2rx A EA rx 7.5 A 10  3 B 10 A 103 B D1 + 0.15(0.075) 0.04330
2
O.K.
Since Pcr 7 P = 10 kN, the column will not buckle. Yielding About Strong Axis. Applying the secant formula. smax =
=
= 10.29 MPa
sec C
10 A 103 B 3.5 ST 2(0.04330) C 10 A 109 B C 7.5 A 10  3 B D Ans.
Since smax 6 sY = 15 MPa , the column will not yield.
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*1356. The wood column is fixed at its base, and its top can be considered pinned. Determine the maximum eccentric force P the column can support without causing it to either buckle or yield. Take E = 10 GPa and sY = 15 MPa .
P 150 mm x y
x 25 mm 25 mm 75 mm 75 mm
5m
Section Properties. A = 0.05(0.15) = 7.5 A 10  3 B m2 Ix = 14.0625 A 10 Ix = C 7.5 A 10  3 B AA 1 (0.05) A 0.153 B = 14.0625 A 10  6 B m4 12
6
rx =
B
= 0.04330 m
Iy =
1 (0.15) A 0.053 B = 1.5625 A 10  6 B m4 12 e = 0.15 m c = 0.075 m
For a column that is fixed at one end and pinned at the other K = 0.7. Then, (KL)x = (KL)y = 0.7(5) = 3.5 m Buckling About the Weak Axis. Applying Eulers formula, Pcr = p2EIy (KL)y
2
=
p2 C 10 A 109 B D C 1.5625 A 10  6 B D 3.52
= 12.59 kN = 12.6 kN
Ans.
Eulers formula is valid if scr 6 sY. scr
12.59 A 103 B Pcr = = = 1.68 MPa 6 sY = 15 MPa A 7.5 A 10  3 B (KL)x P P ec C B 1 + 2 sec B RS A 2rx A EA rx 12.59 A 103 B 7.5 A 10  3 B D1 + 0.15(0.075) 0.043302 sec C
O.K.
Yielding About Strong Axis. Applying the secant formula with P = Pcr = 12.59 kN, smax =
=
= 13.31 MPa 6 sY = 15 MPa
12.59 A 103 B B 3.5 ST 2(0.04330) C 10 A 109 B C 7.5 A 10  3 B D
O.K.
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1357.
The W250 * 28 A36steel column is fixed at its base. Its top is constrained to rotate about the yy axis and free to move along the yy axis. If e = 350 mm, determine the allowable eccentric force P that can be applied without causing the column either to buckle or yield. Use F.S. = 2 against buckling and F.S. = 1.5 against yielding.
Pe x
y
x y
6m
Section Properties. From the table listed in the appendix, necessary section properties for a W250 * 28 are A = 3620 mm2 = 3.62 A 10  3 B m2 Iy = 1.78 A 106 B mm4 = 1.78 A 10  6 B m4 e = 0.35 m Buckling About the Strong Axis. Since the column is fixed at the base and pinned at the top, Kx = 0.7. Applying Eulers formula, Pcr = p2EIy (KL)y 2 = p2 C 200 A 109 B D C 1.78 A 10  6 B D [0.7(6)]2 = 199.18 kN rx = 105 mm = 0.105 m c= 260 d = = 130 mm = 0.13 m 2 2
Eulers formula is valid only if scr 6 sY. scr = Thus, Pallow = Pcr 199.18 = = 99.59 kN F.S. 2
199.18 A 103 B Pcr = = 55.02 MPa 6 sY = 250 MPa A 3.62 A 10  3 B
O.K.
Yielding About Strong Axis. Since the column is fixed at its base and free at its top, Kx = 2. Applying the secant formula, smax = (KL)x Pmax Pmax ec C 1 + 2 sec B RS A 2rx A EA rx 3.62 A 10 Pmax
3
250 A 106 B =
B
250 A 106 B =
3.62 A 10  3 B
Pmax
A 1 + 4.1270 sec (0.0021237) 2Pmax B
D1 +
0.35(0.13) 0.1052
sec C
2(6) Pmax ST 2(0.105) A 200 A 109 B C 3.62 A 10  3 B D
Solving by trial and error, Pmax = 133.45 kN Then, Pallow = Pmax 133.45 = = 88.97 kN = 89.0 kN (controls) 1.5 1.5 Ans.
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1358. The W250 * 28 A36steel column is fixed at its base. Its top is constrained to rotate about the yy axis and free to move along the yy axis. Determine the force P and its eccentricity e so that the column will yield and buckle simultaneously.
Pe x
y
x y
6m
Section Properties. From the table listed in the appendix, the necessary section properties for a W250 * 28 are A = 3620 mm2 = 3.62 A 10  3 B m2 Iy = 1.78 A 106 B mm4 = 1.78 A 10  6 B m4 rx = 105 mm = 0.105 m c= d 260 = = 130 mm = 0.13 m 2 2
Buckling About the Weak Axis. Since the column is fixed at the base and pinned at its top, Kx = 0.7. Applying Eulers formula, Pcr = p2EIy (KL)y 2 = p2 C 200 A 109 B D C 1.78 A 10  6 B D [0.7(6)]2 = 199.18 kN = 199 kN Ans.
Eulers formula is valid only if scr 6 sY. scr =
199.18 A 103 B Pcr = = 55.02 MPa 6 sY = 250 MPa A 3.62 A 10  3 B (KL)x P P ec C 1 + 2 sec B RS A 2rx A EA rx 199.18 A 103 B 3.62 A 10  3 B D1 + e(0.13) 0.1052
O.K.
Yielding About Strong Axis. Since the column is fixed at its base and free at its top, Kx = 2. Applying the secant formula with P = Pcr = 199.18 kN, smax =
250 A 106 B =
e = 0.1753 m = 175 mm
sec C
199.18 A 103 B 2(6) ST 2(0.105) C 200 A 109 B C 3.62 A 10  3 B D
Ans.
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1359. The steel column supports the two eccentric loadings. If it is assumed to be pinned at its top, fixed at the bottom, and fully braced against buckling about the yy axis, determine the maximum deflection of the column and the maximum stress in the column. Est = 200 GPa, sY = 360 MPa.
130 kN 50 kN 80 mm 120 mm
100 mm 10 mm 6m y x x 10 mm 100 mm 10 mm y
Section Properties: A = 0.12(0.1)  (0.1)(0.09) = 3.00 A 10  3 B m2 Ix = Ix 6.90(10  6) = = 0.047958 m AA A 3.00(10  3) 1 1 (0.1) A 0.123 B (0.09) A 0.13 B = 6.90 A 10  6 B m4 12 12
rx =
For a column fixed at one end and pinned at the other end, K = 0.7. (KL)x = 0.7(6) = 4.2 m The eccentricity of the two applied loads is, e= 130(0.12)  50(0.08) = 0.06444 m 180
Yielding About xx Axis: Applying the secant formula, smax = 180(103) 4.2 R 2(0.047958) A 200(109)(3.00)(10  3)
=
(KL)x P P ec B 1 + 2 sec R A 2rx A EA rx 180(103) 3.00(10 )
3
= 199 MPa
Since smax 6 sg = 360 MPa, the column does not yield. Maximum Displacement: ymax = e B sec P KL  1R A EI 2
B1 +
0.06444(0.06) 0.047958
2
sec
Ans.
= 0.06444 B sec
= 0.02433 m = 24.3 mm
4.2 180(103) a b  1R A 200(109)[6.90(10  6)] 2
Ans.
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*1360. The steel column supports the two eccentric loadings. If it is assumed to be fixed at its top and bottom, and braced against buckling about the yy axis, determine the maximum deflection of the column and the maximum stress in the column. Est = 200 GPa, sY = 360 MPa.
130 kN 50 kN 80 mm 120 mm
100 mm 10 mm 6m y x x 10 mm 100 mm 10 mm y
Section Properties: A = 0.12(0.1)  (0.1)(0.09) = 3.00 A 10  3 B m2 Ix = 6.90(10  6) Ix = = 0.047958 m AA A 3.00(10  3) 1 1 (0.1) A 0.123 B (0.09) A 0.013 B = 6.90 A 10  6 B m4 12 12
rx =
For a column fixed at both ends, K = 0.5. (KL)x = 0.5(6) = 3.00 m The eccentricity of the two applied loads is, e= 130(0.12)  50(0.08) = 0.06444 m 180
Yielding About xx Axis: Applying the secant formula, smax = 180(103) 3.00 R 2(0.047958) A 200(109)(3.00)(10  3)
=
(KL)x P P ec B 1 + 2 sec R A 2rx A EA rx 180(103) 3.00(10 )
3
= 178 MPa
Since smax 6 sg = 360 MPa, the column does not yield. Maximum Displacement: ymax = e B sec P KL  1R A EI 2
B1 +
0.06444(0.06) 0.047958
2
sec
Ans.
= 0.06444 B sec
= 0.01077 m = 10.8 mm
3 180(103) a b  1R 9 6 A 200(10 )[6.90(10 )] 2
Ans.
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1361. The W250 * 45 A36steel column is pinned at its top and fixed at its base. Also, the column is braced along its weak axis at midheight. If P = 250 kN, investigate whether the column is adequate to support this loading. Use F.S. = 2 against buckling and F.S. = 1.5 against yielding.
P 4 250 mm
P 250 mm
4m
Section Properties. From the table listed in the appendix, the necessary section properties for a W250 * 45 are A = 5700 mm = 5.70 A 10
2 3
Bm
4m
2
rx = 112 mm = 0.112 m c= 266 d = = 133 mm = 0.133 m 2 2 250 = 312.5 kN is 4
Iy = 7.03 A 106 B mm4 = 7.03 A 10  6 B m4
The eccentricity of the equivalent force P = 250 +
250(0.25) e=
250 (0.25) 4 = 0.15 m 250 250 + 4
Buckling About the Weak Axis. The column is braced along the weak axis at midheight and the support provided by the bracing can be considered as a pin. The top portion of the column is critical is since the top is pinned so Ky = 1 and L = 4 m Applying Eulers formula, Pcr = p2EIy (KL)y 2 = p2 C 200 A 109 B D C 7.03 A 10  6 B D [1(4)]2 = 867.29 kN
Eulers equation is valid only if scr 6 sY. scr = Then,
Pallow =
867.29 A 103 B Pcr = = 152.16 MPa 6 sY = 250 MPa A 5.70 A 10  3 B
O.K.
Pcr 867.29 = = 433.65 kN F.S. 2
Since Pallow 7 P , the column does not buckle.
Yielding About Strong Axis. Since the column is fixed at its base and pinned at its top, Kx = 0.7 and L = 8 m. Applying the secant formula with Pmax = P (F.S.) = 312.5(1.5) = 468.75 kN smax =
(KL)x Pmax Pmax ec C 1 + 2 sec B RS A 2rx A EA rx
=
468.75 A 103 B 5.70 A 10  3 B
= 231.84 MPa
C1 +
0.15(0.133) 0.112
2
Since smax 6 sY = 250 MPa, the column does not yield.
sec B
468.75 A 103 B 0.7(8) RS 2(0.112) C 200 A 109 B C 5.70 A 10  3 B D
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1362.
The W250 * 45 A36steel column is pinned at its top and fixed at its base. Also, the column is braced along its weak axis at midheight. Determine the allowable force P that the column can support without causing it either to buckle or yield. Use F.S. = 2 against buckling and F.S. = 1.5 against yielding.
P 4 250 mm
P 250 mm
4m
Section Properties. From the table listed in the appendix, the necessary section properties for a W250 * 45 are A = 5700 mm2 = 5.70 A 10  3 B m2 Iy = 7.03 A 106 B mm4 = 7.03 A 10  6 B m4 rx = 112 mm = 0.112 m c= d 266 = = 133 mm = 0.133 m 2 2 P = 1.25P is 4
4m
The eccentricity of the equivalent force P = P + P(0.25) e=
P (0.25) 4 = 0.15 m P P+ 4
Buckling About the Weak Axis. The column is braced along the weak axis at midheight and the support provided by the bracing can be considered as a pin. The top portion of the column is critical is since the top is pinned so Ky = 1 and L = 4 m. Applying Eulers formula, Pcr = p2EIy (KL)y
2
=
p2 C 200 A 109 B D C 7.03 A 10  6 B D [1(4)]2
= 867.29 kN
Eulers equation is valid only if scr 6 sY. scr Then,
Pallow =
867.29 A 103 B Pcr = = = 152.16 MPa 6 sY = 250 MPa A 5.70 A 10  3 B Pcr F.S.
O.K.
