Problem Set
This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA
Overview
Download & View Problem Set as PDF for free.
More details
- Words: 591
- Pages: 7
1. Determine the largest W, which can be supported by the 2 wires shown. The stress in wires AB and AC are not to exceed 10 MPa and 150 MPa, respectively. The cross sectional areas of the 2 wires are 400 mm² for wire AC.
Σfy = 0 AC sin 75 = W sin 60 AC = .897W ΣFx = 0 - AB cos 30 + AC cos 45 - 0.897W cos 45 = 0 AB = 0.732W P = σS For AC PAC = 200(150) PAC = 30,000N 0.897W = 30000N W = 33,445 For AB PAB = 400(100) PAB = 40000N 0.732W = 40000 W = 54,645 N The largest safe weight is 33.5 kN
2. For the truss shown at the right, determine the cross sectional areas of bars BE, BF, and CF so that the stresses will not exceed 100 MPa in tension or 80 MPa in compression. A reduced stress in compression is specified to avoid the danger of buckling.
ΣMf = 0 - 3/5 BE (4) +| 50(3) = 0 BE = 62.5kN (T) ΣFy = 0 8/ 13 BF + 4/5 (62.5) = 40 + 50 BF = 42.72 kN (T) A=P/σ For CF ACF = 52.5 (1000) 80 = 656 mm² for BE ABE = 62.5 (1000) 100 = 625 mm² for BF ABF = 42.72 (1000) 100 = 427.2 mm²
3. For the truss shown below,calculate the stress in members CE, DE, DF. The cross sectionalareaof each member is 1125 mm².indicate tension (T) and compression (C).
4. A cast-iron column supports an axial compressive load of 250 kN. Determine the inside diameter of the column of its outside diameter is 200 mm and the limiting compressive stress is 150 MPa.
A = π / 4 (D2² - D1²) A = π / 4 (40000 - D1²) P = Aσ 250 (1000) = π / 4 (40000 - D1²) 40000 - D1² = 6366.2 D1 = 183.4mm
5. A steel tube is rigidly attached between an aluminum rod and a bronze rod as shown below. Axial loads are applied at the positions indicated. Find the maximum value of P that will not exceed a stress in aluminum of 80 MPa, in steel of 150 MPa, or in bronze of 100 MPa.
For aluminum: 80 = P / 200 P = 16000N For steel: 150 = 2P / 400 P = 30000N For bronze: 100 = 4P / 500 P= 12500 N
6. A homogeneous 150 kg bar AB carries a 2 kN force as shown. The bar is supported by a pin at B and a 10 mm diameter cable CD. Determine the stress in the cable.
W = 150 (9.8) / 1000kN ΣMB = 0 4/5 T (3) = 150(9.8) / 1000 (3) + 12 T = 6.84 kN σ=P/A σ = 6.84 (1000) π/4 (10)² σ = 87.08 MPa
7. For the lap joint in problem no. 14, determine the maximum safe load P which may beapplied if the shearing stress in the rivets is limited to 60 MPa, the bearing stress in the plates to 110 MPa, and the average tensile stress in the plate to 140 MPa.
(a) from shearing of the rivets P = σA P = π / 4(20)² (60)(3) P = 56.5kN (b) from bearing of the plates P = σA P = 20(25)(110)(3) P = 165kN (c) from tension in the plates A = 130 – 20(25) A = 2750mm² P = σA P = 2750(140) P = 385kN The maximum safe load is 56.5kN
Σfy = 0 AC sin 75 = W sin 60 AC = .897W ΣFx = 0 - AB cos 30 + AC cos 45 - 0.897W cos 45 = 0 AB = 0.732W P = σS For AC PAC = 200(150) PAC = 30,000N 0.897W = 30000N W = 33,445 For AB PAB = 400(100) PAB = 40000N 0.732W = 40000 W = 54,645 N The largest safe weight is 33.5 kN
2. For the truss shown at the right, determine the cross sectional areas of bars BE, BF, and CF so that the stresses will not exceed 100 MPa in tension or 80 MPa in compression. A reduced stress in compression is specified to avoid the danger of buckling.
ΣMf = 0 - 3/5 BE (4) +| 50(3) = 0 BE = 62.5kN (T) ΣFy = 0 8/ 13 BF + 4/5 (62.5) = 40 + 50 BF = 42.72 kN (T) A=P/σ For CF ACF = 52.5 (1000) 80 = 656 mm² for BE ABE = 62.5 (1000) 100 = 625 mm² for BF ABF = 42.72 (1000) 100 = 427.2 mm²
3. For the truss shown below,calculate the stress in members CE, DE, DF. The cross sectionalareaof each member is 1125 mm².indicate tension (T) and compression (C).
4. A cast-iron column supports an axial compressive load of 250 kN. Determine the inside diameter of the column of its outside diameter is 200 mm and the limiting compressive stress is 150 MPa.
A = π / 4 (D2² - D1²) A = π / 4 (40000 - D1²) P = Aσ 250 (1000) = π / 4 (40000 - D1²) 40000 - D1² = 6366.2 D1 = 183.4mm
5. A steel tube is rigidly attached between an aluminum rod and a bronze rod as shown below. Axial loads are applied at the positions indicated. Find the maximum value of P that will not exceed a stress in aluminum of 80 MPa, in steel of 150 MPa, or in bronze of 100 MPa.
For aluminum: 80 = P / 200 P = 16000N For steel: 150 = 2P / 400 P = 30000N For bronze: 100 = 4P / 500 P= 12500 N
6. A homogeneous 150 kg bar AB carries a 2 kN force as shown. The bar is supported by a pin at B and a 10 mm diameter cable CD. Determine the stress in the cable.
W = 150 (9.8) / 1000kN ΣMB = 0 4/5 T (3) = 150(9.8) / 1000 (3) + 12 T = 6.84 kN σ=P/A σ = 6.84 (1000) π/4 (10)² σ = 87.08 MPa
7. For the lap joint in problem no. 14, determine the maximum safe load P which may beapplied if the shearing stress in the rivets is limited to 60 MPa, the bearing stress in the plates to 110 MPa, and the average tensile stress in the plate to 140 MPa.
(a) from shearing of the rivets P = σA P = π / 4(20)² (60)(3) P = 56.5kN (b) from bearing of the plates P = σA P = 20(25)(110)(3) P = 165kN (c) from tension in the plates A = 130 – 20(25) A = 2750mm² P = σA P = 2750(140) P = 385kN The maximum safe load is 56.5kN
Related Documents
Problem Set
May 2020 13
Problem Set
November 2019 37
Problem Set 3, Cs181
July 2019 27
Problem Set 4
June 2020 6
Problem Set 2
June 2020 4