PE Civil: Structural formula sheet

ASCE 7 Load Combinations (LRFD & ASD)

The ASCE 7-16 strength (LRFD) and allowable-stress (ASD) load combinations, how to find the one that governs a member, and the 1.0W wind factor.

LRFD basic gravity combination

$$U = 1.2D + 1.6L + 0.5(L_r\ \text{or}\ S\ \text{or}\ R)$$

ASCE 7-16 Eq. 2.3-2; governs most floor members. $D$=dead, $L$=live, $L_r$=roof live, $S$=snow, $R$=rain (all service-level nominal loads).

LRFD dead-only combination

$$U = 1.4D$$

ASCE 7-16 Eq. 2.3-1; controls only when dead load dominates (roughly $D > 8L$).

LRFD roof-load-led combination

$$U = 1.2D + 1.6(L_r\ \text{or}\ S\ \text{or}\ R) + (L\ \text{or}\ 0.5W)$$

ASCE 7-16 Eq. 2.3-3; lead action is roof live, snow, or rain. Companion $L$ may reduce to $0.5L$ where permitted.

LRFD wind combination

$$U = 1.2D + 1.0W + L + 0.5(L_r\ \text{or}\ S\ \text{or}\ R)$$

ASCE 7-16 Eq. 2.3-4. Wind factor is $1.0$ because 7-16 $W$ is strength-level. $L$ may reduce to $0.5L$ where $L_0\le100$ psf and not assembly.

LRFD uplift/overturning combination

$$U = 0.9D + 1.0W$$

ASCE 7-16 §2.3.1 combination (5); minimum dead resisting maximum wind — governs net uplift and anchor/connection tension.

ASD service combination with live

$$\textstyle\sum = D + L$$

ASCE 7-16 Eq. 2.4-2; basic gravity service combination compared to $R_n/\Omega$.

ASD combined live + roof

$$\textstyle\sum = D + 0.75L + 0.75(L_r\ \text{or}\ S\ \text{or}\ R)$$

ASCE 7-16 Eq. 2.4-4; the $0.75$ accounts for non-coincident transient maxima.

ASD wind combination

$$\textstyle\sum = D + 0.6W$$

ASCE 7-16 Eq. 2.4-5; $0.6W$ converts the strength-level map back to a service-level wind effect.

ASD combined live + wind + roof

$$\textstyle\sum = D + 0.75L + 0.75(0.6W) + 0.75(L_r\ \text{or}\ S\ \text{or}\ R)$$

ASCE 7-16 Eq. 2.4-6; companion-factored gravity plus reduced wind.

ASD uplift combination

$$\textstyle\sum = 0.6D + 0.6W$$

ASCE 7-16 §2.4.1 combination 7; $0.6D$ (not $1.0D$) resists overturning and uplift conservatively. The seismic ASD uplift case is not in §2.4.1 in 7-16: it is §2.4.5 combination 10, $0.6D - 0.7E_v + 0.7E_h$.

Where it lives: ASCE/SEI 7-16, Minimum Design Loads and Associated Criteria for Buildings and Other Structures — §2.3 (Strength Design / LRFD) and §2.4 (Allowable Stress Design) · ASCE/SEI 7-16 — Commentary C2 (basis of the strength-level wind map and the 1.0W / 0.6W factors) · AISC 360-16 / Steel Construction Manual, 15th ed. — application of LRFD ($\phi$) and ASD ($\Omega$) with the ASCE 7 combinations

Seismic Base Shear (Equivalent Lateral Force)

The ASCE 7-16 equivalent lateral force base shear V = Cs·W, the seismic response coefficient and its caps, the approximate period, and how SDS, SD1, R and Ie set the demand.

Seismic base shear

$$V = C_s\,W$$

$V$ = total design base shear (kip), $C_s$ = seismic response coefficient (dimensionless), $W$ = effective seismic weight (kip) per ASCE 7-16 §12.7.2.

Seismic response coefficient (base form)

$$C_s = \dfrac{S_{DS}}{R/I_e}$$

Short-period (constant-acceleration) value. $S_{DS}$ = design short-period spectral accel (g), $R$ = response modification factor, $I_e$ = importance factor.

Upper limit, T ≤ TL

$$C_{s,\max} = \dfrac{S_{D1}}{T\,(R/I_e)}$$

Velocity-region cap; governs for most moderate-to-long-period buildings. $S_{D1}$ = design 1-second spectral accel (g), $T$ = fundamental period (s).

Upper limit, T > TL

$$C_{s,\max} = \dfrac{S_{D1}\,T_L}{T^2\,(R/I_e)}$$

Long-period (displacement-region) cap beyond the corner period $T_L$ (s, mapped).

Lower limit (general)

$$C_{s,\min} = 0.044\,S_{DS}\,I_e \ge 0.01$$

Floor on $C_s$ for all buildings; the result may not fall below this regardless of $T$ or $R$.

