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MECHANICS OF MATERIALS
Department of Civil Engineering and Electromechanical Engineering
Midterm Exam
Name: Student ID:
Date: 2021/04/28 Time: 08:10-09:
- The force applied at the handle of the rigid lever causes the lever to rotate clockwise about the pin B through an angle of 5ยฐ. Determine the average normal strain in each wire. The wires are unstretched when the lever is in the horizontal position. (20%)
200 mm
100 mm 400 mm 200 mm 400 mm
Figure (1)
Solution: Geometry
The lever arm rotates through an angle of ๐ = ( 1805 ยฐ) ฯ rad = 0.8727 rad. Since ฮธ is small, the displacements of points A, C, and D can be approximated by
๐ฟ๐ด = 100 ร ( 1805 ยฐ) ฯ = 8.7266 mm
๐ฟ๐ถ = 400 ร ( 1805 ยฐ) ฯ = 34.9066 mm
๐ฟ๐ท = 600 ร ( 1805 ยฐ) ฯ = 52.3599 mm Average Normal Strain The unstretched length of wires AH, CG, and DF are LAH = 200 mm , LCG = 400 mm , and LDF = 400 mm. We obtain (๐๐๐ฃ๐)๐ด๐ป = (^) ๐ฟ๐ฟ๐ด๐ป๐ด = 8.7267 200 = ๐. ๐๐๐๐ mm/mm Ans.
(๐๐๐ฃ๐)๐ถ๐บ = (^) ๐ฟ๐ฟ๐ถ๐บ๐ถ = 34.9066 400 = ๐. ๐๐๐๐ mm/mm Ans.
(๐๐๐ฃ๐)๐ท๐น = (^) ๐ฟ๐ฟ๐ท๐น๐ท = 52.3599 400 = ๐. ๐๐๐๐ mm/mm Ans.
- The weight of the kentledge exerts an axial force of P = 1750 kN on the 300-mm diameter high-strength concrete bore pile. If the distribution of the resisting skin friction developed from the interaction between the soil and the surface of the pile is approximated as shown, determine the resisting bearing force F for equilibrium. Take po = 188 kN/m. Also, find the corresponding elastic shortening of the pile. Neglect the weight of the pile. E = 29 GPa (20%)
15 m
Figure (2)
Solution: Internal Loading By considering the equilibrium of the pile with reference to its entire free-body diagram. We have +โ โ ๐น๐ฆ = 0; ๐น + 1882 ร 15 โ 1750 = 0 โน ๐น = ๐๐๐ kN Ans. Also, ๐(๐ฆ) = 18815 ๐ฆ = 12.53๐ฆ kN/m The normal force developed in the pile as a function of y can be determined by considering the equilibrium of the sectional of the pile with reference to its free-body diagram. +โ โ ๐น๐ฆ = 0; 340 + 12 (12.53๐ฆ)๐ฆ โ ๐(๐ฆ) = 0 โน ๐(๐ฆ) = 6.267๐ฆ^2 + 340 kN Displacement The cross-sectional area of the pile is ๐ด = ฯ 4 ร 0.3^2 = 0.0225ฯ m2. We have ฮด = โซ 0 L A(y)EP(y) dy= โซ 0 15 (6.267y0.0225ฯ(29.0)(10^2 +340)(10 (^93) )) dy= โซ 0 15 [3.057(10โ6)๐ฆ^2 + 165.863(10โ6)]๐๐ฆ
= [1.019(10โ6)๐ฆ^3 + 165.863(10โ6)๐ฆ] 015 = ๐. ๐๐๐ mm Ans.
- Elements (1) and (2) were supposed to fit exactly between the rigid walls at A and C in Figure (4). However, element (1) was manufactured a small amount, ฮด โช L , too long. However, an ingenious technician was able to compress element (1) and insert it in the space between A and B (indicated by dashed lines). (a) Determine expressions for the stress ฯ 1 induced in element (1) and the stress ฯ 2 induced in element (2). (10%) (b) Determine the amount uB by which element (2) is shortened when element (1) is forced into the dashed-line position. (10%) Assume ฮด โช L in calculating the flexibility coefficient f 1. Where A 1 = A , A 2 = 1.5A ; E 1 = 2E , E 2 = E.