1.25Pallow = Pallow
867.29 2 = 346.92 kN
Yielding About Strong Axis. Since the column is fixed at its base and pinned at its top, Kx = 0.7 and L = 8 m. Applying the secant formula, smax =
(KL)x Pmax Pmax ec C 1 + 2 sec B RS A 2rx A EA rx
250 A 106 B = 250 A 106 B =
5.70 A 10
1.25Pmax
3
B
Solving by trial and error, Pmax = 401.75 kN Then, Pallow = 401.75 = 267.83 kN = 268 kN (controls) 1.5 Ans.
5.70 A 10  3 B
1.25Pmax
A 1 + 1.5904 sec (0.00082783) 2Pmax B
0.1122
C1 +
0.15(0.133)
sec B
0.7(8) 1.25Pmax RS 2(0.112) A 200 A 109 B C 5.70 A 10  3 B D
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1363. The W14 * 26 structural A36 steel member is used as a 20ftlong column that is assumed to be fixed at its top and fixed at its bottom. If the 15kip load is applied at an eccentric distance of 10 in., determine the maximum stress in the column.
15 kip
10 in.
Section Properties for W 14 * 26 A = 7.69 in2 d = 13.91 in. rx = 5.65 in.
20 ft
Yielding about x  x axis: smax =
P ec KL P b d; c 1 + 2 sec a A 2 r AE A r
K = 0.5
15 P = = 1.9506 ksi; A 7.69
0.5 (20)(12) 15 KL P = = 0.087094 2 r A EA 2(5.65) A 29 (103)(7.69)
10 A 13.91 B ec 2 = = 2.178714 r2 (5.65)2
smax = 1.9506[1 + 2.178714 sec (0.087094)] = 6.22 ksi 6 sg = 36 ksi O.K. Ans.
*1364. The W14 * 26 structural A36 steel member is used as a column that is assumed to be fixed at its top and pinned at its bottom. If the 15kip load is applied at an eccentric distance of 10 in., determine the maximum stress in the column.
15 kip
10 in.
Section Properties for W 14 * 26 A = 7.69 in2 d = 13.91 in. rx = 5.65 in.
20 ft
Yielding about x  x axis: smax =
P ec KL P c 1 + 2 sec a b d; A 2 r AE A r
K = 0.7
15 P = = 1.9506 ksi ; A 7.69
0.7 (20)(12) 15 KL P = = 0.121931 2 r A EA 2(5.65) A 29 (103)(7.69)
10 A 13.91 B ec 2 = = 2.178714 r2 (5.65)2
smax = 1.9506[1 + 2.178714 sec (0.121931)] = 6.24 ksi 6 sg = 36 ksi O.K. Ans.
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1365. Determine the maximum eccentric load P the 2014T6aluminumalloy strut can support without causing it either to buckle or yield. The ends of the strut are pinconnected.
P 150 mm
100 mm
a a 3m 50 mm
P 150 mm
100 mm Section a a
Section Properties. The necessary section properties are A = 0.05(0.1) = 5 A 10  3 B m2 Iy = 4.1667 A 10 Ix = C 5 A 10  3 B AA 1 (0.1) A 0.053 B = 1.04167 A 10  6 B m4 12
6
rx =
B
= 0.02887 m
For a column that is pinned at both of its ends K = 1. Thus, (KL)x = (KL)y = 1(3) = 3 m Buckling About the Weak Axis. Applying Eulers formula, Pcr = p2EIy (KL)y 2 = p2 C 73.1 A 109 B D C 1.04167 A 10  6 B D 32 = 83.50 kN = 83.5 kN Ans.
Critical Stress: Eulers formula is valid only if scr 6 sY. scr =
83.50 A 103 B Pcr = = 16.70 MPa 6 sY = 414 MPa A 5 A 10  3 B (KL)x P P ec C 1 + 2 sec B RS A 2rx A EA rx D1 + 0.15(0.05) 0.028872 sec C
O.K.
Yielding About Strong Axis. Applying the secant formula, smax =
=
83.50 A 103 B 5 A 10  3 B
= 229.27 MPa 6 sY = 414 MPa
83.50 A 10 B 3 ST 2(0.02887) C 73.1 A 109 B C 5 A 10  3 B D
3
O.K.
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1366. The W8 * 48 structural A36 steel column is fixed at its bottom and free at its top. If it is subjected to the eccentric load of 75 kip, determine the factor of safety with respect to either the initiation of buckling or yielding.
75 kip
8 in. y x
y
12 ft
Section Properties: For a wide flange section W8 * 48, A = 14.1 in2 rx = 3.61 in. Iy = 60.9 in4 d = 8.50 in.
For a column fixed at one end and free at the other and, K = 2. (KL)y = (KL)x = 2(12)(12) = 288 in. Buckling About yy Axis: Applying Eulers formula, P = Pcr = p2EIy (KL)2 y p2 (29.0)(103)(60.9) = 2882
= 210.15 kip Critical Stress: Eulers formula is only valid if scr 6 sg. scr = Pcr 210.15 = = 14.90 ksi 6 sg = 36 ksi A 14.1 O. K.
Yielding About xx Axis: Applying the secant formula, smax = 8 A 1.50 B Pmax Pmax 288 2 sec B1 + R 14.1 2(3.61) A 29.0(103)(14.1) 3.612 Pmax = 117.0 kip (Controls!) Factor of Safety: F.S. = Pmax 117.0 = = 1.56 P 75 Ans.
36(14.1) = Pmax A 1 + 2.608943 sec 0.0623802 2Pmax B 36 = Solving by trial and error,
(KL)x Pmax Pmax ec B 1 + 2 sec R A 2rx A EA rx
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1367. The W8 * 48 structural A36 steel column is fixed at its bottom and pinned at its top. If it is subjected to the eccentric load of 75 kip, determine if the column fails by yielding. The column is braced so that it does not buckle about the yy axis.
75 kip
8 in. y x
y
12 ft
Section Properties: For a wide flange section W8 * 48, A = 14.1 in2 rx = 3.61 in. d = 8.50 in.
For a column fixed at one end and pinned at the other end, K = 0.7. (KL)x = 0.7(12)(12) = 100.8 in. Yielding About xx Axis: Applying the secant formula, smax = 8 A 1.50 B 75 100.8 75 2 sec B1 + R 2 14.1 2(3.61) A 29.0(103)(14.1) 3.61 (KL)x P P ec B 1 + 2 sec R A 2rx A EA rx
=
= 19.45 ksi 6 sg = 36 ksi
O.K. Ans.
Hence, the column does not fail by yielding.
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*1368. Determine the load P required to cause the steel W12 * 50 structural A36 steel column to fail either by buckling or by yielding. The column is fixed at its bottom and the cables at its top act as a pin to hold it.
2 in.
P
25 ft
Section Properties: For a wide flange section W12 * 50, A = 14.7 in2 rx = 5.18 in. Iy = 56.3 in4 d = 12.19 in.
For a column fixed at one end and pinned at the other end, K = 0.7. (KL)y = (KL)x = 0.7(25)(12) = 210 in. Buckling About yy Axis: Applying Eulers formula, P = Pcr = p2EIy (KL)2 y p2 (29.0)(103)(56.3) = 2102
= 365.40 kip Critical Stress: Eulers formula is only valid if scr 6 sg. scr = Pcr 365.4 = = 24.86 ksi 6 sg = 36 ksi A 14.7 O. K.
Yielding About xx Axis: Applying the secant formula, smax = 2 A 12.19 B P 210 P 2 sec B1 + R 14.7 2(5.18) A 29.0(103)(14.7) 5.182
36(14.7) = P A 1 + 0.454302 sec 0.0310457 2P B 36 = Solving by trial and error,
(KL)x P P ec B 1 + 2 sec R A 2rx A EA rx
Pmax = 343.3 kip = 343 kip (Controls!)
Ans.
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1369.
Solve Prob. 1368 if the column is an A36 steel W12 * 16 section.
2 in.
P
25 ft
Section Properties: For a wide flange section W12 * 16, A = 4.71 in2 rx = 4.67 in. Iy = 2.82 in4 d = 11.99 in.
For a column fixed at one end and pinned at the other end, K = 0.7. (KL)y = (KL)x = 0.7(25)(12) = 210 in. Buckling About yy Axis: Applying Eulers formula, P = Pcr = p 2EIy (KL)2 y p2 (29.0)(103)(2.82) = 2102 Ans.
= 18.30 kip = 18.3 kip Critical Stress: Eulers formula is only valid if scr 6 sg. scr = Pcr 18.30 = = 3.89 ksi 6 sg = 36 ksi A 4.71
Yielding About xx Axis: Applying the secant formula, smax = 2 A 11.99 B 18.30 210 18.30 2 sec a bR B1 + 4.71 2(4.67) A 29.0(103)(4.71) 4.672
=
(KL)x P P ec B 1 + 2 sec R A 2rx A EA rx
= 6.10 ksi 6 sg = 36 ksi
O.K.
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1370. A column of intermediate length buckles when the compressive stress is 40 ksi. If the slenderness ratio is 60, determine the tangent modulus. p2 Et a KL b = 60 r
scr =
40 =
p2 Et (60)2
A KrL B 2
;
Et = 14590 ksi = 14.6 (103) ksi
Ans
1371. The 6ftlong column has the cross section shown and is made of material which has a stressstrain diagram that can be approximated as shown. If the column is pinned at both ends, determine critical the load Pcr for the column.
s(ksi)
0.5 in. 55 0.5 in. 5 in.
25 3 in.
0.5 in.
Section Properties: The neccessary section properties are A = 2[0.5(3)] + 5(0.5) = 5.5 in2 I = 2B ry = 1 1 (0.5) A 33 B R + (5) A 0.53 B = 2.3021 in4 12 12
0.001 0.004 P (in./in.)
2.3021 Iy = = 0.6470 in. AA A 5.5
For the column pinned at both of its ends, K = 1. Thus, 1(6)(12) KL = 111.29 = ry 0.6470 Critical Stress. Applying Engessers equation, p2Et
a KL
scr =
=
p2Et 111.292
r
b
= 0.7969 A 10  3 B Et
(1)
From the stress  strain diagram, the tangent moduli are (Et)1 = 25 ksi = 25 A 103 B ksi 0.001 (55  25) ksi = 10 A 103 B (ksi) 0.004  0.001 0 s 6 25 ksi
(Et)2 =
25ksi 6 s 40 ksi
Substituting (Et)1 = 25 A 103 B into Eq. (1),
scr = 0.7969 A 10  3 B c 25 A 103 B d = 19.92 ksi Ans. 1 095
Since scr 6 sY = 25 ksi, elastic buckling occurs. Thus, Pcr = scr A = 19.92(5.5) = 109.57 kip = 110 kip
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*1372. The 6ftlong column has the cross section shown and is made of material which has a stressstrain diagram that can be approximated as shown. If the column is fixed at both ends, determine the critical load Pcr for the column.
s(ksi)
0.5 in. 55 0.5 in. 5 in.
25 3 in.
0.5 in.
Section Properties. The neccessary section properties are A = 2[0.5(3)] + 5(0.5) = 5.5 in2 I = 2B ry = 1 1 (0.5) A 33 B R + (5) A 0.53 B = 2.3021 in4 12 12
0.001 0.004 P (in./in.)
Iy 2.3021 = = 0.6470 in. AA A 5.5
For the column fixed at its ends, K = 0.5. Thus, 0.5(6)(12) KL = 55.64 = ry 0.6470 Critical Stress. Applying Engessers equation, From the stress  strain diagram, the tangent moduli are (Et)1 = 25 ksi = 25 A 103 B ksi 0.001 (55  25)ksi = 10 A 103 B ksi 0.004  0.001 0 s 6 25 ksi
(Et)2 =
25 ksi 6 s 40 ksi
Substituting (Et)1 = 25 A 103 B ksi into Eq. (1),
scr = 3.1875 A 10  3 B c 25 A 103 B d = 79.69 ksi
Since scr 7 sY = 25 ksi, the inelastic buckling occurs. Substituting (Et)2 into Eq. (1), scr = 3.1875 A 10  3 B c 10 A 103 B d = 31.88 ksi Since 25 ksi 6 scr 6 55 ksi, this result can be used to calculate the critical load. Pcr = scr A = 31.88(5.5) = 175.31 kip = 175 kip Ans.
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1373.
