Question

A tubular steel column CD supports horizontal cantilever arm ABC, as shown in the figure. Column CD has an outside diameter of 9.000 in. and a wall thickness of 0.460 in. The loads are PA = 300 lb and Pb = 540 lb. Dimensions of the structure are a = 6.3 ft, b= 8.0 ft, and c = 14.0 ft. Determine the maximum compression stress at the base of column CD. b a PA C B A D *Part 1 Determine the cross-sectional area and the area moment of inertia for column CD. A= *1 in 2 1, = * 2 in. Answer *1: the tolerance is +/-4% Answer *2: the tolerance is +/-4% *Part 2 Determine the internal axial force in column CD. Use the sign convention for internal axial forces.

Answer 1

Axial Force Moment D

A)

Cross sectional area = $Area=\frac{\prod }{4}D_o^2(1-k^2)$

where

• D(o) = Outer dia of column = 9 in
• $k=\frac{D_i}{D_o}=\frac{8.08}{9}=0.8977\:$
  • D(i) = Inner dia of column = D(o) - 2t = 8.08 in
    • t = thickness of column wall = 0.460 in

Cross sectional area = $Area=\frac{\prod }{4}D_o^2(1-k^2)$

Cross sectional area = $Area=\frac{\prod }{4}\times 9^2\times (1-0.8977^2)=>\mathbf{Area=12.34\:in^2\:(Ans.)}$

Area Moment of Inertia =

D'(1 - 64) 64

Area Moment of Inertia = $I=\frac{\prod }{64}\times 9^4\times (1-0.8977^4)=>\mathbf{I=112.909\:in^4\:(Ans.)}$

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B)

Internal axial force = P(B) + P(A) = 540 + 300 = 840 lb (Compressive)

Internal axial force = -840 lb (Ans.)

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C)

Internal moment = $P_B\times b+P_A\times (a+b)$

Internal moment = $540\times 8+300\times (6.3+8)=\mathbf{8610\:lb.ft\:(Ans.)}$

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D)

Axial stress = $\sigma _{axial}=\frac{Force}{Area}$

where

• Force = -840 lb
• Area = 12.34 in2

$\sigma _{axial}=\frac{Force}{Area}$

$\sigma _{axial}=\frac{-840}{12.34}=>\mathbf{\sigma _{axial}=-68.07\:psi\:(Ans.)}$

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E)

$\sigma _{bend-max}=\frac{M.y}{I}$

where

• M = Moment acting = 8610 lb.ft = 103320 lb-in
• y = distance of the centroid from the outer edge of the column = D(o) / 2 = 9 / 2 = 4.5 in
• I = 112.8 in4

$\sigma _{bend-max}=\frac{M.y}{I}$

$\sigma _{bend-max}=\frac{103320\times 4.5}{112.8}=>\mathbf{\sigma _{bend-max}=-4121.81\:psi\:(Ans.)}$

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F)

Total stress = $\mathbf{\sigma _{max}}=-4121.81-68.07=\mathbf{-4189.88\:psi\:(Ans.)}$

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