Check a simply supported rectangular beam by hand, bending only. Shear adds about \(3(h/L)^2\) on top of that deflection. It’s 2% at \(L/h = 12\) and 5% at 8. At 4 it’s 19%. A rolled W-shape is worse, around \(12(d/L)^2\). The thin web carries all the shear while the flanges do the bending. That is why a 5 m W610x125 transfer girder picks up 17% more deflection than your bending-only hand check. An 8 m floor beam picks up 3%.
This is about shear deformation, meaning deflection, drift and stiffness. Shear strength is a separate check.
What is shear deformation?
Shear deformation is the sideways sliding of a member’s cross-sections under shear force, and the deflection it adds is called shear deflection. Think of a deck of cards: bending curves the stack, shear slides each card over the next. A beam does both at once, so total deflection is the bending part plus the shear part.
Bending curves the beam; shear slides the slices. The two deflections add.
Euler-Bernoulli beam theory keeps only the bending part. Timoshenko beam theory adds the sliding, and most frame programs switch to it as soon as you give a member a shear area. Slender members don’t care. Deep ones do.
How big is the shear part?
For a simply supported beam with a point load \(P\) at midspan:
$$ \delta_b = \frac{PL^3}{48EI} \qquad \delta_s = \frac{PL}{4A_vG} $$Here \(A_v\) is the effective shear area, not the gross area, and \(G = E/(2(1+\nu))\). That is 77 GPa for steel (\(E/G = 2.6\)) and about \(0.42E\) for uncracked concrete with \(\nu = 0.2\) (\(E/G = 2.4\)). Bending grows with \(L^3\), shear with \(L\), so as the member gets shorter, shear takes a bigger slice. The ratio is:
$$ \frac{\delta_s}{\delta_b} = \frac{12EI}{A_vGL^2} $$For a solid rectangle, \(A_v = \tfrac{5}{6}bh\) and \(I = bh^3/12\). That gives \(\delta_s/\delta_b = 1.2\,(E/G)\,(h/L)^2\), about \(3.0\,(h/L)^2\). Now take a rolled W-shape bending about its strong axis, with \(A_v \approx d\,t_w\). The same ratio is \(C\,(d/L)^2\). \(C\) is between about 10 and 15 for the sections used as beams (W200x27 through W1000x296, \(E/G = 2.6\)).
| L/h or L/d | Rectangle, C of about 3 | W-shape, C of about 12 |
|---|---|---|
| 20 | 1% | 3% |
| 12 | 2% | 8% |
| 8 | 5% | 19% |
| 6 | 9% | 33% |
| 4 | 19% | 75% |
Shear deflection as a percentage of bending deflection, simply supported, midspan point load.
Two things to take from the table. First, the old “L/h of 12 or more is safe” rule holds for rectangles only. For a W-shape, shear is still about 8% at 12. It falls below 5% of total deflection only beyond \(L/d\) of about 15. Second, the load case moves the numbers. A uniformly distributed load gives 0.8 times the ratio above. A cantilever with a tip load gives one quarter of it (with \(L\) the cantilever length). A fixed-fixed member swaying in double curvature, like a coupling beam, gives the same ratio as the table.
Same W610x125 as the worked example, other spans. The lines cross at 2.1 m and shear passes 5% of bending below 9.2 m.
Code check: CSA A23.3-24 Cl. 10.7.1 calls a flexural member deep when the clear span is 4d or less, a significant concentrated load sits within 2d of the support face, or a cantilever’s clear span is 2d or less. Cl. 10.7.2 then allows the strut-and-tie model of Cl. 11.4. That is a strength classification, but it tells you plane sections are gone, and for a rectangle at \(L/h = 4\) shear is already about 16% of the total deflection.
Treat shear as significant once it is above about 5 to 10% of the deflection that governs the limit state.
Worked example: W610x125 transfer girder
A simply supported W610x125, \(L = 5\) m, carries a 500 kN service column load at midspan. Section properties from the CISC tables are \(d = 612\) mm, \(t_w = 11.9\) mm and \(I_x = 985 \times 10^6\) mm\(^4\); use \(E = 200\,000\) MPa and \(G = 77\,000\) MPa. To run it for your own girder, pull d, tw and Ix for your own W-section and swap the numbers in.