Lower limit (high seismicity)

$$C_{s,\min} = \dfrac{0.5\,S_1}{R/I_e}$$

Additional floor where $S_1 \ge 0.6g$ (near-fault). $S_1$ = mapped 1-s MCE accel (g).

Approximate fundamental period

$$T_a = C_t\,h_n^{\,x}$$

$h_n$ = structural height (ft). Steel MF $C_t{=}0.028,x{=}0.8$; concrete MF $0.016,0.9$; EBF/BRBF $0.03,0.75$; other $0.02,0.75$.

Design spectral accelerations

$$S_{DS} = \tfrac{2}{3}F_a S_S, \quad S_{D1} = \tfrac{2}{3}F_v S_1$$

From mapped $S_S$, $S_1$ and site coefficients $F_a$, $F_v$; the $\tfrac{2}{3}$ scales MCE to design basis.

Computed-period upper bound

$$T \le C_u\,T_a$$

A rationally computed period may be used for the cap only up to $C_u T_a$ ($C_u$ from ASCE 7-16 Table 12.8-1, ~1.4–1.7).

Where it lives: ASCE/SEI 7-16, Minimum Design Loads and Associated Criteria for Buildings and Other Structures — §12.8 Equivalent Lateral Force Procedure · ASCE/SEI 7-16 — §11.4 Seismic Ground Motion Values (SDS, SD1, Fa, Fv) and §12.7.2 Effective Seismic Weight · ASCE/SEI 7-16 — Tables 12.2-1 (R, Cd, Ω0) and 12.8-1, 12.8-2 (Cu, Ct, x)

Shear & Moment Diagrams

Build shear and moment diagrams from the load-shear-moment relations, locate maximum moment and inflection points, and read the AISC Table 3-23 standard cases.

Load-shear relation

$$\dfrac{dV}{dx} = w(x)$$

Slope of the shear diagram equals the distributed-load term $w$ (force/length), per handbook §1.6.7. A downward applied load enters as a negative $w$, so the shear drops across it. $w$ in $\text{k/ft}$, $V$ in $\text{kip}$.

Shear-moment relation

$$\dfrac{dM}{dx} = V(x)$$

Slope of the moment diagram equals the shear. Maximum moment occurs where $V=0$. $M$ in $\text{kip-ft}$.

Shear from load area

$$V_2 - V_1 = \int_{x_1}^{x_2}\! w\,dx$$

Change in shear between two sections equals the area of the load diagram (handbook §1.6.7). With a downward load taken as negative $w$, the integral is negative, so the shear drops by the magnitude of that area.

Moment from shear area

$$M_2 - M_1 = \int_{x_1}^{x_2}\! V\,dx$$

Change in moment equals the area under the shear diagram between two sections.

Simply supported, full UDL

$$V_{max}=\dfrac{wL}{2},\quad M_{max}=\dfrac{wL^2}{8}$$

$M_{max}$ at midspan. $w$ in $\text{k/ft}$, $L$ in $\text{ft}$. AISC Table 3-23.

Simply supported, central point load

$$V_{max}=\dfrac{P}{2},\quad M_{max}=\dfrac{PL}{4}$$

$M_{max}$ at midspan under the load. $P$ in $\text{kip}$.

Simply supported, off-center point load

$$M_{max}=\dfrac{P\,a\,b}{L}$$

Load $P$ at distance $a$ from left, $b$ from right, $a+b=L$. Max moment under the load.

Cantilever, tip point load

$$V_{max}=P,\quad M_{max}=PL$$

Maximum (hogging) moment at the fixed end. $L$ = cantilever length.

Cantilever, full UDL

$$V_{max}=wL,\quad M_{max}=\dfrac{wL^2}{2}$$

Hogging at the fixed end. Note the factor is $1/2$, not $1/8$.

Overhang back-moment

$$M_{support}= -\dfrac{w\,L_o^{2}}{2}$$

Hogging moment delivered to the support by an overhang of length $L_o$ carrying UDL $w$.

Where it lives: NCEES PE Civil Reference Handbook — §4.1.7 Moment, Shear, and Deflection Diagrams · NCEES PE Civil Reference Handbook — §1.6.7 Beams (load-shear-moment relations and sign conventions) · AISC Steel Construction Manual, 15th ed. — Table 3-23, Shears, Moments and Deflections

Deflection by Virtual Work (Unit Load)

The unit-load (virtual-work) method for the deflection of any truss joint ($\Sigma nNL/AE$) or beam/frame point ($\int mM/EI\,dx$), and how to choose the virtual system.

External virtual work = internal virtual work

$$1\cdot\Delta = \sum n\,\delta_{\text{real}}$$

The defining statement: a unit load does external work $1\cdot\Delta$ equal to the internal work of virtual forces through real deformations.

