CIV 207 Winter For practice

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1 CIV 07 Winter 009 Assignment #10 Friday, March 0 th Complete the first three questions. Submit your work to Box #5 on the th floor of the MacDonald building by 1 noon on Tuesday March 31 st. No late submissions will be accepted. The second set of three questions are for practice only. Groups of up to 3 members are permitted. Clearly print your names and student numbers on your submission. Ensure that the submission is properly stapled The aluminum column shown has a rectangular cross section and supports an axial load of P. The base of the column is fixed. The support at the top s rotation of the column in the x-y plane (i.e., bending about the strong axis) but prevents rotation in the x-z plane (i.e., bending about the weak axis). (a) Determine the critical buckling load of the column for the following parameters: L = 50 in., b = 0.50 in., h = in., and E = 10,000 ksi. (b) Determine the ratio b/h for which the critical buckling load about both the strong and weak axes is the same A compression chord of a small truss consists of two L steel angles arranged with long legs back-to-back as shown. The angles are separated at intervals by spacer blocks. (a) Determine the spacer thickness required so that the moments of inertia for the section about the two principal axes are equal. (b) For a compression chord with an effective length of KL = 7 m, determine the able axial load P that may be supported by the column using the spacer thickness determined in part (a). Use the AISC equations and assume E = 00 GPa and σy = 30 MPa A 6061-T6 aluminum-alloy rectangular tube shape has the dimensions shown. The rectangular tube is used as a compression member that is.5-m long. Both ends of the compression member are fixed. Determine the able axial load P that may be supported by the rectangular tube. Use the Aluminum Association column design formulas. For practice

2 16.18 The assembly shown in Fig. P16.18 consists of two solid circular steel [E = 00 GPa] rods (1) and (). Assume that the rods are pin-connected and that joint B is restrained against translation in the z direction. If a load of P = 60 kn is applied to the assembly, determine the minimum rod diameters required if a factor of safety of 3.0 is specified for each rod. Fig. P16.18 Method of joints:.0 m tanθab = = θab = m 1. m tanθbc = = θbc = m Equilibrium of joint B: Σ Fx = FBCcos( ) FABcos( ) = 0 Σ Fy = FBCsin( ) FABsin( ) P = 0 Note: Tension assumed in each truss member. Solve these equations simultaneously to obtain: FAB = kn FBC = kn Since members AB and BC are compression members, the Euler buckling loads for these two members must be investigated. Euler buckling load for member AB: L AB = (1. m) + (.0 m) =.1311 m =, mm To attain a factor of safety of 3.0, the critical buckling load of member AB must be: Pcr = 3FAB = 3( kn) = kn = 17, N Therefore, the required moment of inertia for member AB is: π EI 17, N L (17, N)(,1.311 mm) I =, mm π (00,000 N/mm ) Therefore, solid rod AB must have a minimum diameter of: π, mm 5.8 mm 6 D D

3 Euler buckling load for member BC: L BC = (1.7 m) + (1. m) = m =, mm To attain a factor of safety of 3.0, the critical buckling load of member BC must be: Pcr = 3FBC = 3( kn) = kn = 103,.0 N Therefore, the required moment of inertia for member AB is: π EI 103,.0 N L (103,.0 N)(, mm) I = 6, mm π (00,000 N/mm ) Therefore, solid rod BC must have a minimum diameter of: π 6, mm 6.3 mm 6 D D

4 16.75 A Select Structural grade Hem-Fir (F c = 1,500 psi; E = 1,600,000 psi) wood column of rectangular cross section has finished dimensions of b =.50 in. and h = 9.5 in. The length of the column is L = 18 ft. The column is fixed at base A. Pinconnected lateral bracing is present at B so that deflection in the x-z plane is restrained at the upper end of the column; however, the column is free to deflect in the x-y plane at B (see Fig. P16.75). Use the NFPA NDS column design formula to determine the able compressive load P that the column can support. In your analysis, consider the possibility that buckling could occur about either the strong axis (i.e., the z axis) or the weak axis (i.e., the y axis) of the wood column. Fig. P16.75 KL (0.7)(18 ft)(1 in./ft) KL (.0)(18 ft)(1 in./ft) = = 33.6 = = d1.5 in. d 9.5 in. KcEE (0.3)(1,600,000 psi) FcE psi FcE = = = psi = = ( KL / d) (6.707) Fc 1,500 psi 1 + ( FcE / Fc ) 1+ ( FcE / Fc ) FcE / F c σ = Fc c c c 1 + (0.167) 1 + (0.167) = (1,500 psi) = psi (0.8) (0.8) 0.8 P = ( psi)(.5 in.)(7.5 in.) = 8,866.8 lb = 8,870 lb