Bending and shear deflection of a W610x125 transfer girder
Simply supported, midspan point load, service level; \(A_v = d\,t_w\)
7 input values\(P = 500\) kN, \(L = 5\) m, \(d = 612\) mm ...
Step 1: Shear area\(A_v = 7280\ \text{mm}^2\)
Step 2: Bending deflection\(\delta_b = 6.6\ \text{mm}\)
Step 3: Shear deflection\(\delta_s = 1.1\ \text{mm}\)
Step 4: Total deflection\(\delta = 7.7\ \text{mm}\)
Step 5: Shear share\(\delta_s / \delta = 14\,\%\)
At 5 m the shear part adds 1.1 mm to the 6.6 mm of bending, so the girder deflects 7.7 mm (\(L/647\)), 14% of it from shear.
Treated as a spring, the girder is about 14% softer than a bending-only model says. A frame model sees that as extra rotation at the columns, not just extra sag.
Now a typical floor beam: W460x74 (\(d = 457\) mm, \(t_w = 9.0\) mm, \(I_x = 332 \times 10^6\) mm\(^4\)), \(L = 8\) m, 15 kN/m uniformly distributed. \(A_v = 4\,110\) mm\(^2\), \(\delta_b = 5wL^4/(384EI) = 12.0\) mm and \(\delta_s = wL^2/(8A_vG) = 0.4\) mm. Shear is 3% of the total at \(L/d = 17.5\), which is why nobody includes it for floor beams.
The same equation gave 14% on the girder and 3% on the floor beam. Run it on any member with \(L/d\) under 15 before you decide to skip shear.
Where it shows up in a real model
| Member | Typical L/h | What to do |
|---|---|---|
| Transfer girders, deep plate girders | 3 to 8 | Include shear; input \(A_v\) from the web |
| Coupling beams | 1 to 3 | Include shear; check the Table 21.1 modifiers below |
| Perimeter spandrels in moment frames | 4 to 8 | Include shear where they are part of the lateral system |
| Heavy point load near a support | any | Include shear; high shear for little moment |
| Squat walls modelled as columns | \(H/l_w\) under 2 | See the wall section below |
Spandrels matter for drift. NBCC 4.1.8.13 limits interstorey drift, and the number you compare against comes from your model. If a deep spandrel is stiff only because shear deformation was off, the drift is under-predicted and so are the joint rotations. The stiffness modifiers for cracked concrete frames are the other half of that same stiffness question.
Code check: CSA A23.3-24 Cl. 21.2.5.2 and Table 21.1 specify analysis properties for seismic design: a coupling beam without diagonal reinforcement (Cl. 21.5.8.1) takes \(A_{ve} = 0.15A_g\) and \(I_e = 0.4I_g\), and one with diagonal reinforcement (Cl. 21.5.8.2) takes \(A_{ve} = 0.45A_g\) and \(I_e = 0.25I_g\). Table 21.1 gives \(A_{ve}\) as a fraction of gross area, not of the program’s shear area, so check what your modifier multiplies.
Take the first case at a clear span of twice the depth. Uncracked, \(\delta_s/\delta_b = 2.9 \times (1/2)^2 = 0.72\) for concrete, which is 42% of the total. Scale shear area to \(0.15A_g\) (which is 0.18 of \(\tfrac{5}{6}A_g\)) and \(I\) to \(0.4I_g\), and the ratio rises by \(0.4/0.18 = 2.2\) to 1.6, so shear is 62% of the deflection. If your coupling beams carry nothing in the shear area field, you are modelling a much stiffer link than the code intends. A what-if run with softer coupling beams shows how much the piers pick up.
Split the deflection in your model
You don’t need to trust the formulas. Run the member three ways:
- As designed, with a real shear area. This is the total.
- Shear area set to zero. In SAP2000 and ETABS that turns shear deformation off, so you see the bending part. Check the convention in your program.
- \(I\) multiplied by 100 to 1000 with the real shear area. Bending nearly vanishes, so you see the shear part.