The stressstrain diagram of the material of a column can be approximated as shown. Plot P A vs. KL r for the column.
s (MPa)
350
200
Tangent Moduli. From the stress  strain diagram, (Et)1 = 200 A 106 B 0.001 = 200 GPa 0 s 6 200MPa
P (in./in.) 0.001 0.004
(Et)2 =
(350  200) A 106 B 0.004  0.001
0
= 50 GPa
200MPa 6 s 350 MPa
Critical Stress. Applying Engessers equation, scr P = A p2Et KL 2 a b r (1)
If Et = (Et)1 = 200 GPa, Eq. (1) becomes p2 C 200 A 109 B D 1.974 A 106 B P = = MPa A KL 2 KL 2 a b a b r r
when scr =
P = sY = 200 MPa, this equation becomes A p2 C 200 A 109 B D a KL 2 b r
200 A 106 B =
KL = 99.346 = 99.3 r
If Et = (Et)2 = 50 GPa, Eq. (1) becomes MPa KL 2 KL 2 r r P when scr = = sY = 200 MPa, this A equation gives P = A = 200 A 106 B = p2 C 50 A 109 B D a KL 2 b r p2 c 50 A 109 B d 0.4935 A 106 B
KL = 49.67 = 49.7 r Using these results, the graphs of P KL vs. is shown in Fig. a can be plotted. r A 1 097
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1374. Construct the buckling curve, P A versus L r, for a column that has a bilinear stressstrain curve in compression as shown. The column is pinned at its ends.
s (MPa) 260
140
0.001
0.004
P (mm/mm)
Tangent modulus: From the stressstrain diagram, (Et)1 = 140(106) = 140 GPa 0.001 (260  140)(106) = 40 GPa 0.004  0.001
(Et)2 =
Critical Stress: Applying Engessers equation, scr p2Et P = A L2 ab r [1]
Substituting (Et)1 = 140 GPa into Eq. [1], we have p2 C 140(109) D P = A ALB2
r
P = A When L P = 140 MPa, = 99.3 r A
1.38(106)
ALB2 r
MPa
Substitute (Et)2 = 40 GPa into Eq. [1], we have p2 C 40(109) D P = A ALB2
r
P = A When L P = 140 MPa, = 53.1 r A
0.395(106)
ALB2 r
MPa
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1375. The stressstrain diagram for a material can be approximated by the two line segments shown. If a bar having a diameter of 80 mm and a length of 1.5 m is made from this material, determine the critical load provided the ends are pinned. Assume that the load acts through the axis of the bar. Use Engessers equation.
s (MPa) 1100
200 0.001 0.007 P (mm/mm)
E1 =
200 (106) = 200 GPa 0.001 1100 (106)  200 (106) = 150 GPa 0.007  0.001
E2 = Section properties: I=
r= Engessers equation:
p4 I 0.04 c 4c = == = 0.02 m AA C p c2 2 2
p4 c; 4
A = pc2
1.0(1.5) KL = = 75 r 0.02 scr =
A
p2 Et
KL 2 r
B
=
p2 Et (75)2
= 1.7546(10  3) Et
Assume Et = E1 = 200 GPa scr = 1.7546 (10  3)(200)(109) = 351 MPa 7 200 MPa Therefore, inelastic buckling occurs: Assume Et = E2 = 150 GPa scr = 1.7546 (10  3)(150)(109) = 263.2 MPa 200 MPa 6 scr 6 1100 MPa Critical load: Pcr = scr A = 263.2 (106)(p)(0.042) = 1323 kN Ans O.K.
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*1376. The stressstrain diagram for a material can be approximated by the two line segments shown. If a bar having a diameter of 80 mm and a length of 1.5 m is made from this material, determine the critical load provided the ends are fixed. Assume that the load acts through the axis of the bar. Use Engessers equation.
s (MPa) 1100
200 0.001 0.007 P (mm/mm)
E1 =
200 (106) = 200 GPa 0.001 1100 (106)  200 (106) = 150 GPa 0.007  0.001
E2 =
Section properties: I=
p4 I 0.04 c 4c = == = 0.02 m AA C p c2 2 2
p4 c; 4
A = pc2
r=
Engessers equation: 0.5 (1.5) KL = = 37.5 r 0.02 scr =
A
p2 Et
KL 2 r
B
=
p2 Et (37.5)2
= 7.018385(10  3) Et
Assume Et = E1 = 200 GPa scr = 7.018385 (10  3)(200)(109) = 1403.7 MPa 7 200 MPa Assume Et = E2 = 150 GPa scr = 7.018385 (10  3)(150)(109) = 1052.8 MPa 200 MPa 6 scr 6 1100 MPa Critical load: Pcr = scr A = 1052.8 (106)(p)(0.042) = 5292 kN Ans. O.K. NG
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1377.
The stressstrain diagram for a material can be approximated by the two line segments shown. If a bar having a diameter of 80 mm and length of 1.5 m is made from this material, determine the critical load provided one end is pinned and the other is fixed. Assume that the load acts through the axis of the bar. Use Engessers equation.
s (MPa) 1100
200 0.001 0.007 P (mm/mm)
E1 =
200 (106) = 200 GPa 0.001 1100 (106)  200 (106) = 150 GPa 0.007  0.001
E2 =
Section properties: I=
p4 I 0.04 c 4c = == = 0.02 m AA C pc2 2 2
p4 c; 4
A = pc2
r=
Engessers equation: 0.7 (1.5) KL = = 52.5 r 0.02 scr =
A
p2 Et
KL 2 r
B
=
p2Et (52.5)2
= 3.58081 (10  3) Et
Assume Et = E1 = 200 GPa scr = 3.58081 (10  3)(200)(109) = 716.2 MPa 7 200 MPa Assume Et = E2 = 150 GPa scr = 3.58081 (10  3)(150)(109) = 537.1 MPa 200 MPa 6 scr 6 1100 MPa Critical load: Pcr = scr A = 537.1 (106)(p)(0.042) = 2700 kN Ans. O.K. NG
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1378. Determine the largest length of a structural A36 steel rod if it is fixed supported and subjected to an axial load of 100 kN. The rod has a diameter of 50 mm. Use the AISC equations.
Section Properties: A = p A 0.0252 B = 0.625 A 10  3 B p m2 I= 97.65625(10  9)p I = = 0.0125 m AA A 0.625(10  3)p p A 0.0254 B = 97.65625 A 10  9 B p m4 4
r=
Slenderness Ratio: For a column fixed at both ends, K = 0.5. Thus, 0.5L KL = = 40.0L r 0.0125 AISC Column Formula: Assume a long column. sallow = 100(103) 0.625(10  3)p = 23 A KL B 2 r 12p2E
12p2 C 200(109) D 23(40.0L)3
Here,
KL KL 2p2E = 40.0(3.555) = 142.2 and for A36 steel, a b= r r e A sg KL KL 2p2[200(109)] = = 125.7. Since a b 200, the assumption is correct. re r A 250(106) L = 3.555 m
Thus, L = 3.56 m Ans.
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1379. Determine the largest length of a W10 * 45 structural steel column if it is pin supported and subjected to an axial load of 290 kip. Est = 29(103) ksi, sY = 50 ksi. Use the AISC equations.
Section Properties: For a W10 * 45 wide flange section, A = 13.3 in2 ry = 2.01 in
Slenderness Ratio: For a column pinned at both ends, K = 1. Thus, a 1(L) KL b= = 0.49751L ry 2.01
AISC Column Formula: Assume a long column, sallow = 23 A KL B 2 r 12p2E
12p2 C 29(103) D 290 = 13.3 23(0.49751L)2 KL KL 2p2E = 0.49751 (166.3) = 82.76 and for grade 50 steel, a b= r r c A sg KL KL 2p2[29(103)] 6a b , the assumption is not correct. = = 107.0. Since r rc 50 A Thus, the column is an intermediate column. L = 166.3 in. Here, Applying Eq. 1323,
sallow =
R (50) 2(107.02) 290 = 13.3 3(0.49751L) (0.49751L)3 5 + 3 8(107.0) 8(107.03)
(0.49751L)2
3(KL> r) (KL> r)3 5 + 3 8(KL> r)c 8(KL> r)3 c
0 = 12.565658 A 10  9 B L3  24.788132 A 10  6 B L2  1.743638 A 10  3 B L + 0.626437
B1 
B1 
2(KL> r)2 c
(KL> r)2
R sg
Solving by trial and error, L = 131.12 in. = 10.9 ft Ans.
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*1380. Determine the largest length of a W10 * 12 structural A36 steel section if it is pin supported and is subjected to an axial load of 28 kip. Use the AISC equations. For a W 10 * 12, ry = 0.785 in. s= Assume a long column: KL 12p2(29)(103) 12p2E = = 137.4 b= r A 23(7.91) A 23sallow KL KL 7a b r rc sallow = 12p2E 23(KL> r)2 A = 3.54 in2
28 P = = 7.91 ksi A 3.54
a a
KL 2p2(29)(103) 2p2E = = 126.1, b= r c A sg A 36
Long column. KL = 137.4 r
L = 137.4 = 8.99 ft
1381.
r 0.785 = 137.4 = 107.86 in. K 1
Ans.
Using the AISC equations, select from Appendix B the lightestweight structural A36 steel column that is 14 ft long and supports an axial load of 40 kip.The ends are pinned. Take sY = 50 ksi. KL 2p2(29)(103) 2p2E b= = = 107 r c A sY 50 A a ry = 1.46 in.)
Try, W6 * 15 (A = 4.43 in2 a a
(1.0)(14)(12) KL b= = 115.1, ry 1.46
KL KL b7a b ry rc
Long column sallow = 12p2(29)(103) 12 p2E = = 11.28 ksi 2 23(KL> r) 23(115.1)2
Pallow = sallowA = 11.28(4.43) = 50.0 kip 7 40 kip Use W6 * 15 O.K. Ans.
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1382. Using the AISC equations, select from Appendix B the lightestweight structural A36 steel column that is 12 ft long and supports an axial load of 40 kip. The ends are fixed. Take sY = 50 ksi.
Try W6 * 9 a
KL 2p2(29)(103) 2p2E b= = = 107 r c A sY A 50 A = 2.68 in2
ry = 0.905 in.
0.5(12)(12) KL = = 79.56 ry 0.905 KL KL 6a b ry rc Intermediate column sallow =
KL C 1  1 A (KL>>rr)c B 2 D sg 2
KL KL C 5 + 3 A (KL>>rr)c B  1 A (KL>>rr)c B 3 D 3 8 8
=
C 5 + 3 A 79.56 B  1 A 79.56 B 3 D 3 8 126.1 8 12.61
C 1  1 A 79.56 B 2 D 36 ksi 2 126.1
= 15.40 ksi
Pallow = sallowA = 15.40(2.68) = 41.3 kip 7 40 kip Use W6 * 9 O.K. Ans.
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1383. Using the AISC equations, select from Appendix B the lightestweight structural A36 steel column that is 24 ft long and supports an axial load of 100 kip.The ends are fixed.
Section Properties: Try a W8 * 24 wide flange section, A = 7.08 in2 ry = 1.61 in
Slenderness Ratio: For a column fixed at both ends, K = 0.5. Thus, a AISC = Column 0.5(24)(12) KL b= = 89.44 ry 1.61 Formula: For A36 steel, a KL 2p2E b= r c A sg
KL KL 2p2[29(103)] 6a b , the column is an intermediate = 126.1. Since r rc A 36 column. Applying Eq. 1323,
sallow =
3(KL> r) (KL> r)3 5 + 3 8(KL> r)c 8(KL> r)3 c
=
3(89.44) (89.443) 5 + 3 8(126.1) 8(126.13)
B1 
B1 
2(KL> r)2 c
(KL> r)2
(89.442)
2(126.12)
R (36)
R sg
= 14.271 ksi The allowable load is Pallow = sallowA = 14.271(7.08) = 101 kip 7 P = 100 kip Thus, Use W8 * 24 O.K. Ans.
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*1384. Using the AISC equations, select from Appendix B the lightestweight structural A36 steel column that is 30 ft long and supports an axial load of 200 kip.The ends are fixed.
Try W8 * 48 a
KL 2 p2 (29)(103) 2 p2E b= = = 126.1 r c A sg A 36 ry = 2.08 in. A = 14.1 in2
0.5 (30)(12) KL = = 86.54 ry 2.08 a KL KL b6a b intermediate column. ry rc
sallow =
=
b5 + 3B 3 8
e5 3 +
3 8
b1  1B 2
KL r 1 2
A KL B c r
KL r
A KL B c r
e1 
C 86.54 D 2 f 36 126.1 D C
R  1B 8
R r sg
2 KL r
A KL B c r D
Rr
3
C
86.54 126.1
1 86.54 3 f 8 126.1
= 14.611 ksi
Pallow = sallow A = 14.611 (14.1) = 206 kip 7 P = 200 kip Use W 8 * 48
O.K. Ans.
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1385.
A W8 * 24 A36steel column of 30ft length is pinned at both ends and braced against its weak axis at midheight. Determine the allowable axial force P that can be safely supported by the column. Use the AISC column design formulas.
Section Properties. From the table listed in the appendix, the necessary section properties for a W8 * 24 are A = 7.08 in2 rx = 3.42 in. ry = 1.61 in.
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Here, Lx = 30(12) = 360 in. and Ly = 15(12) = 180 in. Thus,
1(360) KL = 105.26 = rx 3.42
AISC
1(180) KL = 111.80 (controls) = ry 1.61
=
C
Column
Formulas.