L L
L + ฮด
A
B
uB C
Elements ( 1 )
Elements ( 2 )
Figure (4)
Solution (a) Stresses ฯ 1 and ฯ 2 Equilibrium โ ๐น๐ฅ = 0; ๐น 2 = ๐น 1 (1) Element force - deformation behavior: ๐ 1 = (^) 2๐ด๐ธ๐ฟ ๐ 2 = (^) 3๐ด๐ธ2๐ฟ
Geometry of deformation: ๐ 1 + ๐ 2 = โ๐ฟ (3) Combine (1), (2) and (3) ๐ฟ๐น 2๐ด๐ธ +^
2๐ฟ๐น 3๐ด๐ธ = โ๐ฟ โ ๐น = โ^
6๐ด๐ธ 7๐ฟ ๐ฟ Stresses
๐ 1 = ๐ด๐น^11 =
โ6๐ด๐ธ7๐ฟ ๐ฟ
๐ด =^ โ^
๐๐ฌ ๐๐ณ ๐น^ Ans.
๐ 2 = ๐ด๐น^22 =
โ6๐ด๐ธ7๐ฟ ๐ฟ 3 2 ๐ด^
= โ ๐๐ฌ๐๐ณ ๐น Ans.
(b) Displacement uB
๐ข๐ต = โ๐ 2 = โ๐ 2 ๐น 2 = โ ( 3๐ด๐ธ2๐ฟ) (โ 6๐ด๐ธ7๐ฟ ๐ฟ) = ๐๐ ๐น Ans.
- The composite shaft shown in the Figure (5) is manufactured by shrink-fitting a steel sleeve over a brass core so that the two parts act as a single solid bar in torsion. The outer diameters of the two parts are d 1 = 35 mm for the brass core and d 2 = 50 mm for the steel sleeve. The shear moduli of elasticity are Gb = 36 GPa for the brass and Gs = 80 GPa for the steel. (a) Assuming that the allowable shear stresses in the brass and steel are ฯb = 31 MPa and ฯs = 52 MPa , respectively, determine the maximum permissible torque Tmax that may be applied to the shaft. (10%) (b) If the applied torque T = 1150 Nโm , find the required diameter d 2 so that allowable shear stress s is reached in the steel. (10%)
T
T
Figure (5)
Solution d 1 = 35 mm = 0.035 m ; d2 = 50 mm = 0.05 m ; Gb = 36 GPa ; Gs = 80 GPa ; ฯb = 31 MPa ; ฯs = 52 MPa ; (a) ๐ผ๐๐ = 32 ๐ ๐ 14 = 1.4732 ร 10โ7^ ๐^4 ; ๐ผ๐๐ = 32 ๐ (๐ 24 โ ๐ 14 ) = 4.6627 ร 10โ7^ ๐^4 ๐๐ = ๐ ( (^) ๐บ๐๐ผ๐บ๐๐๐+๐บ๐ผ๐๐๐ ๐ผ๐๐ ) and ๐๐ = ๐ ( (^) ๐บ๐๐ผ๐๐๐บ๐ ๐ผ+๐บ๐๐ ๐ ๐ผ๐๐ ) (1)
๐๐โ๐๐๐ฅ = 2๐๐ ๐๐ผ 1 ๐๐ ; ๐๐ โ๐๐๐ฅ = 2๐ ๐๐ ๐ผ 2 ๐๐ (2) (2) into (1), and solving for Tmax for either brass or steel Tmax,b = 2 ฯb d 1 Ipb (GbIpb Gb+IpbGsIps ) = 2096. 44 Nโm
Tmax,s = 2 ฯ dsI 2 ps (GbIpb Gs+IpsGsIps ) = 1107. 73 Nโm โ(controls)
so Tmax = 1107.73 Nโm Ans. (b) LetT = 1150 Nโm , so Tmax = T Expand Ips and Ipb in above expression for T max,s, then for required diameter d 2 T = 2ฯs[^ 32 ฯ(d^24 โd^14 )] d 2 {
Gb( 32 ฯd 14 )+Gs[ 32 ฯ(d 24 โd 14 )] Gs[ 32 ฯ(d 24 โd 14 )] } โ ๐ 2 = 0.05052 ๐ = 50.52 ๐๐ Take an integer and must be greater than 50.52 mm , so ๐
๐ = ๐๐ ๐๐ Ans.
