You’re setting up a W610x125 in your frame program, and the shear area field is sitting at zero. For a W-shape the right entry is roughly web depth times web thickness, about 7,300 \(\text{mm}^2\). For a solid rectangle it’s \(A_v = \tfrac{5}{6}bd\). Whether it matters comes down to span-to-depth. A 6 m simply supported W610x125 already gives 10% of its midspan deflection to shear.
Zero is accepted and the model runs. This post covers what the field does, how to calculate \(A_v\) for common sections, and where skipping it starts to cost you.
What is shear area?
Shear area (\(A_v\)) is the effective area a beam element uses to compute its shear stiffness, \(GA_v\). It’s smaller than the gross area because shear stress isn’t uniform across a section. A rectangle has \(A_v = \tfrac{5}{6}bd\). An I-shape has roughly its web area, \(d \times t_w\). A value of zero tells the software to ignore shear deformation.
That last point trips people up. Entering zero doesn’t mean “zero shear capacity.” It means “ignore shear deformation entirely.” The program falls back to classical Euler-Bernoulli beam theory. There, plane sections stay plane and perpendicular to the neutral axis. That’s fine for slender beams where bending dominates. For deeper sections or shorter spans, shear deformation adds to the total deflection.
Note: A zero in the shear area field won’t break your model. It just skips the shear flexibility contribution, which is acceptable for most typical beams.
How to calculate shear area for common sections
Shear stress in a rectangle varies parabolically: zero at the top and bottom fibres, 1.5 times the average value at the neutral axis. The effective area comes from matching the strain energy of that parabola to a uniform stress, which gives \(A_v = \kappa A\) with \(\kappa = 5/6\) for a rectangle.
Shear stress over a rectangle is parabolic, not uniform, so the effective shear area is less than the gross area.
| Section | Shear area \(A_v\) | Note |
|---|---|---|
| Solid rectangle | \(\tfrac{5}{6}bd\) | \(bd/1.2\) |
| Solid circle | \(A/1.11\) | about \(0.9A\) |
| I-shape or H-shape, major axis | \(d \times t_w\) | web only, about 46% of gross for a W610x125 |
| Rectangular HSS, box | \(2 \times h \times t\) | the two walls parallel to the shear, \(h\) is the depth in the shear direction |
| Round tube | about \(A/2\) | thin wall |
For I-shapes the flanges carry a little shear too, so web area slightly underestimates \(A_v\). Web area is a safe ballpark for stiffness. If your program’s section library already populates \(A_v\), use its value and skip the hand calc.
Worked example: shear area and the deflection split
Two members, one deflection check. Pull \(d\) and \(t_w\) for the steel section from the CISC section tables .
Shear area of two sections and the deflection split on a W610x125
Simply supported, point load at midspan; shear area from the rectangle and web-area rules above
9 input values\(b \times h = 400 × 800\) mm, \(d = 612\) mm, \(t_w = 11.9\) mm ...
Step 1: Rectangle shear area\(A_v = 266{,}700\ \text{mm}^2\)
Step 2: W-shape shear area\(A_v = 7283\ \text{mm}^2\)
Step 3: Bending deflection\(\delta_b = 4.57\ \text{mm}\)
Step 4: Shear deflection\(\delta_s = 0.53\ \text{mm}\)
Step 5: Shear share of total\(10.5\ \%\)
Shear is \(0.53/5.10 = 10.5\%\) of the total. The span-to-depth ratio is 9.8, below the usual “above 10” rule of thumb. That rule assumes a solid rectangle, and the thin web here leaves only 46% of the gross area for shear, against 83% for a rectangle. So use the calculated 10.5% share to decide, not L/d.
The 400 x 800 concrete beam over 4 m (\(L/d = 5\)) gives a similar answer. With \(f'_c = 35\) MPa, \(E_c = 4500\sqrt{f'_c} = 26{,}600\) MPa (CSA A23.3-24 Cl. 8.6.2.3), \(G = 0.4E_c = 10{,}600\) MPa and gross section properties, shear is 10.7% of the total.
