A fully braced structural member in a building is subjected to several different loads, including roof loads of D = 5 k and L_r
= 7 k; floor loads of D = 6 k and occupancy L = 15 k; S = 18 k; and W = 17 k and E = 12 k (resulting from overturning forces on the lateral-force-resisting system). Consider the ASD load combinations described in Sec. 2.17 and assume the wind and seismic loads act in the same direction as the gravity toads. Find: The critical ASD combination of loads.
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A silicon carbide plate fractures in bending when a blunt load was applied to the plate center. The distance between the fracture origin and the mirror/mist boundary on the fracture surface was 0.796 mm. To determine the stress used to break the plate, three samples of the same material were tested and produced the following. What is the estimate of the stress present at the time of fracture for the original plate?
Mirror Radius (mm) Bending Failure Stress (MPa)
0.603 225
0.203 368
0.162 442
Answer:
191 MPa
Explanation:
Failure stress of bending is Inversely proportional to the mirror radius
At mirror radius 1 = 0.603 mm Bending stress 1 = 225 Mpa..............(1)
At mirror radius 2 = 0.203 mm Bending stress 2 = 368 Mpa...............(2)
At mirror radius 3 = 0.162 mm Bending stress 3 = 442 Mpa...............(3)
comparing case 1 and 2 using the above equation
Taking the natural logarithm of both side
ln(0.6114) = n ln(0.3366)
n₁ = ln(0.6114)/ln(0.3366)
n₁ = 0.452
comparing case 2 and 3 using the above equation
Taking the natural logarithm of both side
ln(0.8326) = n ln(0.7980)
n₂ = ln(0.8326)/ln(0.7980)
n₂ = 0.821
comparing case 1 and 3 using the above equation
Taking the natural logarithm of both side
ln(0.5090) = n ln(0.2687)
n₃ = ln(0.5090)/ln(0.2687)
n₃ = 0.514
average for n
n = 0.596
Hence to get bending stress x at mirror radius 0.796
stress x = 191 MPa