Truss joint deflection

$$\Delta = \sum \dfrac{n\,N\,L}{A\,E}$$

$n$ = member force from unit load, $N$ = member force from real loads (tension +), $L$ = length, $A$ = area, $E$ = modulus. Result is the displacement at and along the unit load.

Beam/frame deflection

$$\Delta = \sum \int_0^{L} \dfrac{m(x)\,M(x)}{E I}\,dx$$

$m(x)$ = moment from a unit load (force for deflection, couple for rotation); $M(x)$ = moment from real loads; integrate over each member.

Real member elongation (load)

$$(\Delta L) = \dfrac{N\,L}{A\,E}$$

Axial deformation of a truss member under the real load; the real-deformation term inside the truss sum.

Real member elongation (temperature)

$$(\Delta L) = \alpha\,\Delta T\,L$$

$\alpha$ = coefficient of thermal expansion, $\Delta T$ = temperature change. Substitutes for $NL/AE$ to get thermal deflection by the same unit-load sum.

Rotation by a unit couple

$$\theta = \sum \int_0^{L} \dfrac{m_\theta(x)\,M(x)}{E I}\,dx$$

Apply a unit moment at the point of desired rotation; $m_\theta(x)$ is the resulting virtual-moment diagram.

Combined axial + flexural work

$$\Delta = \sum \dfrac{n\,N\,L}{A\,E} + \sum \int_0^{L}\dfrac{m\,M}{E I}\,dx$$

Frames with significant axial force keep both terms; usually the bending term dominates and the axial term is dropped.

Diagram multiplication (Vereshchagin)

$$\int_0^L \dfrac{m\,M}{EI}\,dx = \dfrac{A_M\,\bar{m}}{EI}$$

When $m(x)$ is linear, the integral equals the area $A_M$ of the real $M$ diagram times the ordinate $\bar{m}$ of the virtual diagram at the centroid of $A_M$, over $EI$.

Where it lives: NCEES PE Civil Reference Handbook — §4.1.4 Truss Deflection by Unit Load Method · NCEES PE Civil Reference Handbook — §4.1.5 Frame Deflection by Unit Load Method · Hibbeler, Structural Analysis

Euler Column Buckling & Effective Length

The Euler critical load, the effective-length factor K for ideal end conditions, the radius of gyration and slenderness KL/r, and the critical stress.

Euler critical load

$$P_{cr} = \dfrac{\pi^2 E I}{(KL)^2}$$

Elastic buckling load (kips). $E$ = modulus (ksi), $I$ = moment of inertia about the buckling axis (in$^4$), $K$ = effective-length factor, $L$ = unbraced length (in).

Effective length

$$L_e = KL$$

Distance between inflection points of the buckled shape; the equivalent pinned-pinned length.

Theoretical K factors

$$K = 1.0,\ 0.5,\ 0.7,\ 2.0$$

Pinned-pinned, fixed-fixed, fixed-pinned, fixed-free respectively (NCEES handbook §1.6.8 Columns). AISC recommended design values are larger: 0.65, 0.80, 1.0, 2.10 for fixed-fixed, fixed-pinned, pinned-pinned, fixed-free.

Radius of gyration

$$r = \sqrt{\dfrac{I}{A}}$$

Spread of area about the buckling axis (in). Use the smaller (weak-axis) value unless that axis is braced.

Effective slenderness ratio

$$\lambda = \dfrac{KL}{r}$$

Dimensionless governing parameter; the column buckles about the axis with the larger $KL/r$.

Critical (Euler) stress

$$\sigma_{cr} = \dfrac{\pi^2 E}{(KL/r)^2}$$

Buckling stress (ksi); depends only on $E$ and $KL/r$, NOT on $F_y$. Valid only while $\sigma_{cr} < F_y$.

Squash (yield) load

$$P_y = F_y A$$

Upper bound on capacity (kips); buckling governs when $P_{cr} < P_y$.

Inelastic-elastic transition (AISC 360-16)

$$\dfrac{KL}{r}\Big|_{\text{trans}} = 4.71\sqrt{\dfrac{E}{F_y}}$$

Above this slenderness elastic buckling governs ($F_{cr}=0.877F_e$); below it inelastic buckling/yielding governs. About 113 for $F_y=50$ ksi.

Where it lives: NCEES PE Civil Reference Handbook — §1.6.8 Columns (Chapter 1 General Engineering; Euler's Formula and theoretical K) · AISC 360-16, Chapter E — Design of Members for Compression (Steel Construction Manual, 15th ed.) · Hibbeler, Mechanics of Materials, 10th ed. — column buckling background

Concrete Properties, Modulus & Strength

Specified compressive strength f′c, the elastic modulus Ec = 57000√f′c (psi), modulus of rupture, the stress-strain curve, β1, mix proportioning, and reinforcement grades.

Elastic modulus (normalweight)

$$E_c = 57{,}000\sqrt{f'_c}\ \text{(psi)}$$

Secant modulus of normalweight concrete. $f'_c$ and $E_c$ both in psi; e.g., $f'_c=4{,}000\,\text{psi}\Rightarrow E_c\approx 3{,}605\,\text{ksi}$.

