Question

For 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?
1. 20 kN
2. 48 kN
3. 32 kN
4. 8 kN

Answer 1

Correct Answer - Option 2 : 48 kN

Concept:

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

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	DI + 0.LL + 0.8EQ/WL