5 16.36 The aluminum column shown in Fig. P16.36 has a rectangular cross section and supports an axial load of P. The base of the column is fixed. The support at the top s rotation of the column in the x-y plane (i.e., bending about the strong axis) but prevents rotation in the x-z plane (i.e., bending about the weak axis). (a) Determine the critical buckling load of the column for the following parameters: L = 50 in., b = 0.50 in., h = in., and E = 10,000 ksi. (b) Determine the ratio b/h for which the critical buckling load about both the strong and weak axes is the same. Fig. P16.36 Section properties: I (0.50 in.)(0.875 in.) (0.875 in.)(0.50 in.) strong = = in. Iweak = = in. 1 1 (a) Critical buckling load: Consider buckling about strong axis: K = 0.7 (fixed-pinned column) P cr π EI π (10,000,000 psi)( in. ) = = =,8.933 lb [ ] ( KL) (0.7)(50 in.) Consider buckling about weak axis: K = 0.5 (fixed-fixed column) P cr π EI π (10,000,000 psi)( in. ) = = = 1, lb [ ] ( KL) (0.5)(50 in.) Critical load: P = 1, 39 lb cr

6 (b) Ratio of b/h to give same buckling load: Equate the two buckling equations: bh hb π (10,000,000 psi) π (10,000,000 psi) 1 1 = [(0.7)(50 in.) ] [(0.5)(50 in.) ] bh hb = (0.7) (0.5) (0.5) = b / h (0.7) b/ h= 0.71

7 16.39 A stainless steel pipe with an outside diameter of 100 mm and a wall thickness of 8 mm is rigidly attached to fixed supports at A and B. The length of the pipe is L = 8 m, its elastic modulus is E = 190 GPa, and its coefficient of thermal expansion is α = mm/mm/ C. Determine the temperature increase ΔT that will cause the pipe to buckle. Fig. P16.39 Section properties: d = 100 mm (8 mm) = 8 mm π A = (100 mm) (8 mm),31.1 mm = π I = (100 mm) (8 mm) =, 6,818 mm 6 Critical buckling load: K = 0.5 (fixed-fixed column) π EI π (190,000 N/mm )(,6,818 mm ) Pcr = = = 88,880.5 N ( KL) (0.5)(8,000 mm) [ ] Force-Temperature-Deformation Relationship The relationship between internal force, temperature change, and deformation of an axial member is: FL e = TL AE + α Δ Since the rod is attached to rigid supports, e = 0. FL TL 0 AE + α Δ = Set F = P cr PL cr TL 0 AE + α Δ = and solve for ΔT: P cr Δ T = α AE 88,880.5 N = 6 ( in./in./ C)(,31.1 mm )(190,000 N/mm ) = C = 38.0 C

8 16.6 A compression chord of a small truss consists of two L steel angles arranged with long legs back-to-back as shown in Fig. P16.6. The angles are separated at intervals by spacer blocks. (a) Determine the spacer thickness required so that the moments of inertia for the section about the two principal axes are equal. (b) For a compression chord with an effective length of KL = 7 m, determine the able axial load P that may be supported by the column using the spacer thickness determined in part (a). Use the AISC equations and assume E = 00 GPa and σ Y = 30 MPa. Fig. P16.6 The following section properties for a standard steel L shape are given in Appendix B: A =,0 mm, I x = mm, r x = 0.1 mm, I y = mm, x = 18.9 mm (a) Determine spacer thickness b: 6 6 I x = ( mm ) = mm 6 Iy = mm + ( b/ mm) (,0 mm ) Equate these two moment of inertia expressions and solve for the block thickness b: b = mm = 31.1 mm (b) Determine the able axial load: π E π (00,000 MPa) Cc = = = σ Y 30 MPa Slenderness ratio: mm KL 7,000 mm r = = 0.98 mm = = (,0 mm ) r 0.98 mm Allowable compression stress: Since KL / r > C c, the column is classified as a long column, and Eq. (16.5) is used to calculate the able compression stress: 1π E 1 π (00,000 MPa) σ = = = 3.13 MPa 3( KL / r) 3(173.70) Allowable axial load: P = σ A= (3.13 N/mm )(,0 mm ) = 165. kn

9 16.68 A 6061-T6 aluminum-alloy rectangular tube shape has the dimensions shown in Fig. P The rectangular tube is used as a compression member that is.5-m long. Both ends of the compression member are fixed. Determine the able axial load P that may be supported by the rectangular tube. Use the Aluminum Association column design formulas. Fig. P16.68 Section properties: A = (50 mm)(76 mm) (0 mm)(66 mm) = 1,160 mm I z y y (50 mm)(76 mm) (0 mm)(66 mm) = = 870,76.67 mm 1 870,76.67 mm rz = = mm 1,160 mm I r (76 mm)(50 mm) (66 mm)(0 mm) = = 39, mm 1 39, mm = = mm 1,160 mm Effective-slenderness ratios: KL (0.5)(,500 mm) KL (0.5)(,500 mm) = = 5.65 = = 6.06 r mm r mm z Aluminum Association column design formula: KL σ = [ ( KL / r) ] MPa where 9.5 < 66 r = (6.06) MPa [ ] y = MPa Allowable axial load P : P = σ A= (83.69 N/mm )(1,160 mm ) = 96,59 N = 96.6 kn