Elastic modulus (general, by unit weight)

$$E_c = 33\,w_c^{1.5}\sqrt{f'_c}\ \text{(psi)}$$

$w_c$ = unit weight (pcf), $90\le w_c\le 160$. Reduces to the $57{,}000$ form at $w_c=145\,\text{pcf}$; use for lightweight concrete.

Modulus of rupture

$$f_r = 7.5\,\lambda\sqrt{f'_c}\ \text{(psi)}$$

Flexural tensile strength. $\lambda=1.0$ normalweight; smaller for lightweight. Drives the cracking moment and minimum-steel check.

Cracking moment

$$M_{cr} = \dfrac{f_r\,I_g}{y_t}$$

$I_g$ = gross moment of inertia, $y_t$ = distance from centroid to tension face. Used in effective-moment-of-inertia deflection.

Stress-block factor β1

$$\beta_1 = 0.85 - 0.05\dfrac{f'_c - 4000}{1000}\quad (4000<f'_c<8000\ \text{psi})$$

Depth of equivalent block $a=\beta_1 c$. $\beta_1=0.85$ for $f'_c\le4{,}000$, floor $0.65$ at $f'_c\ge8{,}000$ psi.

Equivalent compression stress

$$C = 0.85\,f'_c\,b\,a$$

Resultant compression on the Whitney block; the $0.85$ stress intensity is constant (does not change with $f'_c$).

Design crushing strain

$$\varepsilon_{cu} = 0.003$$

ACI extreme-fiber compression strain at flexural failure; anchors strain compatibility.

Reinforcement modulus

$$E_s = 29{,}000\ \text{ksi}$$

Same as structural steel; gives modular ratio $n=E_s/E_c \approx 7$ to $9$.

Modular ratio

$$n = \dfrac{E_s}{E_c}$$

Used to transform steel area to equivalent concrete in cracked-section (service) analysis.

Bar size to diameter

$$d_b = \dfrac{\#}{8}\ \text{in (imperial bars)}$$

For #3 through #8, a #n bar is n eighth-inches in diameter (#8 = 1.00 in, $A_b=0.79\,\text{in}^2$; #4 = 0.50 in); #9-#11 are larger and no longer follow the eighths rule. From handbook §4.3.1.

Where it lives: NCEES PE Civil Reference Handbook — §4.3 Concrete (Ec, β1 definitions, §4.3.1 bar areas) · ACI 318-14, Building Code Requirements for Structural Concrete — §19.2 (Ec, fr) and §22.2 (equivalent stress block, β1) · ACI 211.1, Standard Practice for Selecting Proportions for Normal, Heavyweight, and Mass Concrete

Soil Phase Relationships & Classification

The weight-volume phase diagram (e, w, S, unit weights) tied to USCS and AASHTO classification with the group index — the foundation every geotech calc rests on.

Void ratio and porosity

$$e = \dfrac{V_v}{V_s}, \qquad n = \dfrac{e}{1+e}, \qquad e = \dfrac{n}{1-n}$$

$e$ = void ratio (used in consolidation), $n$ = porosity. Both dimensionless; voids = water + air.

Bridge identity

$$S\,e = w\,G_s$$

Ties saturation $S$ and void ratio $e$ to water content $w$ and specific gravity $G_s$. Use $S$ and $w$ as decimals.

Dry unit weight

$$\gamma_d = \dfrac{G_s\,\gamma_w}{1+e}$$

Weight of solids per total volume. $\gamma_w = 62.4\,\text{pcf}$ ($9.81\,\text{kN/m}^3$).

Total (moist) unit weight

$$\gamma_t = \gamma_d(1+w) = \dfrac{(G_s + S e)\gamma_w}{1+e}$$

What a field sample weighs at water content $w$ and saturation $S$.

Saturated and buoyant unit weight

$$\gamma_{sat} = \dfrac{(G_s+e)\gamma_w}{1+e}, \qquad \gamma_b = \gamma_{sat} - \gamma_w$$

$\gamma_{sat}$ at $S=1$; $\gamma_b$ (submerged/effective) governs effective stress below the water table.

Grading coefficients

$$C_u = \dfrac{D_{60}}{D_{10}}, \qquad C_c = \dfrac{D_{30}^{\,2}}{D_{10}\,D_{60}}$$

$D_{xx}$ = grain size with $xx\%$ passing. Well-graded sand needs $C_u\ge6$, gravel $C_u\ge4$, both with $1\le C_c\le3$.

Plasticity index and A-line

$$PI = LL - PL, \qquad PI_{A} = 0.73\,(LL-20)$$

On/above A-line with $PI>7\Rightarrow$ clay (C); below $\Rightarrow$ silt (M). $LL<50$ low (L), $\ge50$ high (H).