For
2p2 C 29 A 103 B D 36
= 126.10. Since
intermediate column.
sallow =
3(KL> r) (KL> r)3 5 + 3 8(KL> r)c 8(KL> r)c 3 C1 2 A 126.102 B 111.802 S (36)
B1 
2(KL> r)c 2
(KL> r)2
KL KL 6 , the ry rc
A36
steel
KL 2p2E = r c A sY column is
an
R sY
=
3(111.80) 5 111.803 + 3 8(126.10) 8 A 126.103 B
= 11.428 ksi Thus, the allowable force is Pallow = sallowA = 11.428(7.08) = 80.91 kip = 80.9 kip Ans.
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1386. Check if a W10 * 39 column can safely support an axial force of P = 250 kip. The column is 20 ft long and is pinned at both ends and braced against its weak axis at midheight. It is made of steel having E = 29(103) ksi and sY = 50 ksi. Use the AISC column design formulas.
Section Properties. From the table listed in the appendix, the necessary section properties for a W10 * 39 are A = 11.5 in2 rx = 4.27 in. ry = 1.98 in.
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Here, Lx = 20(12) = 240 in. and Ly = 10(12) = 120 in. Thus,
1(240) KL = 56.21 = rx 4.27
1(120) KL = 60.606 (controls) = ry 1.98 Column 2p2 c 29 A 103 B d 50 Formulas. For A36 steel S = 107.00 . Since
AISC
=
intermediate column.
sallow =
3(KL> r) (KL> r)3 5 + 3 8(KL> r)c 8(KL> r)c 3 C1 2 A 107.002 B 60.6062 S (50)
B1 
KL KL 6 , the column is an ry rc (KL> r)2
KL 2p2E = r c A sY
2(KL> r)c 2
R sY
=
3(60.606) 5 60.6063 + 3 8(107.00) 8 A 107.003 B = 22.614 ksi
Thus, the allowable force is Pallow = sallowA = 22.614(11.5) = 260.06 kip 7 P = 250 kip Thus, a W10 * 39 column is adequate. O.K.
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1387. A 5ftlong rod is used in a machine to transmit an axial compressive load of 3 kip. Determine its smallest diameter if it is pin connected at its ends and is made of a 2014T6 aluminum alloy.
Section properties: A= I d 64 = = AA C p d2 4 4 I=
pd4
p2 d; 4
p d4 pd4 ab= 42 64
r=
sallow =
P = A
p 4
3 3.820 = d2 d2
Assume long column: 1.0 (5)(12) 240 KL = = d r d
4
sallow =
54 000
A
KL 2 r
B
;
3.820 54000 = d2 C 240 D 2
d
d = 1.42 in. KL 240 = = 169 7 55 r 1.42
Ans.
O.K.
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*1388. Check if a W10 * 45 column can safely support an axial force of P = 200 kip. The column is 15 ft long and is pinned at both of its ends. It is made of steel having E = 29(103) ksi and sY = 50 ksi. Use the AISC column design formulas.
Section Properties. Try W10 * 45. From the table listed in the appendix, the necessary section properties are A = 13.3 in2 ry = 2.01 in.
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Thus, a 1(15)(12) KL b= = 89.552 ry 2.01
KL 2p2E AISC Column Formulas. Here, a b= = r c A sY S KL KL Since a b 6a b , the ry rc column is an intermediate column.
2p2 c 29 A 103 B d 50
= 107.00.
sallow =
(KL> r)3 3(KL> r) 5 + 3 8(KL> r)c 8(KL> r)c 3 C1 2 A 107.002 B 89.5522 S (50)
B1 
2(KL> r)c 2
(KL> r)2
R sY
=
3(89.552) 5 89.5523 + 3 8(107.00) 8 A 107.003 B
= 17.034 ksi Thus, the allowable force is Pallow = sallowA = 17.034(13.3) = 226.55 kip 7 P = 200 kip Thus, A W10 * 45 can be used Ans. O.K.
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1389.
Using the AISC equations, check if a column having the cross section shown can support an axial force of 1500 kN. The column has a length of 4 m, is made from A36 steel, and its ends are pinned.
20 mm
350 mm
20 mm 300 mm
10 mm
Section Properties: A = 0.3(0.35)  0.29(0.31) = 0.0151 m2 Iy = Iy 90.02583(10  6) = = 0.077214 m AA A 0.0151 1 1 (0.04) A 0.33 B + (0.31) A 0.013 B = 90.025833 A 10  6 B m4 12 12
ry =
Slenderness Ratio: For a column pinned at both ends, K = 1. Thus, a A 250(10 ) column. Applying Eq. 1323, AISC = Column 2p2[200(109)]
6
1(4) KL b= = 51.80 ry 0.077214 For A36 steel, a
Formula:
KL 2p2E b= r c A sg
= 125.7. Since
KL KL 6a b , the column is an intermediate r rc
sallow =
3(KL> r) (KL> r)3 5 + 3 8(KL> r)c 8(KL> r)3 c
=
B1 
B1 
2(KL> r)2 c
(KL> r)2
(51.802)
2(125.72)
3(51.80) (51.803) 5 + 3 8(125.7) 8(125.73)
R (250)(106)
R sg
= 126.2 MPa The allowable load is Pallow = sallowA = 126.2 A 106 B (0.0151) = 1906 kN 7 P = 1500 kN Thus, the column is adequate. O.K. Ans.
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1390. The A36steel tube is pinned at both ends. If it is subjected to an axial force of 150 kN, determine the maximum length that the tube can safely support using the AISC column design formulas.
100 mm
80 mm
Section Properties. A = p A 0.052  0.042 B = 0.9 A 10  3 B p m2 I= 0.9225 A 10  6 B p I = = 0.03202 m AA C 0.9 A 10  3 B p p A 0.054  0.044 B = 0.9225 A 10  6 B p m4 4
r=
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Thus, 1(L) KL = = 31.23L r 0.03202 AISC Column Formulas. sallow = 12p2E 23(KL> r)2 =
.9 A 10  3 B p
150 A 103 B
12p2 C 200 A 109 B D 23(31.23L)2
L = 4.4607 m = 4.46 m
Here,
250 A 10 B long column is correct. =
6
C
KL = 31.23(4.4607) = 139.33. r 2p2 C 200 A 109 B D = 125.66. Since a
For
A36
steel
a
KL 2p2E b= r c A sY
Ans.
KL KL b6 6 200, the assumption of a rc r
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1391. The bar is made of a 2014T6 aluminum alloy. Determine its smallest thickness b if its width is 5b. Assume that it is pin connected at its ends.
600 lb b 5b
8 ft
Section Properties: A = b(5b) = 5b2
54 Iy 23 12 b = = b AA C 5b2 6
Iy =
1 54 (5b) A b3 B = b 12 12
600 lb
ry =
Slenderness Ratio: For a column pinned at both ends, K = 1. Thus, a 1(8)(12) 332.55 KL = b= 23 b ry 6b
Aluminum (2014  T6 alloy) Column Formulas: Assume a long column and apply Eq. 1326. sallow = 54 000 (KL> r)2
0.600 54 000 = 2 5b A 332.55 B 2
b
b = 0.7041 in. Here, KL KL 332.55 = 7 55, the assumption is correct. Thus, = 472.3. Since r r 0.7041 b = 0.704 in. Ans.
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*1392. The bar is made of a 2014T6 aluminum alloy. Determine its smallest thickness b if its width is 5b. Assume that it is fixed connected at its ends.
600 lb b 5b
8 ft
Section Properties: A = b(5b) = 5b2 Iy = Iy 23 12 b = = b AA C 5b2 6 1 54 (5b) A b3 B = b 12 12
5 4
600 lb
ry =
Slenderness Ratio: For a column fixed at both ends, K = 0.5. Thus, a 0.5(8)(12) 166.28 KL = b= ry b 23 6b
Aluminum (2014  T6 alloy) Column Formulas: Assume a long column and apply Eq. 1326. sallow = 54 000 (KL> r)2
0.600 54 000 = 5b2 A 166.28 B 2
b
b = 0.4979 in. Here, Thus, b = 0.498 in. Ans. KL KL 166.28 = 334.0. Since = 7 55, the assumption is correct. r r 0.4979
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1393.
The 2014T6 aluminum column of 3m length has the cross section shown. If the column is pinned at both ends and braced against the weak axis at its midheight, determine the allowable axial force P that can be safely supported by the column.
15 mm
170 mm
15 mm
15 mm 100 mm
Section Properties. A = 0.1(0.2)  0.085(0.17) = 5.55 A 10  3 B m2 Ix = 1 1 (0.1) A 0.23 B (0.085) A 0.173 B = 31.86625 A 10  6 B m4 12 12 31.86625 A 10  6 B Ix = = 0.07577 AA C 5.55 A 10  3 B 1 1 (0.015) A 0.13 B d + (0.17) A 0.0153 B = 2.5478 A 10  6 B m4 12 12
Iy = 2 c rx =
ry =
2.5478 A 10  6 B Iy = = 0.02143 m AA C 5.55 A 10  3 B
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Here, Lx = 3 m and Ly = 1.5 m. Thus, a a (1)(3) KL b= = 39.592 rx 0.07577 (1)(1.5) KL b= = 70.009 (controls) ry 0.02143 KL b 7 55, the column can ry
2014T6 Alumimum Alloy Column Formulas. Since a be classified a long column, sallow = D =C 373 A 103 B 373 A 103 B 70.0092 a KL 2 b r
T Mpa
= 76.103 MPa Thus, the allowed force is
S MPa
Pallow = sallowA = 76.103 A 106 B C 5.55 A 10  3 B D = 422.37 kN = 422 kN
Ans.
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1394. The 2014T6 aluminum column has the cross section shown. If the column is pinned at both ends and subjected to an axial force P = 100 kN, determine the maximum length the column can have to safely support the loading.
15 mm
170 mm
15 mm
15 mm 100 mm
Section Properties. A = 0.1(0.2)  0.085(0.17) = 5.55 A 10  3 B m2 Iy = 2 c ry = 2.5478 A 10  6 B Iy = = 0.02143 m AA C 5.55 A 10  3 B 1 1 (0.015) A 0.13 B d + (0.17) A 0.0153 B = 2.5478 A 10  6 B m4 12 12
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Then, a 1(L) KL b= = 46.6727L ry 0.02143 T MPa
2014T6 Alumimum Alloy Column Formulas. Assuming a long column, sallow = D 100 A 103 B 373 A 103 B a =C KL 2 b r
5.55 A 10  3 B
(46.672L)2
373 A 103 B
L = 3.083 m = 3.08 m Since a
S A 106 B Pa
Ans.
KL b = 46.6727(3.083) = 143.88 7 55, the assumption is correct. ry
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1395. The 2014T6 aluminum hollow section has the cross section shown. If the column is 10 ft long and is fixed at both ends, determine the allowable axial force P that can be safely supported by the column.
4 in.
Section Properties. A = p A 22  1.52 B = 1.75p in2 r= I 2.734375p = = 1.25 in. AA A 1.75p I= p4 A 2  1.54 B = 2.734375p in4 4
3 in.
Slenderness Ratio. For a column fixed at both of its ends, K = 0.5. Thus, 0.5(10)(12) KL = = 48 r 1.25 2014T6 Aluminum Alloy Column Formulas. Since 12 6 be classified as an intermediate column. sallow = c 30.7  0.23 a KL b d ksi r KL 6 55, the column can r
= [30.7  0.23(48)] ksi = 19.66 ksi Thus, the allowable load is Pallow = sallowA = 19.66 A 106 B (1.75p) = 108.09 kip = 108 kip Ans.
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*1396. The 2014T6 aluminum hollow section has the cross section shown. If the column is fixed at its base and pinned at its top, and is subjected to the axial force P = 100 kip, determine the maximum length of the column for it to safely support the load.
4 in.
Section Properties. A = p A 22  1.52 B = 1.75p in2 r= I 2.734375p = = 1.25 in. AA A 1.75p I= p4 A 2  1.54 B = 2.734375p in4 4
3 in.
Slenderness Ratio. For a column fixed at its base and pinned at its top, K = 0.7. Thus, 0.7(L) KL = = 0.56L r 1.25 2014T6 Aluminum Alloy Column Formulas. Assuming an intermediate column, sallow = c 30.7  0.23 a KL b d ksi r
100 = 30.7  0.23(0.56L) 1.75p L = 97.13 in. = 8.09 ft Ans.
KL = 0.56(97.13) = 54.39 6 55, the assumption of an intermediate column r is correct. Since
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1397.
The tube is 0.25 in. thick, is made of a 2014T6 aluminum alloy, and is fixed at its bottom and pinned at its top. Determine the largest axial load that it can support.
