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WorksheetsRecitation 1 Wilkerson
Total questions: 23
Worksheet time: 41mins
1. The 500-kg engine is suspended from the crane boom. Assume Beam AB is a W610 X 155 beam. Solid bar CD has a diameter of 50 mm. Both members are made of A-36 steel.
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1a. Based on the reactions below, calculate the average normal stress in bar CD. (dCD = 5 cm)
10.56 MPa
6.25 MPa
3.21 MPa
0.63 MPa
1b. Determine the Factor of Safety in bar CD. The yield strength of bar CD is 250 MPa.
3
2
40
20
1c. Select the general shape of the axial force diagram for the beam. (HINT: Compression (-) and Tension (+).
1d. Select the general shape of the shear force diagram for the beam.
1e. Select the general shape of the moment diagram for the beam.
1f. Based on the system, where will failure most likely occur?
Point B
Point C
Point A
None of these.
1g. Calculate the bending moment at Point C (Mz). Counter-clockwise is positive, clockwise is negative.
-507 N*m
-357 N*m
-45 N*m
-4905 N*m
1h. Solve for the maximum von Mises Stress in the beam using the information below.
1.66 MPa
3.45 MPa
2.08 MPa
5.64 MPa
1i. Calculate the Factor of Safety for the beam using the von Mises stress criterion. The yield strength for the beam is 250 MPa.
32
150
59
4
2. The bars of the truss have a cross-sectional area of 1.25 in^2. The yield stress for the bars of the truss is 20 ksi.
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2a. By inspection, which member do you expect to have the highest tensile force?
CD
CE
BD
AB
2b. By inspection, which member do you expect to have the highest compressive force?
CD
CE
BD
AB
2c. Which of these axial forces is incorrect?
FCD = 2.5P (tension)
FCE = 2.7P (compression)
FBD = 1.67P (compression)
FBC = P (compression)
2d. From the image, determine the maximum value of P that can be applied to the truss that will cause yielding failure.
543.9 lbf
9.38 kip
10.94 kip
92.51 kip
3. The uniform beam is supported by two rods, AB and CD, that have the cross-sectional areas shown below. Both rods have a yield strength of 0.3 MPa and the beam has a yield strength of 250 MPa.
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3a. (T/F) The location of the resultant force of the distributed load is a third of the beam length from the left.
True
False
3b. Select the correct values when solving for NAB and NCD in terms of w and LAC.
NAB = (wLAC)/6 N
NCD = (wLAC)/3 N
NCD = (wLAC)/6 N
NAB = (wLAC)/3 N
3c. Select the correct general shape of the bending moment diagram for this beam.
3d. Using the FBD, solve for the shear and moment equations in terms of w, x, and LAC. For moments, CCW: +, CW: - .
Vy = (wLAC)/6 - (wx2)/(2LAC)
Mz = (wLACx)/6 - (wx3)/(6LAC)
Mz = -(wLACx)/6 + (wx3)/(6LAC)
Vy = -(wLAC)/6 + (wx2)/(2LAC)
3e. Using the image below, solve for the maximum bending moment in terms of w and LAC.
Mmaxz = -0.94wLAC2
Mmaxz = -0.51wLAC2
Mmaxz = 0.064wLAC2
Mmaxz = 0.019wLAC2
3f. Determine the max value of w that will cause failure to one of the cables. (Hint: The yield strength of the cables is 0.3 MPa. Use a FoS = 1).
w = 13.5/LAC
w = 18/LAC
w = 13.5LAC
w = 18/LAC
3g. Determine the max value of w that will cause failure to the beam using von Mises failure criterion.
w = (17*109)/LAC2 (N/mm)
w = 1.2/LAC (N/mm)
w = 17LAC2 (N/mm)
w = 12/LAC2 (N/mm)
