The internal loadings at a cross section through the 120-mm-diameter drive shaft of a turbine consist of an axial force of 12.5 kN, a bending moment of 1.2 kN - m, and a torsional moment of 2.25 kN-m (Eigure 1). .
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- Determine the torsional shear stress in the shaft section AB and BC. There is a fixed support at C that can resist a moment. (Image taken from Statics and Mechanics of Materials by Beer, Johnson, DeWolf and Mazurek) T₁ = 300 N-m 30 mm TB = 400 N-m 46 mm 055 Shear Stress in AB = Shear Stress in BC= B 0.9 m 0.75 m units: units:Find the shear force and bending moment at points B and D. Note: B lies just to the right of the 150 lbf force and D is just to the right of the bearing at C. The bearing at A is a thrust bearing, while the bearing at C is a journal bearing. 150 lb A Answer: VB = -100 lbf MB = 750 lbf in VD = 75lbf Mp = -750 lbf in I 15 in. B 15 in. C. D 75 lb 10 in.The bending moment M is applied to the box beam in the x-y plane at the orientation show below. (a) Determine the maximum magnitude of the bending moment M so that the tensile bending stress in the member does not exceed 15 ksi. (b) Determine the angle that the neutral axis makes with the y-axis. (c) The position or positions on the beam where the maximum compressive bending stress occurs given your answer in a. Clearly indicate this on the figure below (also calculate the values) 4 in. B 4 in. M 6 in. 6 in.
- Below Figure shows the section of an angle purlin. A bending moment of 60 5 kN.m is applied to the purlin in a plane at an angle of 30 deg to the vertical y axis. If the sense of the bending moment is such that both its components Mx and My produce tension in the positive xy quadrant, calculate the maximum direct stress in the purlin, stating clearly the point at which it acts. * 100 mm BỊ 10mm 30° C ID 10mm 57 MPa. 89 MPa. Non Above 72 MPa. 125mm11:11 A The wood beam has an allowable shear stress of Tallow = 9.6 MPa. d3 d3 Variable d₁ d₂ d1 d3 d4 d5 wamap.org Value d2 V Values for the figure are given in the following table. Note the figure may not be to scale. 0.05 m 0.125 m 0.05 m 0.225 m P 0.1125 m d1 d5 a. Determine the Q at point P. b. Determine the moment of inertia, I. d4 c. Determine the magnitude of the max shear force that can be applied to the cross-section at point P, Vmax.O A hollow circular shaft of inner radius 10 mm outer radius 20 mm and length 1 m is to be used as a torsional spring. If the shear modulus of the material of the shaft is 150 GPa, the torsional stiffness of the shaft (in kN-m/rad) is (correct to two decimal places).
- 1: Determine the internal normal force and shear force, and the bending moment in the beam at points C and D. Assume the support at B is a roller. Point C is located just to the right of the 8-kip load. For your explanation section, complete a FBD of the other side of the beam from the version you did to solve the problem. 8 kip 40 kip · ft D B 8 ft – 8 ft 8 ftA folding tray mechanism is attached to a wall as shown. Find the internal forces and bending moment in the lower support arm at section s-s, located midway between points B and E, when a force of F = 270 N is applied at an angle of 0 = 28°. 2013 Michael Swanbom Ⓒ030 BY NC SA k s -Ś a b G с h₁ h₂ ка E S- B Values for dimensions on the figure are given in the following table. Note the figure may not be to scale. Variable Value 16 cm 19 cm 25 cm 23 cm 17 cm The internal axial load at section s-s is A = The internal shear load at section s-s is V = The internal bending moment at section s-s is M = N. N. C. N-m. 0A standard truck moves across a 25- meter span. The wheel loads are P1-36 kN and P2=142 kN separated by 4.3 m and P3=142 kN at 7.6 m from P2. Calculate the maximum shear. O 156.54 KN O 204.16 KN O 259.7 KN O 1409.7 KN
- A folding tray mechanism is attached to a wall as shown. Find the internal forces and bending moment in the lower support arm at section s-s, located midway between points B and E, when a force of F = 250 N is applied at an angle of 0 = 32°. A B. 's E D k a- b – Values for dimensions on the figure are given in the following table. Note the figure may not be to scale. Variable Value a 17 cm b 26 cm 35 cm hi 25 cm h2 18 cm The internal axial load at section s-s is A = N. The internal shear load at section s-s is V = N. The internal bending moment at section s-s is M = N-m.For the frame shown in the figure , the bending moment at B equals * 100 kN 1,5 10 kN/m 1,5 A B. +20m20m -3.0 m- -2.0 m- O 102 kN.m c.w. 98 kN.m c.w. 98 kN.m c.c.W. O 102 kN.m c.c.w.A 18-mm-diameter solid steel shaft supports loads PA = 1,400 N and Pc = 2,100 N as shown. Assume L₁ = 100 mm, L₂ = 200 mm, and L3 = 150 mm. The bearing at B can be idealized as a roller support and the bearing at D can be idealized as a pin support. The moment of inertia of the shaft is 5153 mm4. Determine the magnitude of the maximum bending stress in the shaft. B O 150.2 MPa O 317.3 MPa O 205.6 MPa O 244.5 MPa O 266.8 MPa L₂ Pc L3