Liquidity index

$$LI = \dfrac{w - PL}{PI}$$

Locates in-situ water content between the plastic ($LI=0$) and liquid ($LI=1$) limits; $LI>1$ flags sensitive clay.

AASHTO group index

$$GI = (F-35)\big[0.2+0.005(LL-40)\big] + 0.01(F-15)(PI-10)$$

$F$ = % passing No. 200. Round to nearest integer; report 0 if negative. A-2-6/A-2-7 use only the last term.

Relative density

$$D_r = \dfrac{e_{max}-e}{e_{max}-e_{min}} \times 100\%$$

Density state of a coarse soil between its loosest ($e_{max}$) and densest ($e_{min}$) states.

Where it lives: NCEES PE Civil Reference Handbook — §3.7 Soil Classification and Boring Log Interpretation · ASTM D2487 — Unified Soil Classification System (USCS) · AASHTO M145 — Classification of Soils and Soil-Aggregate Mixtures (group index) · FHWA-NHI-06-088 Soils and Foundations, Vol. I

ACI Flexural Design of RC Beams

The Whitney stress block, nominal moment $M_n$ of a singly-reinforced beam, the tension-controlled $\phi=0.90$, and the $\rho_{min}$/$\rho_{max}$ limits that keep failures ductile.

Stress-block depth

$$a = \dfrac{A_s f_y}{0.85\, f'_c\, b}$$

Depth of the equivalent rectangular compression block. $A_s$ (in²), $f_y$, $f'_c$ (psi), $b$ (in); $a$ in inches. From force balance $C=T$.

Block factor β₁

$$\beta_1 = \begin{cases} 0.85 & f'_c \le 4000 \\ 0.85 - 0.05\dfrac{f'_c-4000}{1000} & 4000 < f'_c \le 8000 \\ 0.65 & f'_c > 8000 \end{cases}$$

Ratio of block depth $a$ to neutral-axis depth $c$ (ACI 318-14 Table 22.2.2.4.3). $f'_c$ in psi.

Nominal moment

$$M_n = A_s f_y\!\left(d - \dfrac{a}{2}\right)$$

Internal couple of the section. Lever arm $d-a/2$ in inches → $M_n$ in $\text{lb}\cdot\text{in}$; divide by 12,000 for $\text{kip}\cdot\text{ft}$.

Design strength requirement

$$\phi M_n \ge M_u, \qquad \phi = 0.90\ \text{(tension-controlled)}$$

Strength condition. $\phi=0.90$ when $\varepsilon_t \ge 0.005$; transition between 0.65/0.75 and 0.90 otherwise.

Net tensile strain

$$\varepsilon_t = 0.003\,\dfrac{d_t - c}{c}$$

Strain in extreme tension steel from the linear strain diagram with $\varepsilon_{cu}=0.003$. $\ge 0.005$ → tension-controlled.

Minimum steel

$$A_{s,min} = \max\!\left(\dfrac{3\sqrt{f'_c}}{f_y},\ \dfrac{200}{f_y}\right) b_w d$$

ACI 318-14 §9.6.1; prevents brittle failure at cracking. $f'_c$, $f_y$ in psi; $\sqrt{f'_c}$ needs psi.

Reinforcement ratio

$$\rho = \dfrac{A_s}{b\,d}$$

Tension steel ratio; compare against $\rho_{min}=A_{s,min}/(b_w d)$ and the strain-limited $\rho_{max}$.

Maximum steel ratio (strain limit)

$$\rho_{max} = 0.85\,\beta_1\dfrac{f'_c}{f_y}\cdot\dfrac{0.003}{0.003+\varepsilon_{t,limit}}$$

From $c$ at the strain limit ($\varepsilon_{t,limit}=0.004$ minimum, 0.005 to keep $\phi=0.90$). Caps the steel indirectly.

Resistance-factor (design) coefficient form

$$M_u = \phi A_s f_y\!\left(d-\dfrac{a}{2}\right)$$

Design equation solved for $A_s$ given $M_u$; quadratic in $A_s$ since $a$ depends on $A_s$.

Concrete modulus

$$E_c = 57{,}000\sqrt{f'_c}\ \ (\text{psi})$$

Normalweight-concrete modulus (ACI 318-14 §19.2.2). $f'_c$ in psi; needed for service/deflection checks.

Where it lives: NCEES PE Civil Reference Handbook — §4.3.2.2 Beams—Flexure Strength (equivalent rectangular stress block, β₁) · ACI 318-14, Building Code Requirements for Structural Concrete — §22.2 (flexural strength), §21.2 (φ factors), §9.6.1 (minimum reinforcement) · ACI 318-14 Table 22.2.2.4.3 — β₁ as a function of f′c

AISC Compression Member Strength

Compute the flexural-buckling strength of a steel column from KL/r: the elastic Fe, the inelastic/elastic Fcr branches at the 4.71 transition, and the LRFD/ASD strengths.

