# 2.2: Load Combinations for Structural Design

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$

( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\id}{\mathrm{id}}$$

$$\newcommand{\Span}{\mathrm{span}}$$

$$\newcommand{\kernel}{\mathrm{null}\,}$$

$$\newcommand{\range}{\mathrm{range}\,}$$

$$\newcommand{\RealPart}{\mathrm{Re}}$$

$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$

$$\newcommand{\Argument}{\mathrm{Arg}}$$

$$\newcommand{\norm}[1]{\| #1 \|}$$

$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$

$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$

$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$

$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vectorC}[1]{\textbf{#1}}$$

$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$

$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$

$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

Structures are designed to satisfy both strength and serviceability requirements. The strength requirement ensures the safety of life and property, while the serviceability requirement guarantees the comfortability of occupancy (people) and the aesthetics of the structure. To meet the afore-stated requirements, structures are designed for the critical or the largest load that would act on them. The critical load for a given structure is found by combining all the various possible loads that a structure may carry during its lifetime. Sections 2.3.1 and 2.4.1 of ASCE 7-16 provide the following load combinations for use when designing structures by the Load and Resistance Factor Design (LRFD) and the Allowable Strength Design (ASD) methods.

For LRFD, the load combinations are as follows:

1.1.4$$D$$

2.1.2 $$D+1.6 L+0.5\left(L_{r} \text { or } S \text { or } R\right)$$

3.1.2$$D+1.6\left(L_{r} \text { or } \text { S or } R\right)+(L \text { or } 0.5 \mathrm{W})$$

4.1.2$$D+1.0 W+L+0.5\left(L_{r} \text { or } S \text { or } R\right)$$

5.0.9$$D + 1.0W$$

For ASD, the load combinations are as follows:

1.$$D$$

2.$$D + L$$

3.$$D + (Lr or S or R) 4.\(D + 0.75 L + 0.75\left(L_{r} \text { or } S \text { or } R\right)$$

5.$$D + (0.6 W)$$

where

• $$D$$ = dead load.
• $$L$$ = live load due to occupancy.
• $$L_{r}$$ = roof live load.
• $$S$$ = snow load.
• $$R$$ = nominal load due to initial rainwater or ice, exclusive of the ponding contributions.
• $$W$$ = wind load.
• $$E$$ = earthquake load.

Example $$\PageIndex{1}$$

A floor system consisting of wooden joists spaced 6 ft apart on the center and a tongue and groove wood boarding, as shown in Figure 2.8, supports a dead load (including the weight of the beam and boarding) of 20 psf and a live load of 30 psf. Determine the maximum factored load in lb/ft that each floor joist must support using the LRFD load combinations.

$$Fig. 2.8$$. Floor system.

Solution

Dead load $$D=(6)(20)=120 \mathrm{lb} / \mathrm{ft}$$

Live load $$L=(6)(30)=180 \mathrm{lb} / \mathrm{ft}$$

Determining the maximum factored loads Wu using the LRFD load combinations and neglecting the terms that have no values, yields the following:

$$W_{u}=(1.4)(120)=168 \mathrm{lb} / \mathrm{ft}$$

$$W_{u}=(1.2)(120)+(1.6)(180)=288 \mathrm{lb} / \mathrm{ft}$$

$$W_{u}=(1.2)(120)+(0.5)(180)=234 \mathrm{lb} / \mathrm{ft}$$

$$W_{u}=(1.2)(120)+(0.5)(180)=234 \mathrm{lb} / \mathrm{ft}$$

$$W_{u}=(1.2)(120)+(0.5)(180)=234 \mathrm{lb} / \mathrm{ft}$$

$$W_{u}=(0.9)(120)=108 \mathrm{lb} / \mathrm{ft}$$

The governing factored load = $$288 \mathrm{lb} / \mathrm{ft}$$

This page titled 2.2: Load Combinations for Structural Design is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by René Alderliesten (TU Delft Open) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.