CIV-331 · Lesson 7 of 12 · 19 min

Shear, web strength and concentrated loads

The web of a beam carries the shear and resists the concentrated forces at supports and under point loads. In most rolled beams under uniform load the web is so generous that shear barely registers, but there are important exceptions: short heavily loaded beams, coped beam ends, deep plate girders with thin webs, and beams resting on walls or carrying columns. This lesson covers shear yielding and shear buckling, the local limit states of the web under concentrated loads, and the stiffeners that strengthen webs when they are needed.

Shear stress in an I-section

Elementary beam theory gives the shear stress at a level in a cross-section as \(\tau = VQ/(It)\), where \(Q\) is the first moment of the area beyond that level and \(t\) is the width there. In a rectangle the stress is parabolic, with a maximum of 1.5 times the average. In an I-section the flanges are wide, so the shear stress in them is small, while the web carries almost the whole shear at a nearly uniform stress. The maximum, at the neutral axis, is only 5 to 15 per cent above the average over the web area. Design therefore treats the web as carrying the shear uniformly over an area equal to the full depth times the web thickness.

Shear yielding occurs when the shear stress reaches the shear yield stress, which by the von Mises criterion is \(F_y/\sqrt 3 = 0.577F_y\). Strain hardening and the small contribution of the flanges allow a little more in practice.

Shear resistance of rolled shapes

S16 gives the factored shear resistance of a web as

\[V_r = \phi A_w F_s \qquad A_w = dw \qquad F_s = 0.66F_y \ \ \text{for}\ \ \frac{h}{w} \le 439\sqrt{\frac{k_v}{F_y}}\]

The shear buckling coefficient \(k_v\) is 5.34 for a web without transverse stiffeners, which gives a limit of \(h/w = 54.6\) at \(F_y = 345\) MPa. Almost all rolled W shapes are below it. Above the limit the shear stress \(F_s\) is reduced for inelastic and then elastic shear buckling, and transverse stiffeners, which raise \(k_v\), may be added. The value 0.66 is above the von Mises 0.577 because of strain hardening; with \(\phi = 0.90\) it gives about \(0.6F_y\) on the web area.

Worked problem

Shear in the mezzanine beams and girders

Check the W460x60 beam (\(d = 455\), \(w = 8.0\), \(t = 13.3\) mm) for \(V_f = 152\) kN and the W530x74 girder (\(d = 529\), \(w = 9.7\), \(t = 13.6\) mm) for \(V_f = 156\) kN.

Show solution
  1. Beam: \(h/w = (455 - 26.6)/8.0 = 53.5 < 54.6\), so \(F_s = 0.66F_y\):
  2. \[V_r = 0.90(455)(8.0)(0.66)(345) = 746\ \text{kN} \ge 152\]
  3. Girder: \(h/w = 51.7 < 54.6\):
  4. \[V_r = 0.90(529)(9.7)(0.66)(345) = 1052\ \text{kN} \ge 156\]

Answer: Shear uses only 20 per cent of the beam's resistance and 15 per cent of the girder's. This is typical of rolled beams under distributed loads.

Shear matters more in short spans with heavy loads, in transfer beams carrying columns, in coped ends where part of the web is removed, and at connections, where the web is checked again for block shear and for bearing of the bolts (lesson 9).

Coped beam ends

When a beam frames into a girder of the same depth, the top flange of the beam, and sometimes the bottom, is cut away so that it clears the girder flange. The cope removes the flange and part of the web at the end of the beam. The reduced section must carry the end shear and the small moment caused by the shear acting over the length of the cope, and a long cope can let the remaining web buckle locally. The block shear check in lesson 9 often governs coped ends, and stiffening or a deeper connection is used for long, deep copes.

For the mezzanine beams a 40 mm cope leaves a depth of 415 mm and a shear resistance of 680 kN, still far above the 152 kN reaction.