Governing slenderness

$$\dfrac{KL}{r} = \max\left[\dfrac{K_xL_x}{r_x},\ \dfrac{K_yL_y}{r_y}\right]$$

Larger of the two principal-axis values governs. $K$ = effective-length factor, $L$ = unbraced length, $r=\sqrt{I/A}$ (in). Keep $KL/r$ preferably below 200.

Elastic (Euler) buckling stress

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

$E = 29{,}000$ ksi for steel. $F_e$ in ksi. Independent of $F_y$. AISC 360-16 Eq. E3-4.

Branch transition (inelastic/elastic)

$$\dfrac{KL}{r}\le 4.71\sqrt{\dfrac{E}{F_y}} \iff \dfrac{F_y}{F_e}\le 2.25$$

Limit = 113 for $F_y=50$ ksi, 134 for $F_y=36$ ksi. At or below: inelastic branch; above: elastic branch.

Inelastic critical stress

$$F_{cr} = \left(0.658^{\,F_y/F_e}\right)F_y$$

AISC 360-16 Eq. E3-2, for $KL/r \le 4.71\sqrt{E/F_y}$. Approaches $F_y$ as the column gets stocky.

Elastic critical stress

$$F_{cr} = 0.877\,F_e$$

AISC 360-16 Eq. E3-3, for slender columns above the transition. The 0.877 covers initial out-of-straightness.

Nominal compressive strength

$$P_n = F_{cr}\,A_g$$

$A_g$ = gross area (in²); $P_n$ in kips. Use gross area — bolt holes do not reduce compression. AISC 360-16 Eq. E3-1.

LRFD design strength

$$\phi_c P_n,\qquad \phi_c = 0.90$$

Compare to factored load $P_u$ from ASCE 7-16 LRFD combinations. Require $\phi_c P_n \ge P_u$.

ASD allowable strength

$$\dfrac{P_n}{\Omega_c},\qquad \Omega_c = 1.67$$

Compare to service load $P_a$ from ASD combinations. Require $P_n/\Omega_c \ge P_a$. Never mix with LRFD.

Equivalent weak-axis length for Table 4-1

$$(KL)_{\text{eff},y} = \dfrac{K_x L_x}{r_x/r_y}$$

Converts a strong-axis effective length to the $y$-axis basis the table uses, so you enter Table 4-1 with the larger of $K_yL_y$ and this value.

Where it lives: AISC 360-16 — Specification for Structural Steel Buildings, Chapter E (Design of Members for Compression) · AISC Steel Construction Manual, 15th ed. — Table 4-1 (Available Strength in Axial Compression) · NCEES PE Civil Reference Handbook v2.2 — §4.2 Steel

NDS Timber Design: Adjustment Factors

How the NDS reference design values become adjusted (allowable) values F' through the chain of adjustment factors — CD, CM, Ct, CF, CL, CP, Cr and the rest.

Adjusted bending value

$$F'_b = F_b\,C_D\,C_M\,C_t\,C_L\,C_F\,C_{fu}\,C_i\,C_r$$

Reference $F_b$ (psi) times every applicable factor; $C_L$ is the beam-stability factor, $C_F$ the size factor (sawn) or $C_V$ the volume factor (glulam — use only the smaller of $C_L$, $C_V$).

Adjusted compression (parallel)

$$F'_c = F_c\,C_D\,C_M\,C_t\,C_F\,C_i\,C_P = F^{*}_c\,C_P$$

$F^{*}_c$ is every factor except $C_P$; the column-stability factor $C_P$ is computed last from $F_{cE}/F^{*}_c$.

Adjusted shear value

$$F'_v = F_v\,C_D\,C_M\,C_t\,C_i$$

Horizontal shear; takes duration and service factors but NOT $C_F$, $C_L$, or $C_r$.

Adjusted bearing (perp. to grain)

$$F'_{c\perp} = F_{c\perp}\,C_M\,C_t\,C_i\,C_b$$

Bearing perpendicular to grain; $C_b$ is the bearing-area factor. Note: NO load-duration factor $C_D$ and NO size factor.

Adjusted modulus (stability)

$$E'_{min} = E_{min}\,C_M\,C_t\,C_i$$

Feeds $F_{cE}$ and $F_{bE}$. Elasticity is duration-independent, so no $C_D$ and no $C_F$ ever apply to $E$ or $E_{min}$.

Starred reference (pre-stability)

$$F^{*}_b = F_b\,C_D\,C_M\,C_t\,C_F\,C_{fu}\,C_i\,C_r$$

All factors except $C_L$; the ratio $F_{bE}/F^{*}_b$ then yields $C_L$. Same idea for $F^{*}_c$ and $C_P$.

Load-duration factor values

$$C_D = 0.9,\ 1.0,\ 1.15,\ 1.25,\ 1.6,\ 2.0$$

Permanent, 10-yr (normal), snow, 7-day/construction, wind/seismic, impact. Use the shortest-duration load in each combination.