P x 6 in. y y 6 in. x
10 ft
Section Properties: A = 6(6)  5.5(5.5) = 5.75 in2 I= I 31.7448 = = 2.3496 in. A 5.75 AA 1 1 (6) A 63 B (5.5) A 5.53 B = 31.7448 in4 12 12
P
r=
Slenderness Ratio: For a column fixed at one end and pinned at the other end, K = 0.7. Thus, 0.7(10)(12) KL = = 35.75 r 2.3496 Aluminium (2014 T6 alloy) Column Formulas: Since 12 6 column is classified as an intermediate column. Applying Eq. 1325, sallow = c 30.7  0.23 a KL b d ksi r KL 6 55, the r
= [30.7  0.23(33.75)] = 24.48 ksi The allowable load is Pallow = sallowA = 22.48(5.75) = 129 kip Ans.
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1398. The tube is 0.25 in. thick, is made of a 2014T6 aluminum alloy, and is fixed connected at its ends. Determine the largest axial load that it can support.
P x 6 in. y y 6 in. x
10 ft
Section Properties: A = 6(6)  5.5(5.5) = 5.75 in2 I= I 31.7448 = = 2.3496 in. AA A 5.75 1 1 (6) A 63 B (5.5) A 5.53 B = 31.7448 in4 12 12
P
r=
Slenderness Ratio: For column fixed at both ends, K = 0.5. Thus, 0.5(10)(12) KL = = 25.54 r 2.3496 Aluminium (2014 T6 alloy) Column Formulas: Since 12 6 column is classified as an intermediate column. Applying Eq. 1325, sallow = c 30.7  0.23 a KL b d ksi r KL 6 55, the r
= [30.7  0.23(25.54)] = 24.83 ksi The allowable load is Pallow = sallowA = 24.83(5.75) = 143 kip Ans.
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1399. The tube is 0.25 in. thick, is made of 2014T6 aluminum alloy and is pin connected at its ends. Determine the largest axial load it can support. Section Properties: A = 6(6)  5.5(5.5) = 5.75 in2 I= I 31.7448 = = 2.3496 in. AA A 5.75 1 1 (6) A 63 B (5.5) A 5.53 B = 31.7448 in4 12 12
P x 6 in. y y 6 in. x
10 ft
r=
P
Slenderness Ratio: For a column pinned as both ends, K = 1. Thus, 1(10)(12) KL = = 51.07 r 2.3496 Aluminum (2014 T6 alloy) Column Formulas: Since 12 6 column is classified as an intermediate column. Applying Eq. 1325, sallow = c 30.7  0.23 a KL b d ksi r KL 6 55, the r
= [30.7  0.23(51.07)] = 18.95 ksi The allowable load is Pallow = sallowA = 18.95(5.75) = 109 kip Ans.
*13100. A rectangular wooden column has the cross section shown. If the column is 6 ft long and subjected to an axial force of P = 15 kip, determine the required minimum 1 dimension a of its crosssectional area to the nearest 16 in. so that the column can safely support the loading. The column is pinned at both ends. Slenderness Ratio. For a column pinned at both of its ends, K = 1. Then, (1)(6)(12) KL 72 = = a a d NFPA Timber Column Formula. Assuming an intermediate column, sallow = 1.20 c 1 1 KL> d 2 a b d ksi 3 26.0
a
2a
15 1 72> a 2 = 1.20 c 1  a bd 2a(a) 3 26.0 a = 2.968 in. Use a = 3 in. KL 72 KL = = 24. Since 11 6 6 26, the assumption is correct. d 3 d Ans.
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A rectangular wooden column has the cross section shown. If a = 3 in. and the column is 12 ft long, determine the allowable axial force P that can be safely supported by the column if it is pinned at its top and fixed at its base. Slenderness Ratio. For a column fixed at its base and pinned at its top K = 0.7. Then, 0.7(12)(12) KL = = 33.6 d 3 NFPA Timer Column Formula. Since 26 6 as a long column. 540 ksi 540 = = 0.4783 ksi 2 (KL> d) 33.62 KL 6 50, the column can be classified d
13101.
a
2a
sallow = The allowable force is
Pallow = sallowA = 0.4783(3)(6) = 8.61 kip
Ans.
13102. A rectangular wooden column has the cross section shown. If a = 3 in. and the column is subjected to an axial force of P = 15 kip, determine the maximum length the column can have to safely support the load. The column is pinned at its top and fixed at its base. Slenderness Ratio. For a column fixed at its base and pinned at its top, K = 0.7. Then, KL 0.7L = = 0.2333L d 3 NFPA Timber Column Formula. Assuming an intermediate column, sallow = 1.20 c 1 1 KL> d 2 a b d ksi 3 26.0
a
2a
15 1 0.2333L 2 = 1.20 c 1  a bd 3(6) 3 26.0 L = 106.68 in. = 8.89 ft Ans.
KL KL = 0.2333(106.68) = 24.89. Since 11 6 6 26, the assumption is d d correct. Here,
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13103. The timber column has a square cross section and is assumed to be pin connected at its top and bottom. If it supports an axial load of 50 kip, determine its smallest side dimension a to the nearest 1 in. Use the NFPA formulas. 2
14 ft a
Section properties: A = a2 sallow = s = 50 P =2 A a
Assume long column: sallow =
A KL B 2 d
540
540
50 = a2
C
(1.0)(14)(12) a
D
2
a = 7.15 in. (1.0)(14)(12) KL KL = = 23.5, 6 26 d 7.15 d Assume intermediate column: sallow = 1.20 B 1 a = 7.46 in. 1 KL> d 2 a bR 3 26.0
1.0(14)(12)
Assumption NG
1.0(14)(12) KL KL = = 22.53, 11 6 6 26 d 7.46 d 1 Use a = 7 in. 2
2 50 1 a bR = 1.20 B 1  a 2 a 26.0 3
Assumption O.K.
Ans.
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*13104. The wooden column shown is formed by gluing together the 6 in. * 0.5 in. boards. If the column is pinned at both ends and is subjected to an axial load P = 20 kip, determine the required number of boards needed to form the column in order to safely support the loading.
P 0.5 in. 6 in.
9 ft
Slenderness Ratio. For a column pinned at both of its ends, K = 1. If the number of the boards required is n and assuming that n(0.5) 6 6 in. Then, d = n(0.5). Thus, (1)(9)(12) KL 216 = = n d n(0.5) NFPA Timber Column Formula. Assuming an intermediate column, sallow = 1.20 B 1 1 KL> d 2 a b R ksi 3 26.0
P
n2  5.5556n  23.01 = 0
Solving for the positive root, n = 8.32 Use n = 9 Here,
1 216> n 2 20 = 1.20 B 1  a bR [n(0.5)](6) 3 26.0
Ans.
KL KL 216 = = 24. Since n(0.5) = 9(0.5) = 4.5 in. 6 6 in. and 11 6 6 26, d 9 d the assumptions made are correct.
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13105.
The column is made of wood. It is fixed at its bottom and free at its top. Use the NFPA formulas to determine its greatest allowable length if it supports an axial load of P = 2 kip.
P 2 in. y x y x 4 in.
Slenderness Ratio: For a column fixed at one end and free at the other end, K = 2. Thus, 2(L) KL = = 1.00L d 2 NFPA Timber Column Formulas: Assume a long column. Apply Eq. 1329, sallow = 540 ksi (KL> d)2
L
2 540 = 2(4) (1.00L)2 L = 46.48 in Here, Thus, L = 46.48 in. = 3.87 ft Ans. KL KL = 1.00(46.48) = 46.48. Since 26 6 6 50, the assumption is correct. d d
13106. The column is made of wood. It is fixed at its bottom and free at its top. Use the NFPA formulas to determine the largest allowable axial load P that it can support if it has a length L = 4 ft. Slenderness Ratio: For a column fixed at one end and free at the other end, K = 2. Thus,
P 2 in. y x y x 4 in.
L
2(4)(12) KL = = 48.0 d 2 NFPA Timber Column Formulas: Since 26 6 Eq. 1329, 540 ksi (KL> d)2 540 48.02 KL 6 50, it is a long column. Apply d
sallow =
=
= 0.234375 ksi The allowable axial force is Pallow = sallowA = 0.234375[2(4)] = 1.875 kip Ans.
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13107. The W14 * 53 structural A36 steel column supports an axial load of 80 kip in addition to an eccentric load P. Determine the maximum allowable value of P based on the AISC equations of Sec. 13.6 and Eq. 1330. Assume the column is fixed at its base, and at its top it is free to sway in the xz plane while it is pinned in the yz plane.
z 80 kip x y
P
y x 10 in.
12 ft
Section Properties: For a W14 * 53 wide flange section. A = 15.6 in2 ry = 1.92 in. Slenderness Ratio: By observation, the largest slenderness ratio is about y  y axis. For a column fixed at one end and free at the other end, K = 2. Thus, a 2(12)(12) KL b= = 150 ry 1.92 d = 13.92 in. Ix = 541 in4 rx = 5.89 in.
Allowable Stress: The allowable stress can be determined using AISC Column 2p2[29(103)] 2p2E KL b= = = 126.1. Since Formulas. For A36 steel, a rc B sY B 36 KL KL b 200, the column is a long column. Applying Eq. 1321, a rc r sallow = 12p2E 23(KL> r)2 12p2(29.0)(103) = 23(1502)
= 6.637 ksi Maximum Stress: Bending is about x  x axis. Applying we have smax = sallow = Mc P + A I
6.637 =
P(10) A 13.92 B P + 80 2 + 15.6 541 Ans.
P = 7.83 kip
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*13108. The W12 * 45 structural A36 steel column supports an axial load of 80 kip in addition to an eccentric load of P = 60 kip. Determine if the column fails based on the AISC equations of Sec. 13.6 and Eq. 1330. Assume that the column is fixed at its base, and at its top it is free to sway in the xz plane while it is pinned in the yz plane.
z 80 kip x y
P
y x 10 in.
12 ft
Section Properties: For a W12 * 45 wide flange section, A = 13.2 in2 ry = 1.94 in. Slenderness Ratio: By observation, the largest slenderness ratio is about y  y axis. For a column fixed at one end and free at the other end, K = 2. Thus, a 2(12)(12) KL b= = 148.45 ry 1.94 d2 = 12.06 in. Ix = 350 in4 rx = 5.15 in.
Allowable Stress: The allowable stress can be determined using AISC Column 2p2[29(103)] 2p2E KL b= = = 126.1. Since Formulas. For A36 steel, a rc B sY B 36 KL KL b 200, the column is a long column. Applying Eq. 1321, a rc r sallow = 12p2E 23(KL> r)2 12p2(29.0)(103) = 23(148.452)
= 6.776 ksi Maximum Stress: Bending is about x  x axis. Applying Eq. 1 we have smax = Mc P + A I
=
60(10) A 12.06 B 140 2 + 13.2 350
= 20.94 ksi Since smax 7 sallow, the column is not adequate.
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13109.
The W14 * 22 structural A36 steel column is fixed at its top and bottom. If a horizontal load (not shown) causes it to support end moments of M = 10 kip # ft, determine the maximum allowable axial force P that can be applied. Bending is about the xx axis. Use the AISC equations of Sec. 13.6 and Eq. 1330.
P x y
M y x
12 ft
Section properties for W14 * 22: A = 6.49 in2 d = 13.74 in2 Ix = 199 in4 ry = 1.04 in.
M P
Allowable stress method: 0.5(12)(12) KL = = 69.231 ry 1.04 a KL KL 2p2E KL 2p2(29)(103) b= = = 126.1, 6a b rc ry rc B sY B 36
Hence,
(sa)allow =
smax = (sa)allow =
B5 3
B1 +
3 8
1 2
A KL B c r
KL r
A KL B 2 r A KL B 2 rc

1 8
R sY
A KL B 3 r A KL B 3 rc
My c P + A Ix
R
c1 = c5 + 3
3 8
1 2
A 69.231 B 2 d 36 126.1
A 69.231 B  1 A 69.231 B 3 d 126.1 8 126.1
= 16.510 ksi
16.510 =
10(12)(13.74) P 2 + 6.49 199 Ans.
P = 80.3 kip
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13110. The W14 * 22 column is fixed at its top and bottom. If a horizontal load (not shown) causes it to support end moments of M = 15 kip # ft, determine the maximum allowable axial force P that can be applied. Bending is about the xx axis. Use the interaction formula with 1sb2allow = 24 ksi.
P x y
M y x
12 ft
Section Properties for W 14 * 22: A = 6.49 in2 d = 13.74 in2 Ix = 199 in4 ry = 1.04 in.
