Structural engineering

This chapter covers the following:

• Stability and determinacy of structures
• Snow and wind loads
• Load combinations for steel member design: ASD vs LRFD
• Tension member design: ASD vs LRFD
• Compression member design: ASD vs LRFD
• Flexural member design: ASD vs LRFD
• Shear member design: ASD vs LRFD

Stability and determinacy of structures

Stability and determinacy tell you whether a structure can be analyzed using statics and whether it will hold its shape under load.

Stability

Definitions

Stable structure

A structure is stable if it maintains its shape and position under applied loads and does not undergo rigid body motion (translation or rotation).

A stable structure resists rigid body motion (translation or rotation) and also avoids internal collapse.

• Externally stable: The entire structure doesn’t move as a rigid body.
• Internally stable: The members are arranged so the structure can’t collapse internally.

Determinacy

Definitions

Statically determinate structure

A structure is statically determinate if all support reactions and internal forces can be found using only the equations of static equilibrium.

A statically determinate structure has just enough unknowns that the equilibrium equations are sufficient to solve for all reactions and internal forces.

Equilibrium equations (2D)

$$\sum F_x=0,\sum F_y=0,\sum M=0$$

Equilibrium equations (3D)

$$\sum F_x=0,\sum F_y=0,\sum F_z=0,\sum M_x=0,\sum M_y=0,\sum M_z=0$$

Determinacy and stability criteria

Beams (2D)

For a simply supported beam:

• Number of unknown support reactions: 3 (1 pin + 1 roller)
• Number of equilibrium equations: 3

$$\text{If }r=3\Rightarrow \text{statically determinate}$$

If $r>3$ $\to$ Indeterminate
If $r<3$ $\to$ Unstable

Frames (2D)

Let:

• $r$: number of unknown reactions
• $m$: number of members
• $j$: number of joints
• $c$: equations of condition - one for each internal pin, two for each internal roller. $c=0$ when the frame has no internal releases.

Each member of a rigid frame carries three internal force components (axial force, shear, and bending moment), so the unknowns number $3m+r$. Three equilibrium equations are available at each joint, plus one for each equation of condition, giving $3j+c$.

Determinacy condition:

$$3m+r=3j+c$$

• If $3m+r=3j+c$: Statically determinate
• If $3m+r>3j+c$: Indeterminate
• If $3m+r<3j+c$: Unstable

Trusses (2D)

Let:

• $m$: number of members
• $j$: number of joints
• $r$: number of support reactions

Determinacy condition:

$$m+r=2j$$

• If $m+r=2j$: Statically determinate, and stable provided the members and supports are arranged so no part can move as a mechanism (the count is necessary but not sufficient)
• If $m+r<2j$: Unstable
• If $m+r>2j$: Indeterminate

Examples

Example 1: truss

Example 2: frame

Given:

• $m=5$ members
• $j=4$ joints
• $r=3$ reactions

Check:

$$\begin{gathered} 3m+r=3(5)+3=18 \\ 3j+c=3(4)+0=12 \end{gathered}$$

Statically indeterminate to the 6th degree (since $3m+r>3j+c$)

Summary table

Snow and wind loads

Flat roof snow loads

The flat roof snow load is given by:

$$p_f=0.7C_e{C}_t{I}_s{P}_g$$

Where:

• $C_e$ = exposure factor
• $C_t$ = thermal factor
• $I_s$ = importance factor
• $P_g$ = ground snow load ($lb/f{t}^2$)

Exposure factor, $C_e$

• Fully exposed: Roofs exposed on all sides with no shelter afforded by terrain, higher structures, or trees
• Sheltered: Roofs located tightly in among conifers that qualify as obstructions
• Partially exposed: All other cases

Thermal factor, $C_t$

Importance factor, $I_s$

Wind loads

The velocity pressure at height $z$ is given by:

Velocity pressure exposure coefficient, $K_z$

Load combinations for steel member design: ASD vs LRFD

In steel design, ASD (Allowable strength design) and LRFD (Load and resistance factor design) use different load levels and different safety formats.

Definitions

Allowable strength design (ASD)

ASD checks service-level (unfactored) loads against an allowable strength.

ASD:

$$R_a\le \frac{R_n}{\Omega }$$

• $R_a$: Actual (required) strength due to service loads
• $R_n$: Nominal strength
• $\Omega$: Safety factor (typically $>1.0$)

Definitions

Load and resistance factor design (LRFD)

LRFD checks factored (ultimate) loads against a reduced nominal strength.

LRFD:

$$P_u\le \varphi R_n$$

• $P_u$: Factored (ultimate) load
• $R_n$: Nominal strength
• $\varphi$: Resistance factor (typically $<1.0$)

Common mistake: Don’t mix load levels between the two methods. ASD strength checks use service (unfactored) loads on the left side of the inequality, while LRFD checks use factored (ultimate) loads. Plugging factored loads into the ASD equation, or service loads into the LRFD equation, gives a nonsense comparison. In the FE Reference Handbook, the ASD and LRFD strength equations for each member type (tension, compression, flexure, shear) are listed side by side, directly below that member type’s load combination tables - match the equation to the load set you started from.

