Problem_12_35

# Problem_12_35 - 2 ⋅ + 2.306 10 6 × mm 4 ⋅ = := I y t b...

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Problem 12.35 Given: A stainless steel column has the tangent-modulus curve shown in Figure (a) and a cross section illustrated in Figure (b). Its length is 2.20 m and it has a clamped end at the top and pinned end at the bottom. Find: Determine the design load for the column based on a factor of safety SF=2.50. Solution: b 100 mm := t2 0 m m := h7 0 m m := L 2.20 m := SF 2.5 := Area of section: Ab t ht + 3400 mm 2 = := Location of centroid: y c 1 A bt t 2 ht t h 2 + + 28.529 mm = := I x bt 3 12 bt y c t 2 2 + th 3 12 + th h 2 t + y c
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Unformatted text preview: 2 ⋅ + 2.306 10 6 × mm 4 ⋅ = := I y t b 3 ⋅ 12 h t 3 ⋅ 12 + 1.713 10 6 × mm 4 ⋅ = := x y y c r x I x A 26.043 mm ⋅ = := r y I y A 22.448 mm ⋅ = := From Table 12.1 k 0.7 := From the figure at: k L ⋅ r x 59.133 = σ cr 300 MPa ⋅ := approximately P crY σ cr A ⋅ SF 408 kN ⋅ = := From the figure at: k L ⋅ r y 68.602 = σ cr 270 MPa ⋅ := approximately P crX σ cr A ⋅ SF 367.2 kN ⋅ = := Since P crX P crY < then the design load is: P crX 367.2 kN ⋅ =...
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## This note was uploaded on 01/11/2011 for the course MAE 3040 taught by Professor Mechanicsofsolids during the Spring '10 term at Utah State University.

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