M P
Interaction method: 0.5(12)(12) KL = = 69.231 ry 1.04 a KL 2p2E KL KL 2p2(29)(103) b= = = 126.1, 6a b rc ry rc B sY B 36
Hence,
(sa)allow =
sa =
15(12) A 13.74 B Mxc 2 = = 6.214 ksi sb = Ix 199 sb sa + = 1.0 (sa)allow (sb)allow 0.15408 P 6.2141 + = 1.0 16.510 24 P = 79.4 kip Ans.
P P = = 0.15408 P A 6.49
B5 3
B1 +
3 8
1 2
A KL B c r
KL r
A KL B 2 r A KL B 2 rc

1 8
R sY
A KL B 3 r A KL B 3 rc
R
c1 = c5 + 3
3 8
1 2
A 69.231 B 2 d 36 126.1 B
1 8
A
69.231 126.1
A
69.231 3 d 126.1
B
= 16.510 ksi
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13111. The W14 * 43 structural A36 steel column is fixed at its bottom and free at its top. Determine the greatest eccentric load P that can be applied using Eq. 1330 and the AISC equations of Sec. 13.6.
40 kip
P 16 in.
10 ft
Section properties for W14 * 43: A = 12.6 in2 Iy = 45.2 in4 b = 7.995 Allowable stress method: 2(10)(12) KL = = 126.98 ry 1.89 a 2p2E KL KL KL 2p2 (29)(103) b= = = 126.1, 200 7 7a b rc ry rc B sY B 36 12p2(29)(103) 12p2E = = 9.26 ksi 2 23(KL> r) 23(126.98)2 My c P + A Iy d = 13.66 in. ry = 1.89 in.
(sa)allow =
smax = (sa)allow =
P(16) A 7.995 B P + 40 A 9.26 = + 12.6 45.2 P = 4.07 kip Ans.
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*13112. The W10 * 45 structural A36 steel column is fixed at its bottom and free at its top. If it is subjected to a load of P = 2 kip, determine if it is safe based on the AISC equations of Sec. 13.6 and Eq. 1330.
40 kip
P 16 in.
10 ft
Section Properties for W10 * 45: A = 13.3 in2 Iy = 53.4 in4 b = 8.020 in. Allowable stress method: 2.0(10)(12) KL = = 119.4 ry 2.01 a KL 2p2E 2p2(29)(103) b= = = 126.1 rc B sY B 36 d = 10.10 in. ry = 2.01 in.
KL KL 6a b r rc c1 5 3 1 (KL> r) 2 (KL> r)3 d sY
2 c
(sa)allow
+
3 KL> r 8 KL> rc
(sa)allow =
10.37
2(16) A 8.020 B 42 2 + 13.3 53.4 O.K.
My c P + A Iy
A
B
1 (KL> r) 8 (KL> rc)3
3
=
5 3
C 1  1 A 119.4 B 2 D 36 2 126.1
+
3 8
A 119.4 B  1 A 119.4 B 3 136.1 8 126.1
= 10.37 ksi
10.37 5.56 Column is safe. Yes.
Ans.
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13113.
The A36steel W10 * 45 column is fixed at its base. Its top is constrained to move along the xx axis but free to rotate about and move along the yy axis. Determine the maximum eccentric force P that can be safely supported by the column using the allowable stress method.
24 ft
12 in. x y
P y
x
Section Properties. From the table listed in the appendix, the section properties for a W10 * 45 are A = 13.3 in2 bf = 8.02 in. rx = 4.32 in. Iy = 53.4 in4 ry = 2.01 in.
Slenderness Ratio. Here, Lx = 24(12) = 288 in. and for a column fixed at its base and free at its top, Kx = 2. Thus, a 2(288) KL b= = 133.33 (controls) rx 4.32
Since the column is fixed at its base and pinned at its top, Ky = 0.7 and Ly = 24(12) = 288 in. Then, a 0.7(288) KL b= = 100.30 ry 2.01
Allowable Stress. The allowable stress will be determined using the AISC column formulas. For A36 steel, a
2p2 C 29 A 103 B D KL 2p2E KL KL b= = = 126.10. Since a b 6a b 6 200, rc rc rx B sY C 36 the column is classified as a long column. 12p2E 23(KL> r)2
sallow =
=
12p2 C 29 A 103 B D 23(133.332)
= 8.400 ksi
Maximum Stress. Bending is about the weak axis. Since M = P(12) and bf 8.02 = = 4.01 in, c= 2 2 sallow = Mc P + A I [P(12)](4.01) P + 13.3 53.4 Ans.
8.400 =
P = 8.604 kip = 8.60 kip
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13114. The A36steel W10 * 45 column is fixed at its base. Its top is constrained to move along the xx axis but free to rotate about and move along the yy axis. Determine the maximum eccentric force P that can be safely supported by the column using an interaction formula. The allowable bending stress is (sb)allow = 15 ksi.
12 in. x y 24 ft
P y
x
Section Properties. From the table listed in the appendix, the section properties for a W10 * 45 are A = 13.3 in2 bf = 8.02 in. rx = 4.32 in. Iy = 53.4 in4 ry = 2.01 in.
Slenderness Ratio. Here, Lx = 24(12) = 288 in and for a column fixed at its base and free at its top, Kx = 2. Thus, a 2(288) KL b= = 133.33 (controls) rx 4.32
Since the column is fixed at its base and pinned at its top, Ky = 0.7 and Ly = 24(12) = 288 in. Then, a 0.7(288) KL b= = 100.30 ry 2.01
Allowable Stress. The allowable stress will be determined using the AISC column 2p2 C 29 A 103 B D 2p2E KL = 126.10. Since formulas. For A36 steel, a b= = C rc B sY 36 a KL KL b 6a b 6 200, the column is classified as a long column. rc rx sallow = 12p2E 23(KL> r)2
=
12p2 C 29 A 103 B D 23 A 133.332 B
= 8.400 ksi
Interaction Formula. Bending is about the weak axis. Here, M = P(12) and bf 8.02 = = 4.01 in. c= 2 2 P> A Mc> Ar2 + =1 (sa)allow (sb)allow P> 13.3 + 8.400 P(12)(4.01) n C 13.3 A 2.012 B D 15
=1 Ans. O.K.
P = 14.57 kip = 14.6 kip 14.57> 13.3 sa = = 0.1304 6 0.15 (sa)allow 8.400
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13115. The A36steel W12 * 50 column is fixed at its base. Its top is constrained to move along the xx axis but free to rotate about and move along the yy axis. If the eccentric force P = 15 kip is applied to the column, investigate if the column is adequate to support the loading. Use the allowable stress method. Section Properties. From the table listed in the appendix, the section properties for a W12 * 50 are A = 14.7 in2 ry = 1.96 in. Slenderness Ratio. Here, Lx = 24(12) = 288 in. and for a column fixed at its base and pinned at its top, Kx = 2. Thus, a 2(288) KL b= = 111.20 (controls) rx 5.18 bf = 8.08 in. rx = 5.18 in. Iy = 56.3 in4
12 in. x y 24 ft
P y
x
Since the column is fixed at its base and pinned at its top, Ky = 0.7 and Ly = 24(12) = 288 in. Thus, a 0.7(288) KL b= = 102.86 ry 1.96 2p2 C 29 A 103 B D KL 2p2E = 126.10. Since b= = C r c A sY 36
Allowable Stress. The allowable stress will be determined using the AISC column formulas. For A36 steel, a a
KL KL b 6a b , the column can be classified as an intermediate column. rx rc
sallow =
(KL> r)3 3(KL> r) 5 + 3 8(KL> r)C 8(KL> r)C 3 C1 2 A 126.102 B 111.202 S (36)
B1 
2(KL> r)C 2
(KL> r)2
R sY
=
3(111.20) 5 111.203 + 3 8(126.10) 8 A 126.103 B
= 11.51 ksi Maximum Stress. Bending is about the weak axis. Since, M = 15(12) = 180 kip # in. bf 8.08 and c = = = 4.04 in., 2 2 smax = 180(4.04) P Mc 15 + = + = 13.94 ksi A I 14.7 56.3
Since smax 7 sallow, the W12 * 50 column is inadequate according to the allowable stress method.
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*13116. The A36steel W12 * 50 column is fixed at its base. Its top is constrained to move along the xx axis but free to rotate about and move along the yy axis. If the eccentric force P = 15 kip is applied to the column, investigate if the column is adequate to support the loading. Use the interaction formula. The allowable bending stress is (sb)allow = 15 ksi. Section Properties. From the table listed in the appendix, the section properties for a W12 * 50 are A = 14.7 in2 bf = 8.08 in. rx = 5.18 in. ry = 1.96 in.
12 in. x y 24 ft
P y
x
Slenderness Ratio. Here, Lx = 24(12) = 288 in and for a column fixed at its base and pinned at its top, Kx = 2. Thus, a 2(288) KL b= = 111.20 (controls) rx 5.18
Since the column is fixed at its base and pinned at its top, Ky = 0.7 and Ly = 24(12) = 288 in. Then, a Allowable = 0.7(288) KL b= = 102.86 ry 1.96 Stress. For A36 steel, a KL 2p2E b= r c A sY
C
Axial
2p2 C 29 A 103 B D 36
= 126.10. Since a C1 
KL KL b 6a b , the column can be rx rc S sY
classified as an intermediate column. 2(KL> r)c 2 (KL> r)2
sallow =
3(KL> r) (KL> r)3 5 + 3 8(KL> r)c 8(KL> r)c 3 C1 2 A 126.102 B 111.202 S (36)
=
3(111.20) 5 111.203 + 3 8(126.10) 8 A 126.103 B
= 11.51 ksi Interaction Formula. Bending is about the weak axis. Here, M = 15(12) bf 8.08 = 180 kip # in. and c = = = 4.04 in. 2 2 P> A Mc> Ar2 15> 14.7 + + = (sa)allow (sb)allow 11.51 P = 14.57 kip = 14.6 kip 15> 14.7 sa = = 0.089 6 0.15 (sa)allow 11.51 Thus, a W12 * 50 column is adequate according to the interaction formula. 180(4.04) n C 14.7 A 1.962 B D 15
= 0.9471 6 1 Ans. O.K.
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A 16ftlong column is made of aluminum alloy 2014T6. If it is fixed at its top and bottom, and a compressive load P is applied at point A, determine the maximum allowable magnitude of P using the equations of Sec. 13.6 and Eq. 1330.
13117.
P A 4.25 in. x 0.5 in. y 8 in. y x 8 in. 0.5 in. 0.5 in.
Section properties: A = 2(0.5)(8) + 8(0.5) = 12 in2 Ix = 1 1 (8)(93) (7.5)(83) = 166 in4 12 12 Iy 42.75 = = 1.8875 in. AA A 12 1 1 b (0.5)(83) + (8)(0.53) = 42.75 in4 12 12
Iy = 2 a ry =
Allowable stress method: 0.5(16)(12) KL KL = 50.86, 12 6 = 6 55 ry ry 1.8875 sallow = c 30.7  0.23 a KL bd r
= [30.7  0.23(50.86)] = 19.00 ksi smax = sallow = Mx c P + A Ix
19.00 =
P(4.25)(4.5) P + 12 166 Ans.
P = 95.7 kip
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13118. A 16ftlong column is made of aluminum alloy 2014T6. If it is fixed at its top and bottom, and a compressive load P is applied at point A, determine the maximum allowable magnitude of P using the equations of Sec. 13.6 and the interaction formula with 1sb2allow = 20 ksi.
P A 4.25 in. x 0.5 in. y 8 in. y x 8 in. 0.5 in. 0.5 in.
Section Properties: A = 2(0.5)(8) + 8(0.5) = 12 in2 Ix = 1 1 (8)(93) (7.5)(83) = 166 in4 12 12 Iy 42.75 = = 1.8875 in. AA A 12 1 1 b (0.5)(83) + (8)(0.53) = 42.75 in4 12 12
Iy = 2 a ry =
Interaction method: 0.5(16)(12) KL KL = 50.86, 12 6 = 6 55 ry ry 1.8875 sallow = c 30.7  0.23 a KL bd r
= [30.7  0.23(50.86)] = 19.00 ksi sa = P P = = 0.08333P A 12 P(4.25)(4.50) Mc = = 0.1152P Ix 166 sb sa + = 1.0 (sa)allow (sb)allow 0.08333P 0.1152P + =1 19.00 20 P = 98.6 kip Ans.
sb =
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13119. The 2014T6 hollow column is fixed at its base and free at its top. Determine the maximum eccentric force P that can be safely supported by the column. Use the allowable stress method. The thickness of the wall for the section is t = 0.5 in.
6 in. 3 in. 6 in.
