Question

Question 21 A W12x14 A36 steel section is used as a column. If the column is 15 ft long with pin supports at each end, then the largest permissible compressive axial load is most nearly... 0 783 kips O 150 kips 0 241 kips O 21 kips Question 22 AW12x14 A36 steel section is used as a column. If the column is 5 ft long with pin supports at each end, then the largest permissible compressive axial load is most nearly... O 188 kips O 7044 kips 0 241 kips 150 kips

Answer 1

Solution 21:-

for pin-pin end condtions about the minor axis (y-axis), we have

$K_{y}=1.0 \ \ \left ( Recommended \ design \ value \right )$

for pin-pin end condtions about the major axis (x-axis), we have

$K_{x}=1.0 \ \ \left ( Recommended \ design \ value \right )$

According to the problem statement, the unsupported length for buckling about the major axis = Lx = 15 ft.

= 1.0 x 15 ft = 180 in

= 1.0 x 15 ft = 180 in

For W12 x 14: elastic modulus = E = 29000 ksi (constant for all steels)

$For \ W12\times 14: \ \ I_{x}=88.6 \ in^{4} \ \ I_{y}=2.4 \ in^{4}$

$Critical \ load \ for \ buckling \ about \ x -axis = P_{cr-x}=\frac{\pi ^{2}EI_{x}}{\left ( K_{x}L_{x} \right )^{2}}$

$P_{cr-x}=\frac{\pi ^{2}\times 29000\times 88.6}{\left ( 180 \right )^{2}}=783 \ ksi$

$Critical \ load \ for \ buckling \ about \ y -axis = P_{cr-y}=\frac{\pi ^{2}EI_{y}}{\left ( K_{y}L_{y} \right )^{2}}$

$P_{cr-y}=\frac{\pi ^{2}\times 29000\times 2.4}{\left ( 180 \right )^{2}}=21.2 \ ksi$​​​​​​​

so the largest permissible compressive load is 783 ksi..........................Answer.

Correct option is A.

Solution 22:-

for pin-pin end condtions about the minor axis (y-axis), we have

$K_{y}=1.0 \ \ \left ( Recommended \ design \ value \right )$

for pin-pin end condtions about the major axis (x-axis), we have

$K_{x}=1.0 \ \ \left ( Recommended \ design \ value \right )$

According to the problem statement, the unsupported length for buckling about the major axis = Lx = 5 ft.

$=1.0\times 5 \ ft = 60 \ in$

$=1.0\times 5 \ ft = 60 \ in$​​​​​​​

For W12 x 14: elastic modulus = E = 29000 ksi (constant for all steels)

$For \ W12\times 14: \ \ I_{x}=88.6 \ in^{4} \ \ I_{y}=2.4 \ in^{4}$

$Critical \ load \ for \ buckling \ about \ x -axis = P_{cr-x}=\frac{\pi ^{2}EI_{x}}{\left ( K_{x}L_{x} \right )^{2}}$

$P_{cr-x}=\frac{\pi ^{2}\times 29000\times 88.6}{\left ( 60 \right )^{2}}=7044.15 \ ksi$​​​​​​​

$Critical \ load \ for \ buckling \ about \ y -axis = P_{cr-y}=\frac{\pi ^{2}EI_{y}}{\left ( K_{y}L_{y} \right )^{2}}$

$P_{cr-y}=\frac{\pi ^{2}\times 29000\times 2.4}{\left ( 60 \right )^{2}}=190.8 \ ksi$​​​​​​​

so the largest permissible compressive load is 7044 ksi..........................Answer.

Correct option is B.

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