Question
Download Solution PDFFor a structural member, dead load = 20 kN and live load = 12 kN. What will be its design load as per the limit state of collapse philosophy?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Values of the factor of safety (partial) for load combination:
|
Load combination |
Limit state of collapse |
|
1) Dead load & live load |
1.5(DL + LL) |
|
2) Dead seismic/wind load a) Dead load contributes to the stability b) Dead load assists overturning |
0.9 DL + 1.5 (EL/WL) 1.5 (DL + EL/WL) |
|
3) Dead, live load, and Seismic/wind load |
1.2 (DL + LL + EL/WL) |
Where, DL = Dead load, LL = Live load WL = Wind load EL = Earthquake load
Calculation:
Given: Dead load (DL) = 20 KN and Live load (LL) = 12 kN
Partial factor of safety = 1.5 (DL + LL) = 1.5 (20 + 12) = 48 kN
Additional Information
Values of the factor of safety (partial) for load combination:
|
Load combination |
Serviceability limit state |
|
1) Dead load & live load |
DL + LL |
|
2) Dead seismic/wind load a) Dead load contributes to stability b) Dead load assists overturning |
DL + EQ/WL DL + EQ/WL |
|
3) Dead, live load and Seismic/wind load |
DL + 0.8LL + 0.8EQ/WL |
Last updated on May 28, 2025
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