P
8 ft
Section Properties. A = 6(3)  5(2) = 8 in2 Ix = 1 1 (3) A 63 B (2) A 53 B = 33.1667 in4 12 12 1 1 (6) A 33 B (5) A 23 B = 10.1667 in4 12 12 rx = 33.1667 Ix = = 2.036 in. AA A 8
Iy =
ry =
Iy 10.1667 = = 1.127 in. AA A 8
Slenderness Ratio. For a column fixed at its base and free at its top, K = 2. Thus, a 2(8)(12) KL b= = 170.32 ry 1.127 KL b 7 55, the column can be classified as a long ry
Allowable Stress. Since a column.
sallow =
54 000 ksi 54 000 ksi = = 1.862 ksi (KL> r)2 170.312
Maximum Stress. Bending occurs about the strong axis so that M = P(6) and 6 c = = 3 in. 2 sallow = Mc P + A I
1.862 =
C P(6) D (3) P + 8 33.1667
Ans.
P = 2.788 kip = 2.79 kip
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*13120. The 2014T6 hollow column is fixed at its base and free at its top. Determine the maximum eccentric force P that can be safely supported by the column. Use the interaction formula. The allowable bending stress is (sb)allow = 30 ksi. The thickness of the wall for the section is t = 0.5 in.
6 in. 3 in. 6 in.
P
8 ft
Section Properties. A = 6(3)  5(2) = 8 in2 Ix = 1 1 (3) A 63 B (2) A 53 B = 33.1667 in4 12 12 1 1 (6) A 33 B (5) A 23 B = 10.1667 in4 12 12 rx = 33.1667 Ix = = 2.036 in. AA A 8
Iy =
ry =
Iy 10.1667 = = 1.127 in. AA A 8
Slenderness Ratio. For a column fixed at its base and pinned at its top, K = 2. Thus, a 2(8)(12) KL b= = 170.32 ry 1.127
KL b 7 55, the column can be classified as the column is ry classified as a long column. Allowable Stress. Since a 54000 ksi 54000 ksi = = 1.862 ksi (KL> r)2 170.312
sallow =
Interaction Formula. Bending is about the strong axis. Since M = P(6) and 6 c = = 3 in, 2 P> A Mc> Ar2 + =1 (sa)allow (sb)allow P> 8 + 1.862 [P(6)](3) n C 8 A 2.0362 B D 30
=1 Ans.
P = 11.73 kip = 11.7 kip
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13121. The 10ftlong bar is made of aluminum alloy 2014T6. If it is fixed at its bottom and pinned at the top, determine the maximum allowable eccentric load P that can be applied using the formulas in Sec. 13.6 and Eq. 1330.
P x 1.5 in. 1.5 in. x y 3 in. 2 in. y 2 in.
Section Properties: A = 6(4) = 24.0 in2 Ix = 1 (4) A 63 B = 72.0 in4 12 Iy 32.0 = = 1.155 in. AA A 24 1 (6) A 43 B = 32.0 in4 12
Iy =
ry =
Slenderness Ratio: The largest slenderness ratio is about y  y axis. For a column pinned at one end fixed at the other end, K = 0.7. Thus, a 0.7(10)(12) KL b= = 72.75 ry 1.155
Allowable Stress: The allowable stress can be determined using aluminum KL (2014 T6 alloy) column formulas. Since 7 55, the column is classified as a long r column. Applying Eq. 1326, sallow = c = 54 000 d ksi (KL> r)2
54 000 72.752
= 10.204 ksi Maximum Stress: Bending is about x  x axis. Applying Eq. 1330, we have smax = sallow = P Mc + A I P(1.5)(3) P + 24.0 72.0 Ans.
10.204 =
P = 98.0 kip
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13122. The 10ftlong bar is made of aluminum alloy 2014T6. If it is fixed at its bottom and pinned at the top, determine the maximum allowable eccentric load P that can be applied using the equations of Sec. 13.6 and the interaction formula with 1sb2allow = 18 ksi.
P x 1.5 in. 1.5 in. x y 3 in. 2 in. y 2 in.
Section Properties: A = 6(4) = 24.0 in2 Ix = 1 (4) A 63 B = 72.0 in4 12 Ix 72.0 = = 1.732 in. AA A 24.0 Iy 32.0 = = 1.155 in. AA A 24.0 1 (6) A 43 B = 32.0 in4 12
Iy =
rx =
ry =
Slenderness Ratio: The largest slenderness radio is about y  y axis. For a column pinned at one end and fixed at the other end, K = 0.7. Thus a 0.7(10)(12) KL b= = 72.75 ry 1.155
Allowable Stress: The allowable stress can be determined using aluminum KL (2014 T6 alloy) column formulas. Since 7 55, the column is classified as a long r column. Applying Eq. 1326, (sa)allow = c = 54 000 d ksi (KL> r)2
54 000 72.752
= 10.204 ksi Interaction Formula: Bending is about x  x axis. Applying Eq. 1331, we have Mc> Ar2 P> A + =1 (sa)allow (sb)allow P(1.5)(3)> 24.0(1.7322) P> 24.0 + =1 10.204 18 P = 132 kip Ans.
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13123. The rectangular wooden column can be considered fixed at its base and pinned at its top. Also, the column is braced at its midheight against the weak axis. Determine the maximum eccentric force P that can be safely supported by the column using the allowable stress method.
5 ft
6 in.
P 6 in. 3 in. 6 in.
Section Properties. A = 6(3) = 18 in2 Ix = 1 (3) A 63 B = 54 in4 12
5 ft
dx = 6 in.
dy = 3 in.
Slenderness Ratio. Here, Lx = 10(12) = 120 in. and for a column fixed at its base and pinned at its top, K = 0.7. Thus, a 0.7(120) KL b= = 14 dx 6
Since the bracing provides support equivalent to a pin, Ky = 1 and Ly = 5(12) = 60 in. Then a 1(60) KL b= = 20 (controls) dy 3
KL 6 26, the column can be classified as the column d is classified as an intermediate column. Allowable Stress. Since 11 6 sallow = 1.20 c 1 = 1.20 c 1 1 KL> d 2 a b d ksi 3 26.0 1 20 2 a b d ksi = 0.9633 ksi 3 26.0
Maximum Stress. Bending occurs about the strong axis. Here, M = P(6) and 6 c = = 3 in. 2 sallow = P Mc + A I [P(6)](3) P + 18 54 Ans.
0.9633 =
P = 2.477 kip = 2.48 kip
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*13124. The rectangular wooden column can be considered fixed at its base and pinned at its top. Also, the column is braced at its midheight against the weak axis. Determine the maximum eccentric force P that can be safely supported by the column using the interaction formula. The allowable bending stress is (sb)allow = 1.5 ksi.
5 ft
6 in.
P 6 in. 3 in. 6 in.
5 ft
Section Properties. A = 6(3) = 18 in2 Ix 54 = = 1.732 in. AA A 18 Ix = 1 (3) A 63 B = 54 in4 12 dx = 6 in. dy = 3 in.
rx =
Slenderness Ratio. Here, Lx = 10(12) = 120 in. and for a column fixed at its base pinned at its top, K = 0.7. Thus, a 0.7(120) KL b= = 14 dx 6
Since the bracing provides support equivalent to a pin, Ky = 1 and Ly = 5(12) = 60 in. Then a 1(60) KL b= = 20 (controls) dy 3
KL 6 26, the column can be classified as the d column is classified as an intermediate column. Allowable Axial Stress. Since 11 6 sallow = 1.20 c 1 = 1.20 c 1 1 KL> d 2 a b d ksi 3 26.0 1 20 2 a b d ksi = 0.9633 ksi 3 26.0
Interaction Formula. Bending occurs about the strong axis. Since M = P(6) and 6 c = = 3 in. 2 P> A Mc> Ar2 + =1 (sa)allow (sb)allow P> 18 + 0.9633 [P(6)](3) n C 18 A 1.7322 B D 1.5
=1 Ans.
P = 3.573 kip = 3.57 kip
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13125.
The 10in.diameter utility pole supports the transformer that has a weight of 600 lb and center of gravity at G. If the pole is fixed to the ground and free at its top, determine if it is adequate according to the NFPA equations of Sec. 13.6 and Eq. 1330.
G
15 in.
18 ft
2(18)(12) KL = = 43.2 in. d 10 26 6 43.2 50 Use Eq. 1329, sallow = 540 540 = = 0.2894 ksi (KL> d) (43.2)2 Mc P + A I (600)(15)(5) 600 + 2 p (5) A p B (5)4 4 O.K. Ans.
smax =
smax =
smax = 99.31 psi 6 0.289 ksi Yes.
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13126. Using the NFPA equations of Sec. 13.6 and Eq. 1330, determine the maximum allowable eccentric load P that can be applied to the wood column. Assume that the column is pinned at both its top and bottom.
3 in.
P 0.75 in. 6 in.
12 ft
Section Properties: A = 6(3) = 18.0 in2 1 (6) A 33 B = 13.5 in4 12
Iy =
Slenderness Ratio: For a column pinned at both ends, K = 1.0. Thus, a 1.0(12)(12) KL b= = 48.0 dy 3
Allowable Stress: The allowable stress can be determined using NFPA timber KL column formulas. Since 26 6 6 50, it is a long column. Applying Eq. 1329, d sallow = 540 ksi (KL> d)2 540 = 0.234375 ksi 48.02
=
Maximum Stress: Bending is about y  y axis. Applying Eq. 1330, we have smax = sallow = P Mc + A I P(0.75)(1.5) P + 18.0 13.5 Ans.
0.234375 =
P = 1.69 kip
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13127. Using the NFPA equations of Sec. 13.6 and Eq. 1330, determine the maximum allowable eccentric load P that can be applied to the wood column. Assume that the column is pinned at the top and fixed at the bottom.
3 in.
P 0.75 in. 6 in.
12 ft
Section Properties: A = 6(3) = 18.0 in2 Iy = 1 (6) A 33 B = 13.5 in4 12
Slenderness Ratio: For a column pinned at one end and fixed at the other end, K = 0.7. Thus, a 0.7(12)(12) KL b= = 33.6 dy 3
Allowable Stress: The allowable stress can be determined using NFPA timber KL column formulas. Since 26 6 6 50, it is a long column. Applying Eq. 1329, d sallow = 540 ksi (KL> d)2 540 = 0.4783 ksi 33.62
=
Maximum Stress: Bending is about y  y axis. Applying Eq. 1330, we have smax = sallow = Mc P + A I P(0.75)(1.5) P + 18.0 13.5 Ans.
0.4783 =
P = 3.44 kip
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*13128. The wood column is 4 m long and is required to support the axial load of 25 kN. If the cross section is square, determine the dimension a of each of its sides using a factor of safety against buckling of F.S. = 2.5. The column is assumed to be pinned at its top and bottom. Use the Euler equation. Ew = 11 GPa, and sY = 10 MPa.
25 kN
4m a
a
1 a4 (a) A a3 B = , P = (2.5)25 = 62.5 kN and K = 1 12 12 cr for pin supported ends column. Applying Eulers formula, Critical Buckling Load: I = p2EI (KL)2
Pcr = 62.5 A 10
3
B=
a p2(11)(109) A 12 B
4
[1(4)]2
a = 0.1025 m = 103 mm Critical Stress: Eulers formula is only valid if scr 6 sY. scr = 62.5(103) Pcr = = 5.94 MPa 6 s Y = 10 MPa A 0.1025(0.1025)
Ans.
O.K.
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13129. If the torsional springs attached to ends A and C of the rigid members AB and BC have a stiffness k, determine the critical load Pcr.
P
k L 2
A
B
Equilibrium. When the system is given a slight lateral disturbance, the configuration shown in Fig. a is formed. The couple moment M can be related to P by considering the equilibrium of members AB and BC. Member AB + c Fy = 0; a + MA = 0; Member BC a + MC = 0;  By a L L sin u b + Bx a cos u b + M = 0 2 2 (3) By  P = 0 By a L L sin u b + Bx a cos u b  M = 0 2 2 (1) (2)
L 2 k C
Solving Eqs. (1), (2), and (3), we obtain Bx = 0 By = 2M L sin u M= PL sin u 2 u. Thus, (4)
Since u is very small, the small angle analysis gives sin u M= PL u 2
Torslonal Spring Moment. The restoring couple moment Msp can be determined using the torsional spring formula, M = ku. Thus, Msp = ku Critical Buckling Load. When the mechanism is on the verge of bucklling M must equal Msp. M = Msp Pcr L u = ku 2 Pcr = 2k L Ans.
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13129.