When shear deformation matters
The total deflection has a bending part that scales with \(L^3\) and a shear part that scales with \(L\). For a simply supported beam with a midspan point load, the ratio is:
$$\frac{\delta_s}{\delta_b} = \frac{12EI}{A_vGL^2}$$For a tip-loaded cantilever it’s \(3EI/(A_vGL^2)\), a quarter as much at the same \(L\). With \(I = bd^3/12\), \(A_v = \tfrac{5}{6}bd\) and \(E/G \approx 2.6\), a rectangle works out to:
- Simply supported, midspan load: \(\delta_s/\delta_b \approx 3.1(d/L)^2\)
- Cantilever, tip load: \(\delta_s/\delta_b \approx 0.78(d/L)^2\)
A cantilever of length \(L\) behaves like half of a simply supported span of \(2L\), which is why the cantilever column below lags by a factor of two in \(L/d\). Shear as a share of total deflection, rectangular section:
| L/d | Simply supported | Cantilever |
|---|---|---|
| 2 | 43.8% | 16.3% |
| 3 | 25.7% | 8.0% |
| 4 | 16.3% | 4.6% |
| 6 | 8.0% | 2.1% |
| 8 | 4.6% | 1.2% |
| 10 | 3.0% | 0.8% |
| 12 | 2.1% | 0.5% |
| 15 | 1.4% | 0.3% |
So for a rectangle on simple supports, shear is about 2% at \(L/d = 12\), under 5% down to about 8, and 5 to 10% or more below that. A cantilever doesn’t cross 5% until \(L/d\) is about 4. Below \(L/d\) of 4 (2 for a cantilever), shear is over 15%. You’re past ordinary beam theory entirely. CSA A23.3-24 Cl. 10.7.1 draws the deep flexural member line at the same place. That means a clear span of at most 4d, or at most 2d for a cantilever. It also covers a significant concentrated load within 2d of the support. Cl. 10.7.2 then points to strut-and-tie modelling under Cl. 11.4.
Shear share of total deflection for a point load. The W610x125 crosses 5% near L/d of 15, not 8.
Thin-webbed sections are the catch. The W610x125 curve sits about 3.6 times above the rectangle’s, so steel I-shapes need roughly \(L/d \approx 15\) before shear drops under 5%. Use the general ratio with your own \(A_v\) rather than the rectangular rule.
Comparison of deflected shapes for a cantilever beam with and without shear deformation, showing the additional “sliding” displacement from shear.
I see this come up most often when engineers model plate girders, spandrel beams or transfer beams as standard beam elements without a shear area. For how Timoshenko elements treat those members, see shear deformation in deep members.
A quick check before you decide:
- Compute \(\delta_s/\delta_b\) for your support and load case with your own \(A_v\), \(E/G\) and \(L\).
- Under about 5% of total deflection, enter zero or ignore it.
- At 10 to 15% or more, include the shear area.
What the Canadian codes say about shear stiffness
Seismic analysis is where shear area stops being optional for concrete. CSA A23.3-24 Cl. 21.2.5.2 (Table 21.1) tells you to analyze with reduced section properties. For coupling beams it gives an effective shear area as a fraction of gross. That’s 0.15\(A_g\) for beams without diagonal reinforcement (Cl. 21.5.8.1) and 0.45\(A_g\) with it (Cl. 21.5.8.2). The matching \(I_e\) values are 0.4\(I_g\) and 0.25\(I_g\). That’s far below the \(\tfrac{5}{6}bd\) you’d use uncracked, so a cracked coupling beam is much softer in shear than your first guess. In ETABS or SAP2000 it goes in as the shear area property modifier. Check the table in the edition your project references. The values above are from the 2024 edition.
Drift is checked against NBCC 4.1.8.13, and the deflections come from that same model. A coupling beam or spandrel that’s too stiff in shear under-predicts drift.
Stiffness matrix and software notes
With shear included, the beam stiffness matrix picks up a factor \(\Gamma = 12EI/(A_vGL^2)\), the same expression as the simply supported ratio above. The translational terms scale by \(1/(1+\Gamma)\), and the rotational terms become \((4+\Gamma)/(1+\Gamma)\) and \((2-\Gamma)/(1+\Gamma)\) times \(EI/L\). At \(\Gamma = 0\) (infinite shear area) you recover Euler-Bernoulli. Shear makes the structure more flexible and can shift load distribution in indeterminate frames.
- SAP2000 and ETABS: a shear area of zero excludes shear deformation in that direction. Standard database sections come with \(A_v\) already filled in.
- S-Frame and RISA: check the manual. Some programs control shear deformation with a global solver option instead of a per-section zero.
If your program never asks for shear area, find out whether its beam element is Timoshenko (includes shear) or Euler-Bernoulli (excludes it).
Back to that field in your model
Three rules for Monday:
- Rectangles get \(\tfrac{5}{6}bd\). I-shapes get the web area, \(d \times t_w\).
- Run the ratio with your own \(A_v\). Rectangles are fine at \(L/d\) above about 8, steel I-shapes want 15 or more, and cantilevers get half those values.
- For coupling beams and deep spandrels in seismic models, use the A23.3 Table 21.1 shear area and run the model once with shear on and once with it off. If drifts or force distribution move, keep it on.