Choose the combination that results in the highest U, use U in your calculations.

$$U = 1.4(D+F)$$

$$U = 1.2(D+F+T) + 1.6(L+H) + 0.5(L_r or S or R)$$

$$U = 1.2D + 1.6(L_r or S or R) + (1.0L or 0.8W)$$

$$U = 1.2D + 1.6W + 1.0L + 0.5(L_r or S or R)$$

$$U = 1.2DL + 1.0E + 1.0L + 0.2S$$

$$U = 0.9D + (1.6W or 1.0E) + 1.6H$$

D = Dead Load, L = Live Load, T = self-straining, H = earth pressure, $$L_r$$ = roof live load, S = Snow Load, R = rain load, E = earthquake load

In some sample problems I have seen $$\gamma$$ and Q variables. Individual load factors like 1.2 and 1.4 are sometimes represented as $$\gamma$$ and the loads themselves are represented as $$Q$$. The factoring for a single load would then be $$Q \gamma$$, and all of them $$U = \sum{Q \gamma}$$.

## Strength Reduction Factor ($$\phi$$)

Strength reduction factors are applied to the nominal (design) strength:

$$\phi R_n \geq P_u$$

or (this is seen pretty often in practice)

$$R_n \geq \frac{P_u}{\phi}$$

Where $$P_u$$ is determined with U from the factored loads. Strength reduction factors vary for concrete and steel.

### Concrete

0.9 for Tension controlled

0.7 compression with spiral steel

0.65 compression with tied steel

0.75 shear and torsion

0.65 bearing on concrete

Furthermore there is a range based on stress ($$\varepsilon$$) value for beams ($$0.48 + 83 \varepsilon$$) and the same equation applies to tied transition members. Spiral transition members use $$0.57 + 67 \varepsilon$$.

### Steel

0.9 for yield and 0.75 fracture