Web strength under concentrated loads

Where a beam sits on a support or carries a point load on its flange, the force passes through the flange into the web over a short length. Three things can happen to an unstiffened web:

  • Web yielding: the web just below the flange yields in compression over the length on which the force spreads.
  • Web crippling: the thin web buckles locally and folds just below the loaded flange.
  • Web compression buckling and sidesway web buckling: under forces applied to both flanges, or with the compression flange free to move sideways, the whole web depth buckles. These are checked for deep beams and for columns in moment connections.
A beam end resting on a bearing plate of length N on a concrete wall, with the force spreading through the flange and fillet at a slope so that it acts on a web length of N plus 4t at the end; an interior point load on a bearing of length N spreads to N plus 10t. A small inset shows the folded shape of web crippling beneath the flange.
Web yielding and crippling under end reactions and interior loads (S16 notation).

S16 gives the bearing resistance of the web, with \(\phi_{bi} = 0.80\), as the smaller of a yielding and a crippling expression:

\[\text{interior:}\ \ B_r = \phi_{bi}w(N + 10t)F_y \ \ \text{and}\ \ 1.45\phi_{bi}w^2\sqrt{F_yE} \qquad \text{end:}\ \ B_r = \phi_{bi}w(N + 4t)F_y \ \ \text{and}\ \ 0.60\phi_{bi}w^2\sqrt{F_yE}\]

where \(N\) is the length of bearing along the beam and \(t\) the flange thickness. The yielding expressions assume the force spreads at a slope through the flange; the crippling expressions come from tests and depend mainly on the web thickness squared.

Worked problem

Beam bearing on the concrete wall

At the end wall of the maintenance bay, the mezzanine beams rest on 150 mm steel bearing plates on the concrete wall. Check the W460x60 web for the end reaction \(V_f = 152\) kN, and check an interior storage rack post that delivers 120 kN through a 100 mm cap plate.

Show solution
  1. End, yielding:
  2. \[B_r = 0.80(8.0)(150 + 4 \times 13.3)(345) = 449\ \text{kN}\]
  3. End, crippling:
  4. \[B_r = 0.60(0.80)(8.0)^2\sqrt{345(200\,000)} = 30.7(8307) = 255\ \text{kN} \ge 152\]
  5. Interior, yielding and crippling:
  6. \[B_r = 0.80(8.0)(100 + 133)(345) = 514\ \text{kN} \qquad B_r = 1.45(0.80)(64)(8307) = 617\ \text{kN} \ge 120\]

Answer: Web crippling at the end, 255 kN, governs, and the web needs no stiffeners. The bearing plate is sized for the concrete: \(A_1 = 152\,000/(0.85 \times 0.65 \times 25) = 11\,000\) mm², so a 150 × 200 mm plate is ample.

Sofia's program runs every shear and bearing check in this lesson so they can be rerun if the reactions change.

Shear resistance, coped end, web bearing and bearing plate checks.
import math
# Shear and web bearing checks, CSA S16 clauses 13.4.1 and 14.3.2
E, Fy, phi, phibi = 200000.0, 345.0, 0.90, 0.80

def Vr(d, w, t):
    h = d - 2 * t
    kv = 5.34                                    # unstiffened web
    limit = 439 * math.sqrt(kv / Fy)
    assert h / w <= limit, "web needs the reduced shear stress"
    return phi * d * w * 0.66 * Fy / 1e3, h / w, limit

for name, (d, w, t, Vf) in {"W460x60 beam": (455, 8.0, 13.3, 151.5),
                             "W530x74 girder": (529, 9.7, 13.6, 151.5 + 1.25 * 0.73 * 4.5)}.items():
    v, hw, lim = Vr(d, w, t)
    print(f"{name}: h/w = {hw:.1f} <= {lim:.1f}, Vr = {v:.0f} kN, Vf = {Vf:.1f} kN, ratio {Vf/v:.2f}")

