1.5 in. 30° A B 1.5 in. 1.5 in The A-36 steel bar AB has a square cross section, and is pin connected on both ends. Esteel = 29×103 ksi. Determine: The maximum allowable load P that can be applied to the frame before AB buckles. Use a SF=2.28 Por SF = = π2E1 (K2)2 Per P => Solve for inertia =>plug in to Solve for P 10 ft | = bh³ (1.5) (1.5) ³ I = Pc₁ = 1 12 P =0.4218 -π² EI _ πT² (29×103) (0.4218)=8.3838 Cr (KL)2 SF = Per P = 42.28 = 8.3838 P P=3.6771 (1-10-12)² here Im not sure what im doing wrong 56 of 59

Mechanics of Materials (MindTap Course List)
9th Edition
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
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Chapter11: Columns
Section: Chapter Questions
Problem 11.9.30P
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please help, i am not sure what i am doing wrong

1.5 in.
30°
A
B
1.5 in.
1.5 in
The A-36 steel bar AB has a square cross section, and is
pin connected on both ends. Esteel = 29×103 ksi.
Determine:
The maximum allowable load P that can be applied to
the frame before AB buckles. Use a SF=2.28
Por
SF =
=
π2E1
(K2)2
Per
P
=> Solve for inertia
=>plug in to Solve for P
10 ft
| = bh³ (1.5) (1.5) ³
I =
Pc₁ = 1
12
P
=0.4218
-π² EI _ πT² (29×103) (0.4218)=8.3838
Cr (KL)2
SF = Per
P
=
42.28 = 8.3838
P
P=3.6771
(1-10-12)²
here
Im not sure what im
doing wrong
56 of 59
Transcribed Image Text:1.5 in. 30° A B 1.5 in. 1.5 in The A-36 steel bar AB has a square cross section, and is pin connected on both ends. Esteel = 29×103 ksi. Determine: The maximum allowable load P that can be applied to the frame before AB buckles. Use a SF=2.28 Por SF = = π2E1 (K2)2 Per P => Solve for inertia =>plug in to Solve for P 10 ft | = bh³ (1.5) (1.5) ³ I = Pc₁ = 1 12 P =0.4218 -π² EI _ πT² (29×103) (0.4218)=8.3838 Cr (KL)2 SF = Per P = 42.28 = 8.3838 P P=3.6771 (1-10-12)² here Im not sure what im doing wrong 56 of 59
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