Question

Chapter 3, Reserve Problem 122 The rigid bar AC is supported by two axial bars (1) and (2). Both axial bars are made of bronze [E = 100 GPa; a = 18 × 10-mm/mm/°C]. The cross-sectional area of bar (1) is A1 = 236 mm2 and the cross- sectional area of bar (2) is Az = 389 mm2. After load P has been applied and the temperature of the entire assembly has increased by 26°C, the total strain in bar (2) is measured as 1450 ue (elongation). Assume L1 = 1490 mm, a = 590 mm, b = 910 mm, and L2 = 2130 mm. Determine: (a) the magnitude of load P. (b) the vertical displacement of pin A.

Answer 1

Given: E=100 GP a=18x10 mm/mm/°C 4 = 236 mm A2 = 389 mm? AT = 26°C Ez = 1450 uɛ 4 =1490 mm a= 590 mm b=910 mm L = 2130 mm

(a) Free body diagram: 2130 1490 590 910 All dimensions are in mm

The thermal strain: Et =AAT =18x10- 26 = 4.68x10-4 mm/mm Axial strain in member 2: E2,0 = &z – E2,1 = (1450x10-)-4.68x104 = 0.000982 mm/mm The stress in member (2) is: 02 = E 2.0 = 100x10 MPa 0.000982 = 98.2 MPa = 98.2 N/mm2

Hence, the force in the member 2 is: F, = 0242 =98.2x389 = 38199.8 N Consider moment about A: M = 0 -P(590)+F (910+590)=0 -P(590)+38199.8(910+590)=0 P=97118.13 N P=97.118 kN F, = 0 F +F2 - P=0 F. +38199.8-97118.13=0 F = 58918.34 N

(b) The stress in member (1): 58918.34 236 = 249.654 N/mm2 The normal strain in member (1): €1,0 = 249.654 100x103 = 0.0024965 mm/mm

The thermal strain in member (1): EL1 = WAT = 18x10-6x26 = 4.68x10+ mm/mm The total strain member (1): & = €1,0 + ELI = 0.0024965+4.68x104 = 2.9645x10 mm/mm The elongation in the bar (1) is: e = &L = 2.9645x10²x1490 = 4.4171 mm Therefore, the displacement at joint A is: VA= e VA = 4.4171 mm(1)

Answers:

P = 97.118 kN

Displacement = 4.417 mm