AISC 360 — Shear (Chapter G)

Web shear design per AISC 360-16: yielding, buckling, Cv factors, tension field action, and stiffeners

Shear rarely governs the design of standard rolled W shapes — the webs are thick enough that shear yielding capacity far exceeds the demand from typical loading. However, shear becomes the controlling limit state for plate girders with slender webs, coped beams, short heavily loaded beams, and HSS members. Chapter G of AISC 360-16 provides shear design provisions ranging from the simple (rolled shapes) to the sophisticated (tension field action in stiffened plate girders).

1. Shear Yielding — Section G2

The nominal shear strength for I-shaped members is:

V_n = 0.60 \, F_y \, A_w \, C_{v1}

where $A_w = d \cdot t_w$ is the shear area (overall depth times web thickness), and $C_{v1}$ is the web shear coefficient that accounts for web buckling.

1.1. Resistance Factor for Shear

AISC 360 uses two different resistance factors for shear, depending on the web slenderness:

Condition	$\phi_v$ (LRFD)	$\Omega_v$ (ASD)
$h/t_w \leq 2.24\sqrt{E/F_y}$ (all standard W shapes in A992)	1.00	1.50
$h/t_w > 2.24\sqrt{E/F_y}$	0.90	1.67

For A992 ( $F_y = 50$ ksi): $2.24\sqrt{29{,}000/50} = 53.9$ . Every standard W shape has $h/t_w \leq 53.9$ , so $\phi_v = 1.00$ for all rolled wide-flange beams. This is the only limit state in AISC 360 where $\phi = 1.00$ , reflecting the very low variability and high ductility of web shear yielding.

2. Web Shear Coefficient C_v1

The $C_{v1}$ coefficient reduces the shear capacity when the web is prone to buckling:

Web Slenderness	C_v1	Behavior
$h/t_w \leq 1.10\sqrt{k_v E/F_y}$	1.0	Full shear yielding — no web buckling
$1.10\sqrt{k_v E/F_y} < h/t_w \leq 1.37\sqrt{k_v E/F_y}$	$\dfrac{1.10\sqrt{k_v E/F_y}}{h/t_w}$	Inelastic web buckling
$h/t_w > 1.37\sqrt{k_v E/F_y}$	$\dfrac{1.51 \, k_v \, E}{(h/t_w)^2 \, F_y}$	Elastic web buckling

The plate buckling coefficient $k_v$ depends on the presence of transverse stiffeners:

$k_v = 5.34$ for webs without transverse stiffeners (unstiffened)
$k_v = 5 + 5/(a/h)^2$ for webs with transverse stiffeners at spacing $a$

For unstiffened webs with A992 steel and $k_v = 5.34$ :

1.10\sqrt{k_v E/F_y} = 1.10\sqrt{5.34 \times 29{,}000/50} = 61.2

Since all standard W shapes have $h/t_w < 61.2$ , we get $C_{v1} = 1.0$ for every rolled wide-flange beam. This simplifies the shear check dramatically.

3. Shear in HSS — Section G4

For rectangular HSS, the shear area is based on two webs:

V_n = 0.60 \, F_y \, A_w \, C_{v2} \quad \text{where } A_w = 2 h t_{\text{des}}

For round HSS and pipes:

A_w = 0.50 \, A_g

The $C_{v2}$ coefficient uses the same three-branch formulas as $C_{v1}$ but with $k_v = 5.0$ for the HSS webs and $\phi_v = 0.90$ .

4. Transverse Stiffeners — Section G2.2

Transverse stiffeners are required when $h/t_w > 2.46\sqrt{E/F_y}$ (= 59.2 for A992) and the required shear strength exceeds the available strength of the unstiffened web. Stiffener proportioning rules:

Moment of inertia about web: $I_{st} \geq b t_w^3 j$ where $j = 2.5/(a/h)^2 - 2 \geq 0.5$
Width: at least $b_f/3 - t_w/2$ but not less than 4 times the stiffener thickness
May be single-sided or pairs (pairs preferred for fatigue-critical applications)
Need not extend to the tension flange, but must be welded to the compression flange

5. Tension Field Action — Section G3

After a slender web panel buckles in shear, it does not lose all capacity. Diagonal tension stresses develop in the web, acting like the diagonals of a Pratt truss. This post-buckling strength is called tension field action (TFA):

V_n = 0.60 \, F_y \, A_w \left[C_{v2} + \frac{1 - C_{v2}}{1.15\sqrt{1 + (a/h)^2}}\right]

TFA is permitted only when the web panel is bounded by stiffeners and flanges on all four sides, and when the panel aspect ratio satisfies $a/h \leq 3.0$ (and $a/h \leq [260/(h/t_w)]^2$ ). End panels (at supports) cannot develop tension field action because there is no adjacent panel to anchor the diagonal tension.

