Problem 7.9. Critical load of frame

Replace the eight-storey frame given in the Figure by a continuous beam. Determine the stiffnesses and the critical force of the replacement beam. Stiffness of the beams and columns are: EI_b = 2.4×10⁷ kNcm², EI_c= 3.2×10⁷ kNcm², EA = 2.6×10⁵kN.
a) Assume concentrated load at the top.
b) Assume identical loads at each level.

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Problem a)

Critical force, N_cr[kN]=

Problem b)

Critical force, N_cr[kN]=

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Steps
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Step 1. Calculate the shear stiffness of the replacement beam.

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Eq.(7-169)

S_{c} = 2 \frac{π^{2} E I_{c}}{h^{2}} = 2 \frac{π^{2} \times 3.2 \times 10^{7}}{300^{2}} = 7018 kN S_{b} = \frac{12 E I_{b}}{d h} = \frac{12 \times 3.2 \times 10^{7}}{600 \times 300} = 1600 kN S = {(\frac{1}{S_{c}} + \frac{1}{S_{b}})}^{- 1} = {(\frac{1}{7018.4} + \frac{1}{1600})}^{- 1} = 1303.0 kN

Step 2. Calculate the replacement bending stiffness.

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Eq.(7-171)

E I = \frac{E A_{c} d^{2}}{2} = \frac{2.6 \times 10^{5} \times 6^{2}}{2} = 4.68 \times 10^{6} {kNm}^{2}

Problem a)

Step 3. Give the smallest buckling load of the replacement beam with concentrated load at the top.

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Eq.(7-165)

N_{c r} = {(\frac{1}{\frac{π^{2} E I}{4 L^{2}}} + \frac{1}{S})}^{- 1} = {(\frac{1}{\frac{π^{2} \times 4.68 \times 10^{6}}{4 \times {(3 \times 8)}^{2}}} + \frac{1}{1303.0})}^{- 1} = 1223.4 kN

Problem b)

Step 3. Give the smallest buckling load of the replacement beam with identical loads at each level.

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Eq.(7-172)

N_{c r} = {(\frac{1}{7.8 \frac{E I}{H^{2}}} + \frac{1}{S})}^{- 1} = {(\frac{1}{7.8 \frac{4.68 \times 10^{6}}{{(3 \times 8)}^{2}}} + \frac{1}{1303.0})}^{- 1} = 1277 kN

Results
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The shear stiffness of the replacement beam is calculated as

Eq.(7-169)

S_{c} = 2 \frac{π^{2} E I_{c}}{h^{2}} = 2 \frac{π^{2} \times 3.2 \times 10^{7}}{300^{2}} = 7018 kN S_{b} = \frac{12 E I_{b}}{d h} = \frac{12 \times 3.2 \times 10^{7}}{600 \times 300} = 1600 kN S = {(\frac{1}{S_{c}} + \frac{1}{S_{b}})}^{- 1} = {(\frac{1}{7018.4} + \frac{1}{1600})}^{- 1} = 1303.0 kN

The replacement bending stiffness is

Eq.(7-171)

E I = \frac{E A_{c} d^{2}}{2} = \frac{2.6 \times 10^{5} \times 6^{2}}{2} = 4.68 \times 10^{6} {kNm}^{2}

Problem a)

The smallest buckling load of the replacement beam with concentrated load at the top is approximated as

Eq.(7-165)

N_{c r} = {(\frac{1}{\frac{π^{2} E I}{4 L^{2}}} + \frac{1}{S})}^{- 1} = {(\frac{1}{\frac{π^{2} \times 4.68 \times 10^{6}}{4 \times {(3 \times 8)}^{2}}} + \frac{1}{1303.0})}^{- 1} = 1223.4 kN

Problem b)

The smallest buckling load of the replacement beam with identical loads at each level is approximated as

Eq.(7-172)

N_{c r} = {(\frac{1}{7.8 \frac{E I}{H^{2}}} + \frac{1}{S})}^{- 1} = {(\frac{1}{7.8 \frac{4.68 \times 10^{6}}{{(3 \times 8)}^{2}}} + \frac{1}{1303.0})}^{- 1} = 1277 kN