Load combinations

Load combinations define how you compute $R_a$ (ASD) and $P_u$ (LRFD) when multiple load types can act together.

ASD load combinations

$$\begin{array}{rl}{R}_a& =D\\ R_a& =D+L\\ R_a& =D+(L_r\text{ or }S\text{ or }R)\\ R_a& =D+0.75L+0.75(L_r\text{ or }S\text{ or }R)\\ R_a& =D+(0.6W\text{ or }0.7E)\\ R_a& =D+0.75L+0.75(0.6W)+0.75(L_r\text{ or }S\text{ or }R)\\ R_a& =D+0.75L+0.75(0.7E)+0.75S\\ R_a& =0.6D+0.6W\\ R_a& =0.6D+0.7E\end{array}$$

LRFD load combinations

$$\begin{array}{rl}{P}_u& =1.4D\\ P_u& =1.2D+1.6L+0.5(L_r\text{ or }S\text{ or }R)\\ P_u& =1.2D+1.6(L_r\text{ or }S\text{ or }R)+(L\text{ or }0.5W)\\ P_u& =1.2D+1.0W+L+0.5(L_r\text{ or }S\text{ or }R)\\ P_u& =1.2D+1.0E+L+0.2S\\ P_u& =0.9D+1.0W\\ P_u& =0.9D+1.0E\end{array}$$

Notation:

Tension member design: ASD vs LRFD

In tension design, you typically check yielding on the gross section and fracture on the effective net section.

ASD:

$$P_a\le \frac{P_n}{{\Omega }_t}$$

LRFD:

$$P_u\le {\varphi }_t{P}_n$$

Nominal strength expressions

Yielding limit state:

$$P_n=F_y{A}_g$$

• $F_y$: Yield strength
• $A_g$: Gross area of the member

Fracture limit state:

$$P_n=F_u{A}_e$$

• $F_u$: Ultimate strength
• $A_e=U{A}_n$: Effective net area
• $U$: Shear lag factor
• $A_n$: Net area

Net area calculation

Example: Net area with staggered holes

A tension member is a flat plate with gross width $b_g=10in$ and thickness $t=0.5in$. It has two staggered rows of $\frac{7}{8}$-inch bolts, with one hole deducted along the critical path in each row, longitudinal spacing $s=3in$, and transverse spacing $g=3in$. Find the net area.

• Deduction per hole: $d_h+\frac{1}{16}=\left(0.875+\frac{1}{16}\right)+\frac{1}{16}=1.0in$
• Two holes lie on the critical path: $\sum \left(d_h+\frac{1}{16}\right)=2(1.0)=2.0in$
• Staggered-pitch add-back term: $\sum \frac{s^2}{4g}=\frac{3^2}{4(3)}=0.75in$
• Net width: $b_g-2.0+0.75=10-2.0+0.75=8.75in$
• Net area: $A_n=8.75\times 0.5=4.375i{n}^2$

Answer: $A_n=4.375i{n}^2$

Effective area $Ae=UAn$

For bolted members:

• Flat bars: $U=1.0$
• Angles: $U=1-\frac{\bar{x}}{L}$

For welded members:

$$U=\{\begin{array}{ll}1.0& \text{(Flat bars/angles with transverse welds)}\\ 1.0& \text{if }L\ge 2w\\ 0.87& \text{if }2w>L\ge 1.5w\\ 0.75& \text{if }1.5w>L>w\\ 1-\frac{\bar{x}}{L}& \text{(Angles with longitudinal welds only)}\end{array}$$

• $w$: Width of flat bar
• $L$: Length of weld
• $\bar{x}$: Distance from centroid to connection

Block shear strength

Resistance factor:

$$\varphi =0.75$$

Shear lag factor:

$$U_{bs}=1.0\text{(flat bars and angles)}$$

Block shear strength:

• $A_{gv}$: Gross area in shear
• $A_{nv}$: Net area in shear
• $A_{nt}$: Net area in tension

$$\varphi T_n=\text{smaller of}\{\begin{array}{l}0.75F_u\left(0.6A_{nv}+U_{bs}{A}_{nt}\right)\\ 0.75\left(0.6F_y{A}_{gv}+U_{bs}{F}_u{A}_{nt}\right)\end{array}$$

Block shear strength is the smaller (governing) of the two expressions.

Compression member design: ASD vs LRFD

Compression members are usually controlled by buckling, so the key step is finding the critical stress $F_{cr}$.