TFA can increase the shear capacity of a stiffened plate girder by 30–80% beyond the initial buckling strength. It is the reason plate girders can have very slender webs (h/tw = 200+) and still carry significant shear — the web works as a tension field, not a shear panel.

6. Moment-Shear Interaction

AISC 360-16 does not have an explicit moment-shear interaction equation for most members. The interaction is considered indirectly: when $V_u > 0.60 \phi_v V_n$ , the moment capacity may be reduced. In practice, this is rarely critical for rolled shapes. For plate girders with tension field action, G3.3 provides a moment-shear interaction check using a circular interaction equation.

Solved Example 1 — W18×35 Simply Supported Beam

Given: W18×35 (W460×52), A992 ( $F_y = 50$ ksi), simply supported span 30 ft (9.14 m), uniform load $w_u = 2.5$ kip/ft (LRFD factored).

Properties: $d = 17.7$ in, $t_w = 0.300$ in, $h/t_w = 53.5$ .

Step 1 — Required Shear

V_u = \frac{w_u \cdot L}{2} = \frac{2.5 \times 30}{2} = 37.5 \text{ kips}

Step 2 — Web Slenderness Check

$h/t_w = 53.5 \leq 2.24\sqrt{E/F_y} = 53.9$ → $\phi_v = 1.00$ , $C_{v1} = 1.0$

Step 3 — Available Shear Strength

A_w = d \cdot t_w = 17.7 \times 0.300 = 5.31 \text{ in}^2

\phi_v V_n = 1.00 \times 0.60 \times 50 \times 5.31 \times 1.0 = 159.3 \text{ kips}

V_u = 37.5 \leq \phi_v V_n = 159.3 \text{ kips} \quad \checkmark \;\; (23.5\%)

At only 23.5% utilization, shear is far from governing. This is typical — for standard rolled W shapes with uniform loads, flexure controls long before shear becomes an issue. Shear typically governs only for short, heavily loaded beams (L/d < 5–8) or beams with concentrated loads near supports.

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Solved Example 2 — Plate Girder with Slender Web

Given: Welded plate girder, A572 Gr 50 ( $F_y = 50$ ksi), web 48 in × 3/8 in ( $h/t_w = 128$ ), no transverse stiffeners. Check shear capacity.

Step 1 — Web Coefficient (Unstiffened)

$k_v = 5.34$ (no stiffeners)

1.10\sqrt{k_v E/F_y} = 1.10\sqrt{5.34 \times 29{,}000/50} = 61.2

1.37\sqrt{k_v E/F_y} = 1.37\sqrt{5.34 \times 29{,}000/50} = 76.2

Since $h/t_w = 128 > 76.2$ → elastic web buckling:

C_{v1} = \frac{1.51 \times 5.34 \times 29{,}000}{128^2 \times 50} = \frac{233{,}827}{819{,}200} = 0.285

Step 2 — Available Shear Strength (Unstiffened)

A_w = 48 \times 0.375 = 18.0 \text{ in}^2

\phi_v V_n = 0.90 \times 0.60 \times 50 \times 18.0 \times 0.285 = 138.5 \text{ kips}

(Note: $\phi_v = 0.90$ because $h/t_w > 53.9$ )

Step 3 — With Stiffeners at a = 60 in (a/h = 1.25)

k_v = 5 + \frac{5}{(60/48)^2} = 5 + \frac{5}{1.5625} = 8.20

1.10\sqrt{8.20 \times 29{,}000/50} = 75.8

1.37\sqrt{8.20 \times 29{,}000/50} = 94.3

Since $128 > 94.3$ → still elastic buckling:

C_{v2} = \frac{1.51 \times 8.20 \times 29{,}000}{128^2 \times 50} = \frac{358{,}878}{819{,}200} = 0.438

Now add tension field action (G3):

V_n = 0.60 \times 50 \times 18.0 \left[0.438 + \frac{1 - 0.438}{1.15\sqrt{1 + 1.25^2}}\right]

= 540 \left[0.438 + \frac{0.562}{1.15 \times 1.601}\right] = 540 \left[0.438 + 0.305\right] = 540 \times 0.743

V_n = 401.2 \text{ kips}

\phi_v V_n = 0.90 \times 401.2 = 361 \text{ kips}

Impact of stiffeners + TFA: The unstiffened capacity is 138 kips. Adding stiffeners at 60 in spacing with tension field action increases it to 361 kips — a 161% increase. For plate girders, transverse stiffeners are not just cosmetic; they are the primary mechanism for achieving adequate shear capacity.