Repetitive-member factor

$$C_r = 1.15$$

Applies to $F_b$ only when 3+ parallel members spaced $\le 24$ in. share load through a deck (joists, studs, rafters).

ASD demand-to-capacity

$$f \le F' \quad\Longleftrightarrow\quad \dfrac{f}{F'} \le 1.0$$

The ASD check: actual stress $f$ (e.g., $f_b = M/S$) must not exceed the adjusted allowable $F'$. Never mix in LRFD $\phi$-factors.

Where it lives: AWC National Design Specification (NDS) for Wood Construction, 2018 edition (ASD method) — Ch. 2 general adjustment factors + Ch. 4/5 applicability tables · NDS Supplement, 2018 — Design Values for Wood Construction (reference $F_b$, $F_v$, $F_c$, $F_{c\perp}$, $E$, $E_{min}$ and size factors $C_F$) · AWC NDS 2018 Table 4.3.1 (sawn lumber) and Table 5.3.1 (glulam) — adjustment-factor applicability grids

TMS 402 Masonry: ASD Flexure & Axial

Allowable-stress flexure and axial design of concrete masonry per TMS 402-16: Fb, the slenderness-reduced axial capacity, the cracked-transformed flexure check, and the combined unity equation.

Allowable flexural compressive stress (reinforced)

$$F_b = 0.45\,f'_m$$

Reinforced masonry ASD bending limit (TMS 402-16 §8.3.4.2.2, Eq. 8-21). $f'_m$ in psi. The $\tfrac{1}{3}f'_m$ coefficient is the unreinforced-flexure allowable, not this value (the axial coefficient is $\tfrac{1}{4}f'_m$).

Allowable steel tensile stress

$$F_s = 32\,\text{ksi}$$

Grade 60 reinforcement, ASD (Grade 40/50 → 20 ksi). Used in $M_s = A_s F_s j d$.

Modular ratio

$$n = \dfrac{E_s}{E_m} = \dfrac{29{,}000\,\text{ksi}}{900\,f'_m}$$

$E_m = 900 f'_m$ for CMU. At $f'_m = 2000\,\text{psi}$, $n \approx 16.1$ (much larger than concrete's ~8).

Radius of gyration, solid wall

$$r = \dfrac{t}{\sqrt{12}}$$

Net thickness $t$ (in). 8-in CMU: $t = 7.625\,\text{in}$, $r \approx 2.20\,\text{in}$.

Allowable axial force (slender)

$$P_a = \left(0.25 f'_m A_n + 0.65 A_{st} F_s\right)\!\left[1-\left(\tfrac{h}{140 r}\right)^2\right]$$

For $h/r \le 99$. $A_n$ = net area, $A_{st}$ = longitudinal steel. (For $h/r > 99$, use $(70r/h)^2$.)

Axial capacity, very slender

$$P_a = \left(0.25 f'_m A_n + 0.65 A_{st} F_s\right)\!\left(\tfrac{70 r}{h}\right)^2$$

Use when $h/r > 99$ (Euler-controlled branch).

Allowable axial stress

$$F_a = \tfrac{1}{4}f'_m\!\left[1-\left(\tfrac{h}{140 r}\right)^2\right]$$

Allowable axial compressive stress, TMS 402-16 §8.2.4.1 ($h/r \le 99$; above 99 use $\tfrac{1}{4}f'_m(70r/h)^2$). The reinforced-masonry check (§8.3.4.2.2) requires $f_a \le F_a$. It is the $f_a/F_a$ denominator in the unity equation, not a bending allowable.

Unreinforced flexure check

$$f_b = \dfrac{M}{S} \le F_t$$

$F_t$ = allowable flexural tension from TMS 402-16 Table 8.2.4.2 (mortar/grout/direction); $S$ = net section modulus. Unreinforced flexural-compression allowable is $\tfrac{1}{3}f'_m$ (§8.2.4.1).

Cracked neutral-axis ratio

$$k = \sqrt{2\rho n + (\rho n)^2} - \rho n, \quad \rho = \dfrac{A_s}{bd}$$

Working-stress cracked transformed section for reinforced flexure.

Lever-arm ratio

$$j = 1 - \dfrac{k}{3}$$

Internal moment arm $= jd$. Often $j \approx 0.9$ for typical wall reinforcement.

Allowable reinforced moment

$$M_{\text{allow}} = \min\!\left(A_s F_s\,j\,d,\ \tfrac{1}{2}F_b\,k\,j\,b\,d^2\right), \quad F_b = 0.45 f'_m$$

Smaller of steel-controlled and masonry-controlled allowable moment; masonry side uses $F_b = 0.45 f'_m$.

Unity (interaction) equation

$$\dfrac{f_a}{F_a} + \dfrac{f_b}{F_b} \le 1.0$$

$f_a = P/A_n$, $F_a = \tfrac{1}{4}f'_m[1-(h/140r)^2]$, $f_b$ = flexural compressive stress, $F_b = 0.45 f'_m$.