ASD:

$$P_a\le \frac{P_n}{{\Omega }_c}$$

LRFD:

$$P_u\le {\varphi }_c{P}_n$$

Where:

• $P_a$: Actual axial service load
• $P_u$: Factored axial load
• $P_n$: Nominal axial compressive strength
• ${\Omega }_c=1.67$: ASD compression safety factor
• ${\varphi }_c=0.9$: LRFD compression resistance factor

Nominal strength expressions

Compressive strength:

$$P_n=F_{cr}{A}_g$$

Where:

• $F_{cr}$: Critical buckling stress
• $A_g$: Gross cross-sectional area

Critical stress: $F_{cr}$

Based on Euler and inelastic buckling criteria:

$$F_{cr}=\{\begin{array}{ll}0.658^{\frac{F_y}{F_e}}{F}_y,& \text{if }\frac{KL}{r}\le 4.71\sqrt{\frac{E}{F_y}}\text{(Inelastic)}\\ 0.877F_e,& \text{if }\frac{KL}{r}>4.71\sqrt{\frac{E}{F_y}}\text{(Elastic)}\end{array}$$

Where:

Elastic buckling stress

For elastic buckling (long slender columns), the Euler stress is:

$$F_e=\frac{{\pi }^{2}E}{{\left(\frac{KL}{r}\right)}^2}$$

Example: Critical buckling stress

A column has $K=1.0$, $L=15ft$, radius of gyration $r=2.0in$, $F_y=50ksi$, and $E=29,000ksi$. Find $F_{cr}$.

• $r$ is given in inches, so $L$ must be converted from feet to inches before computing $KL/r$ - mixing units here (dividing feet by inches) is a common FE slip. Slenderness ratio: $\frac{KL}{r}=\frac{1.0\times (15\times 12)}{2.0}=90$
• Limiting slenderness: $4.71\sqrt{\frac{E}{F_y}}=4.71\sqrt{\frac{29,000}{50}}=113.4$
• Since $90<113.4$, the column buckles inelastically, so the first branch of $F_{cr}$ applies.
• Euler stress: $F_e=\frac{{\pi }^{2}E}{(KL/r{)}^2}=\frac{{\pi }^2(29,000)}{90^2}=35.3ksi$
• Critical stress: $F_{cr}=0.658^{F_y/F_e}{F}_y=0.658^{50/35.3}(50)\approx 27.6ksi$

Answer: $F_{cr}\approx 27.6ksi$

Flexural member design: ASD vs LRFD

Flexural design compares the required moment to the available flexural strength.

ASD:

$$M_a\le \frac{M_n}{{\Omega }_b}$$

LRFD:

$$M_u\le {\varphi }_b{M}_n$$

Where:

• $M_a$: Actual service moment
• $M_u$: Factored design moment
• $M_n$: Nominal flexural strength
• ${\Omega }_b=1.67$: ASD bending safety factor
• ${\varphi }_b=0.9$: LRFD bending resistance factor

Nominal strength expressions $M_n$

$$M_n=\{\begin{array}{ll}{M}_p=F_y{Z}_x,& \text{Plastic (compact, full lateral support)}\\ C_b\left[M_p-\left(M_p-0.7F_y{S}_x\right)\left(\frac{L_b-L_p}{L_r-L_p}\right)\right]\le M_p,& \text{Inelastic (lateral-torsional buckling)}\\ F_{cr}{S}_x,& \text{Elastic buckling}\end{array}$$

Where:

• $F_y$: Yield strength
• $Z_x$: Plastic section modulus
• $S_x$: Elastic section modulus
• $L_b$: Laterally unbraced length
• $L_p$: Limit for full plastic bending
• $L_r$: Limit between inelastic and elastic buckling
• $C_b$: Moment gradient factor

Example: Selecting the governing nominal moment

A compact W-shape beam has $F_y=50ksi$ and plastic section modulus $Z_x=200i{n}^3$. The beam is fully braced, so $L_b$ is well below $L_p$. Find $M_n$.

• Because the section is compact and fully braced, lateral-torsional buckling doesn’t govern - the first case in $M_n$ applies, not the inelastic or elastic buckling cases.
• $M_n=M_p=F_y{Z}_x=50\times 200=10,000in\text{-}kip$

Answer: $M_n=10,000in\text{-}kip$ ($833.3ft\text{-}kip$)

Moment gradient factor $C_b$

$$C_b=\frac{12.5M_{max}}{2.5M_{max}+3M_A+4M_B+3M_C}$$

Where $M_A,M_B,M_C$ are moments at quarter points in the unbraced length, and $M_{max}$ is the maximum moment.

Shear member design: ASD vs LRFD

Shear design compares the required shear to the available shear strength of the web.

ASD:

$$V_a\le \frac{V_n}{{\Omega }_v}$$

LRFD:

$$V_u\le {\varphi }_v{V}_n$$

Nominal shear strength

$$V_n=0.6F_y{A}_w{C}_v$$

Where:

• $F_y$: Yield strength
• $A_w$: Area of the web
• $C_v$: Shear buckling coefficient