Runs 2 and 3 add up to run 1 for a statically determinate member. In an indeterminate frame the split is only approximate, because stiffness changes redistribute forces, so read it as an indicator. If shear is above 5 to 10% at the member and load case that governs, keep it on there. Don’t turn it on for every beam; keep a bending-only reference model so the difference stays visible.
Shear area for common sections
Most programs take \(A_v\) from the section database, but you set it for custom and built-up sections. These are stiffness values, not the strength values in CSA S16 or A23.3.
| Section | \(A_v\) for stiffness |
|---|---|
| Solid rectangle | \(\tfrac{5}{6}bh\) |
| Rolled I or H shape, major axis | \(d\,t_w\) (clear web depth gives about 6% less for the W610x125) |
| Welded plate girder | web depth times \(t_w\) |
| Box girder | sum of the web areas |
| Channel or tee, major axis | web depth times \(t_w\) |
| Rectangular HSS, bending over depth \(h\) | \(2ht = A\,h/(b+h)\), so \(0.5A\) square, up to about \(0.7A\) for deep rectangles |
| Round HSS | about \(0.5A\) |
Watch out: Gross area is the wrong input. For the W610x125 it is 15 900 mm\(^2\) against 7 280 mm\(^2\) of shear area, so using it cuts the shear deflection by more than half. For where the form factors come from, see the shear area input and its validation check.
Squat shear walls
For a cantilever wall with a tip load, \(\delta_s/\delta_b = 3EI/(A_vGH^2) = 0.3\,(E/G)\,(l_w/H)^2\) for a rectangular wall, which is about \(0.72\,(l_w/H)^2\) for concrete (0.78 with \(E/G = 2.6\)).
| \(H/l_w\) | 0.5 | 1 | 2 | 3 |
|---|---|---|---|---|
| Shear share of total deflection | 74% | 42% | 15% | 7% |
Flexure curves, shear is a near-straight ramp, and the total is their sum. Wall with \(H/l_w = 1\) under a tip load, uncracked.
The usual rule is shear dominates below \(H/l_w = 1\), both matter from 1 to 2, and flexure governs above 2. It’s close, but not exact: shear and bending are equal at \(H/l_w\) of about 0.85, shear is still 42% at 1, and it’s still 15% at 2. These are uncracked, tip-load numbers, so use them to decide whether the wall deserves a shear area, not to size it. Shell-element walls normally include shear deformation. Column-type wall elements need \(A_v = \tfrac{5}{6}l_wt_w\) typed in.
Shear in the stiffness matrix
If you assemble matrices yourself, define
$$ \Gamma = \frac{12EI}{A_vGL^2} $$For the midspan point load, \(\Gamma = \delta_s/\delta_b\), and the bending stiffness over the shear stiffness is the same \(\Gamma\). The Timoshenko bending stiffness matrix (two degrees of freedom per node, translation and rotation) is:
$$ \mathbf{k}_b = \frac{EI}{L^3(1+\Gamma)} \begin{bmatrix} 12 & 6L & -12 & 6L \\ 6L & (4+\Gamma)L^2 & -6L & (2-\Gamma)L^2 \\ -12 & -6L & 12 & -6L \\ 6L & (2-\Gamma)L^2 & -6L & (4+\Gamma)L^2 \end{bmatrix} $$As \(A_v \to \infty\), \(\Gamma \to 0\) and you recover the Euler-Bernoulli matrix. As \(A_v\) drops, \(\Gamma\) grows: the translational terms soften by \(1/(1+\Gamma)\), the rotational stiffness \((4+\Gamma)/(1+\Gamma)\) drops more slowly, and the carry-over term \((2-\Gamma)/(1+\Gamma)\) shrinks fastest. In practice you only touch this through the shear area input, so getting \(A_v\) right is getting the shear part of the matrix right.
Monday checklist
- Compute \(C\,(d/L)^2\) for every deep member in the lateral system or carrying transfer loads. Use 3 for rectangles and 10 to 15 for W-shapes.
- Where it is above 5%, give the member a real \(A_v\) from the web and run the zero-shear-area and inflated-\(I\) cases to see the split.
- On coupling beams, confirm the Table 21.1 modifiers are applied to the shear area as well as to \(I\).