Continued
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13130. Determine the maximum intensity w of the uniform distributed load that can be applied on the beam without causing the compressive members of the supporting truss to buckle. The members of the truss are made from A36steel rods having a 60mm diameter. Use F.S. = 2 against buckling.
w
B A 1.5 m
C 2m 3.6 m
D
Equilibrium. The force developed in member BC can be determined by considering the equilibrium of the freebody diagram of the beam AB, Fig. a. a + MA = 0; 3 w(5.6)(2.8)  FBC a b (5.6) = 0 FBC = 4.6667w 5
The Force developed in member CD can be obtained by analyzing the equilibrium of joint C, Fig. b, + c Fy = 0; + : Fx = 0; FAC a 5 3 b  4.6667w a b = 0 13 5 FAC = 7.28w (T)
4 12 4.6667w a b + 7.28 a b w  FCD = 0 5 13
FCD = 10.4533w (C)
Section Properties. The crosssectional area and moment of inertia of the solid circular rod CD are A = p A 0.032 B = 0.9 A 10  3 B p m2 I= p A 0.034 B = 0.2025 A 10  6 B p m4 4
Critical Buckling Load. Since both ends of member CD are pinned, K = 1. The critical buckling load is Pcr = FCD (F.S.) = 10.4533w(2) = 20.9067w Applying Eulers formula, Pcr = p2EI (KL)2
20.9067w =
p2 C 200 A 109 B D C 0.2025 A 10  6 B p D [1(3.6)]2
w = 4634.63 N> m = 4.63 kN> m Critical Stress: Eulers formula is valid only if scr 6 sY. scr = 20.907(4634.63) Pcr = = 34.27 MPa 6 sY = 250 MPa A 0.9 A 10  3 B p
Ans.
O.K.
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13131. The W10 * 45 steel column supports an axial load of 60 kip in addition to an eccentric load P. Determine the maximum allowable value of P based on the AISC equations of Sec. 13.6 and Eq. 1330. Assume that in the xz plane Kx = 1.0 and in the yz plane Ky = 2.0. Est = 2911032 ksi, sY = 50 ksi.
z
P 60 kip
x y
y x 8 in.
10 ft
Section properties for W 10 * 45: A = 13.3 in2 rx = 4.32 in. d = 10.10 in. ry = 2.01 in. Ix = 248 in4
Allowable stress method: a a a 1.0(10)(12) KL b= = 27.8 rx 4.32 2.0(10)(12) KL b= = 119.4 ry 2.01 2p2(29)(103) KL 2p2E b= = = 107 rc B sg B 50 KL KL 7a b r rc (sa)allow = 12p2(29)(103) 12p2E = = 10.47 ksi 2 23(KL> r) 23(119.4)4 Mc P + A I (controls)
smax = (sa)allow =
P(8) A 10.10 B P + 60 2 + 10.47 = 13.3 248 P = 25.0 kip
Ans.
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*13132. The A36steel column can be considered pinned at its top and fixed at its base. Also, it is braced at its midheight along the weak axis. Investigate whether a W250 * 45 section can safely support the loading shown. Use the allowable stress method.
4.5 m
600 mm 10 kN 40 kN
Section Properties. From the table listed in the appendix, the necessary section properties for a W250 * 45 are A = 5700 mm2 = 5.70 A 10  3 B m2 d = 266 mm = 0.266 m rx = 112 mm = 0.112 m ry = 35.1 mm = 0.0351 mm
4.5 m
Ix = 71.1 A 106 B mm4 = 71.1 A 10  6 B m4
Slenderness Ratio. Here, Lx = 9 m and for a column fixed at its base and pinned at its top, Kx = 0.7. Thus,
Since the bracing provides support equivalent to a pin, Ky = 1 and Ly = 4.5 m. Then,
Allowable Stress. For A36 steel, a Since a
1(4.5) KL = 128.21 (controls) = ry 0.0351
0.7(9) KL = 56.25 = rx 0.112
KL KL b 6a b 6 200, the column can be classified as a long column. rc ry sallow = 12p2 C 200 A 109 B D 12p2E = = 62.657 MPa 2 23(KL> r) 23(128.21)2
2p2 C 200 A 109 B D KL 2p2E b= = = 125.66. rc C 250 A 106 B B sY
Maximum Stress. Bending occurs about the strong axis. Here, P = 10 + 40 0.266 d = 50 kN, M = 40(0.6) = 24 kN # m and c = = = 0.133 m, 2 2 smax = 50 A 103 B 24 A 103 B (0.133) P Mc + = + = 53.67 MPa A I 5.70 A 10  3 B 71.1 A 10  6 B
Since smax 6 sallow, the column is adequate according to the allowable stress method.
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13133.
The A36steel column can be considered pinned at its top and fixed at its base. Also, it is braced at its midheight along the weak axis. Investigate whether a W250 * 45 section can safely support the loading shown. Use the interaction formula. The allowable bending stress is (sb)allow = 100 MPa.
4.5 m
600 mm 10 kN 40 kN
Section Properties. From the table listed in the appendix, the necessary section properties for a W250 * 45 are A = 5700 mm2 = 5.70 A 10  3 B m2 Ix = 71.1 A 10 B mm = 71.1 A 10
6 4 6
d = 266 mm = 0.266 m
Bm
4.5 m
4
rx = 112 mm = 0.112 m
ry = 35.1 mm = 0.0351 mm
Slenderness Ratio. Here, Lx = 9 m and for a column fixed at its base and pinned at its top, Kx = 0.7. Thus, a 0.7(9) KL b= = 56.25 rx 0.112
Since the bracing provides support equivalent to a pin, Ky = 1 and Ly = 4..5 m. Then, a 1(4.5) KL b= = 128.21 (controls) ry 0.0351 a
Allowable
=
C
Axial
Stress.
For
A36
steel,
2p2 C 200 A 109 B D 250 A 10
6
KL 2p2E b= rc B sY
classified as a long column. 12p2 C 200 A 109 B D 12p2E = = 62.657 MPa 23(KL> r)2 23(128.21)2
B
= 125.66. Since a
KL KL b 6a b 6 200, the column can be rc ry
sallow =
Interaction Formula. Bending is about the strong axis. Here, P = 10 + 40 = 50 kN, d 0.266 M = 40(0.6) = 24 kN # m and c = = = 0.133 m, 2 2 Mc> Ar2 P> A + = (sa)allow (sb)allow 50 A 103 B n 5.70 A 10  3 B 62.657 A 106 B 24 A 103 B (0.133) n C 5.70 A 10  3 B A 0.1122 B D 100 A 106 B
+
= 0.5864 6 1 sa = (sa)allow 50 A 103 B n 5.7 A 10  3 B 62.657 A 106 B
O.K.
= 0.140 6 0.15
O.K.
Thus, a W250 * 45 column is adequate according to the interaction formula.
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13134. The member has a symmetric cross section. If it is pin connected at its ends, determine the largest force it can support. It is made of 2014T6 aluminum alloy.
0.5 in. 2 in.
P
5 ft
Section properties: A = 4.5(0.5) + 4(0.5) = 4.25 in2
P
1 1 (0.5)(4.53) + (4)(0.5)3 = 3.839 in4 I= 12 12 r= I 3.839 = = 0.9504 in. AA A 4.25
Allowable stress: 1.0(5)(12) KL = = 63.13 r 0.9504 KL 7 55 r Long column sallow = 54000 54000 = = 13.55 ksi (KL> r)2 63.132
Pallow = sallowA = 13.55(4.25) = 57.6 kip Ans.
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13135. The W200 * 46 A36steel column can be considered pinned at its top and fixed at its base. Also, the column is braced at its midheight against the weak axis. Determine the maximum axial load the column can support without causing it to buckle.
6m
Section Properties. From the table listed in the appendix, the section properties for a W200 * 46 are A = 5890 mm2 = 5.89 A 10  3 B m2 Iy = 15.3 A 106 B mm4 = 15.3 A 10  6 B m4 Ix = 45.5 A 106 B mm4 = 45.5 A 10  6 B m4
6m
Critical Buckling Load. For buckling about the strong axis, Kx = 0.7 and Lx = 12 m. Since the column is fixed at its base and pinned at its top, p2EIx (KL)x 2 p2 c 200 A 109 B d c 45.5 A 10  6 B d [0.7(12)]2 = 1.273 A 106 B N = 1.27 MN
Pcr =
=
For buckling about the weak axis, Ky = 1 and Ly = 6 m since the bracing provides a support equivalent to a pin. Applying Eulers formula, p2EIy (KL)y 2 = p2 c 200 A 109 B d c 15.3 A 10  6 B d [1(6)]2
Pcr =
= 838.92 kN = 839 kN (controls)Ans.
Critical Stress. Eulers formula is valid only if scr 6 sY. scr =
838.92 A 103 B Pcr = = 142.43 MPa 6 sY = 250 MPa A 5.89 A 10  3 B
O.K.
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*13136. The structural A36 steel column has the cross section shown. If it is fixed at the bottom and free at the top, determine the maximum force P that can be applied at A without causing it to buckle or yield. Use a factor of safety of 3 with respect to buckling and yielding. Section properties: A = 0.2(0.01) + 0.15 (0.01) + 0.1(0.01) = 4.5(10 ) m
3 2
P 20 mm A 10 mm 100 mm 4m 100 mm 10 mm 150 mm A 10 mm 100 mm
0.005 (0.2)(0.01) + 0.085 (0.15)(0.01) + 0.165 (0.1)(0.01) xA x= = A 4.5(10  3) = 0.06722 m Iy = 1 (0.2)(0.013) + 0.2 (0.01)(0.06722  0.005)2 12 + 1 (0.01)(0.153) + 0.01 (0.15)(0.085  0.06722)2 12 1 (0.1)(0.013) + 0.1 (0.01)(0.165  0.06722)2 12
+
= 20.615278 (10  6) m4 Ix = 1 1 1 (0.01)(0.23) + (0.15)(0.013) + (0.01)(0.13) 12 12 12 Iy 20.615278(10  6) = = 0.0676844 AA A 4.5 (10  3)
= 7.5125 (10  6) m4 ry =
Buckling about x  x axis: Pcr = p2(200)(109)(7.5125)(10  6) p2 EI = 2 (KL) [2.0(4)]2 (controls)
= 231.70 kN scr =
231.7 (103) Pcr = 51.5 MPa 6 sg = 250 MPa = A 4.5 (10  3) P ec KL P c 1 + 2 sec a b d; A 2r A EA r
Yielding about y  y axis: smax =
e = 0.06722  0.02 = 0.04722 m
0.04722 (0.06722) ec = = 0.692919 0.0676844 r2
250(106)(4.5)(10  3) = P[1 + 0.692919 sec (1.96992P (10  3) 2P)] By trial and error: P = 378.45 kN Hence, Pallow = 231.70 = 77.2 kN 3
2.0 (4) P P KL = = 1.96992 P (10  3) 2P 2r A EA 2(0.0676844) A 200 (109)(4.5)(10  3)
Ans.
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13137.
The structural A36 steel column has the cross section shown. If it is fixed at the bottom and free at the top, determine if the column will buckle or yield when the load P = 10 kN. Use a factor of safety of 3 with respect to buckling and yielding.
P 20 mm A 10 mm 100 mm 4m 100 mm 10 mm 150 mm A 10 mm 100 mm
Section properties: A = 0.2 (0.01) + 0.15 (0.01) + 0.1 (0.01) = 4.5 (10  3) m2 x= 0.005 (0.2)(0.01) + 0.085 (0.15)(0.01) + 0.165 (0.1)(0.01) xA = 0.06722 m = A 4.5 (10  3) 1 (0.2)(0.013) + 0.2 (0.01)(0.06722  0.005)2 12 + 1 (0.01)(0.153) + 0.01 (0.15)(0.085  0.06722)2 12 1 (0.1)(0.013) + 0.1 (0.01)(0.165  0.06722)2 = 20.615278 (10  6) m4 12 B
Iy =
+
Ix =
ry =
BA
1 1 1 (0.01)(0.23) + (0.15)(0.013) + (0.01)(0.13) = 7.5125 (10  6) m4 12 12 12 Iy = 20.615278 (10  6) 4.5 (10  3) = 0.067843648 m
Buckling about x  x axis: Pcr = p2(200)(109)(7.5125)(10  6) p2 EI = = 231.70 kN 2 (KL) [2.0(4)]2 231.7 (103) Pcr = 51.5 MPa 6 sg = 250 MPa = A 4.5 (10  3) Pcr 231.7 = = 77.2 kN 7 P = 10 kN FS 3 O.K.
scr =
Pallow =
Hence the column does not buckle. Yielding about y  y axis: smax = P KL ec P bd c 1 + 2 sec a A 2r A EA r
e = 0.06722  0.02 = 0.04722 m
P=
10 = 3.333 kN 3
3.333 (103) P = 0.7407 MPa = A 4.5 (10  3) 0.04722 (0.06722) ec = 0.689815 = (0.067844) r2 2.0 (4) P KL 3.333 (103) = = 0.1134788 2 r AE A 2(0.06783648) A 200 (109)(4.5)(10  3)
smax = 0.7407 [1 + 0.692919 sec (0.1134788)] = 1.25 MPa 6 sg = 250 MPa Hence the column does not yield! No. 1 158 Ans.