Eccentric Euler buckling check

$$P \le \tfrac{1}{4}P_e, \quad P_e = \dfrac{\pi^2 E_m I_n}{h^2}\!\left(1 - 0.577\tfrac{e}{r}\right)^{3}$$

Independent stability check on eccentric axial load; $e$ = load eccentricity, $I_n$ = net moment of inertia.

Where it lives: TMS 402-16 — Building Code Requirements for Masonry Structures, §8.2 (Unreinforced) and §8.3 (Reinforced), Allowable Stress Design · TMS 402-16 §8.3.4.2.2 — Reinforced flexural compressive stress $F_b = 0.45 f'_m$ (Eq. 8-21) and combined unity interaction · TMS 402-16 §8.2.4.1: allowable axial stress $F_a = \tfrac{1}{4}f'_m[1-(h/140r)^2]$; §8.3.4.2.1: allowable axial force $P_a$ · TMS 402-16 §8.2.4: unreinforced allowables (flexural compression $\tfrac{1}{3}f'_m$ in §8.2.4.1, flexural tension table in §8.2.4.2) · ASCE 7-16 — Minimum Design Loads (service-load combinations applied to ASD masonry checks)

AISC Bolted Connections (Shear, Bearing, Slip)

Bolt shear strength, bearing and tearout at the holes, bearing vs slip-critical behavior, and how the bolt-group capacity is the sum of per-bolt limit states.

Bolt shear strength

$$\phi R_n = \phi\,F_{nv}\,A_b\,m$$

Per bolt. $F_{nv}=54$ ksi (A325-N) / $68$ (A325-X or A490-N) / $84$ (A490-X); $A_b=\pi d^2/4$ (in²); $m$ = shear planes; $\phi=0.75$.

Bolt nominal area

$$A_b = \dfrac{\pi d^2}{4}$$

Gross shank area from nominal diameter $d$ (in); a 3/4-in bolt has $A_b=0.442\,\text{in}^2$, a 7/8-in bolt $0.601\,\text{in}^2$.

Bearing strength at a hole

$$\phi R_n = \phi\,2.4\,d\,t\,F_u$$

Deformation-considered upper limit. $d$ = bolt dia, $t$ = ply thickness, $F_u$ = ply tensile strength (ksi); $\phi=0.75$.

Tearout strength at a hole

$$\phi R_n = \phi\,1.2\,l_c\,t\,F_u$$

$l_c$ = clear distance in the line of force: $l_e - d_h/2$ from an edge (end) bolt to the edge, $s - d_h$ between two holes; $d_h$ = standard hole ($d+\tfrac{1}{16}$ in for $d\le\tfrac78$ in, $d+\tfrac18$ in for $d\ge1$ in, AISC Table J3.3).

Controlling bearing/tearout per bolt

$$\phi R_n = \phi\min(2.4\,d\,t\,F_u,\ 1.2\,l_c\,t\,F_u)$$

Take the smaller; tearout usually governs at edge bolts (small $l_c$), bearing at interior bolts.

Slip-critical design resistance

$$\phi R_n = \phi\,\mu\,D_u\,h_f\,T_b\,n_s$$

$\mu=0.30$ (A), $0.50$ (B); $D_u=1.13$; $h_f=1.0$ (no filler); $T_b$ from Table J3.1; $n_s$ = slip planes; $\phi=1.00$ std holes.

Minimum pretension (Table J3.1, sample)

$$T_b\!: \ \tfrac{3}{4}''\,\text{A325}=28,\ \tfrac{7}{8}''\,\text{A325}=39,\ \tfrac{7}{8}''\,\text{A490}=49\ \text{kips}$$

Specified minimum pretension for fully tensioned high-strength bolts (kips); required for SC and for fatigue/seismic joints.

Block shear rupture

$$\phi R_n = \phi\big(0.6 F_u A_{nv} + U_{bs}F_u A_{nt}\big) \le \phi\big(0.6 F_y A_{gv} + U_{bs}F_u A_{nt}\big)$$

$A_{nv},A_{gv}$ net/gross shear area; $A_{nt}$ net tension area; $U_{bs}=1.0$ uniform tension; $\phi=0.75$.

Group capacity (concentric)

$$\phi R_{n,\text{group}} = N_b \times \min(\phi R_n)_{\text{per bolt}}$$

Equal sharing only when the load passes through the group centroid; eccentric groups use the elastic-vector or IC method.

Standard hole diameter

$$d_h = d + \tfrac{1}{16}\ \text{in}\ (d\le\tfrac78''),\quad d+\tfrac18\ \text{in}\ (d\ge1'')$$

Standard round hole per AISC Table J3.3; oversized and slotted holes are larger and reduce SC $\phi$ and bearing/tearout $l_c$.