# Top flange cope 40 mm deep at the beam end
d, w, dc = 455, 8.0, 40
print(f"Coped W460x60: Vr = {phi*(d-dc)*w*0.66*Fy/1e3:.0f} kN on the reduced depth")

# Web bearing at the end support on the concrete wall: N = 150 mm
d, w, t, N = 455, 8.0, 13.3, 150
yield_end = phibi * w * (N + 4 * t) * Fy / 1e3
crip_end = 0.60 * phibi * w**2 * math.sqrt(Fy * E) / 1e3
print(f"End bearing: web yielding {yield_end:.0f} kN, web crippling {crip_end:.0f} kN, Vf = 151.5 kN")

# Interior concentrated load from a storage rack post: 120 kN on N = 100 mm
N, P = 100, 120.0
yield_int = phibi * w * (N + 10 * t) * Fy / 1e3
crip_int = 1.45 * phibi * w**2 * math.sqrt(Fy * E) / 1e3
print(f"Interior load: web yielding {yield_int:.0f} kN, web crippling {crip_int:.0f} kN, Pf = {P:.0f} kN")

# Bearing plate on concrete, f'c = 25 MPa, A2/A1 taken as 1
fc, phic = 25.0, 0.65
A1 = 151.5e3 / (0.85 * phic * fc)
print(f"Bearing plate area needed = {A1:.0f} mm2, e.g. 150 x {A1/150:.0f} mm; use 150 x 200 mm")
Output
W460x60 beam: h/w = 53.5 <= 54.6, Vr = 746 kN, Vf = 151.5 kN, ratio 0.20
W530x74 girder: h/w = 51.7 <= 54.6, Vr = 1052 kN, Vf = 155.6 kN, ratio 0.15
Coped W460x60: Vr = 680 kN on the reduced depth
End bearing: web yielding 449 kN, web crippling 255 kN, Vf = 151.5 kN
Interior load: web yielding 514 kN, web crippling 617 kN, Pf = 120 kN
Bearing plate area needed = 10968 mm2, e.g. 150 x 73 mm; use 150 x 200 mm

Bearing stiffeners

If the web cannot carry a concentrated force, pairs of bearing stiffeners are welded to both sides of the web under the load and fitted tightly to the loaded flange. They act with a strip of the web as a short column. The stiffener outstands must meet a local buckling limit, the bearing area of the stiffener ends must carry the force without crushing, and the stiffener-web column, with an effective length of about three quarters of the web depth, must not buckle. Stiffeners are expensive because of the fitting and welding, so designers often prefer a longer bearing plate, which raises the yielding and crippling resistances, or a heavier beam.

Plate girders and tension field action

When spans or loads are too large for rolled shapes, girders are built up from plates welded together. Their webs are deep and thin to save weight, and shear buckling becomes the governing concern. A thin web panel between transverse stiffeners does not collapse when it buckles: the buckled web carries diagonal tension like the diagonals of a Pratt truss, while the stiffeners act as the verticals. This tension field action raises the post-buckling shear strength well above the buckling load. Both standards allow it for interior panels with stiffeners at suitable spacing, and the shear resistance then depends on the panel aspect ratio and the web slenderness. Plate girders also need checks for flange-induced web buckling and for interaction of shear and moment in some cases.

A plate girder elevation with a thin web divided into panels by transverse stiffeners. In each panel, dashed diagonal lines show the tension field running from the top of one stiffener to the bottom of the next, like the diagonals of a Pratt truss, with the stiffeners acting as compression verticals.
Tension field action in a stiffened plate girder web.

Shear and moment together

A simply supported beam has its largest shear at the supports, where the moment is zero, and its largest moment at midspan, where the shear is small, so the two rarely interact. They meet at the interior supports of continuous beams, at the ends of cantilevers and under heavy point loads close to a support. For rolled shapes with stocky webs, tests show little reduction in plastic moment until the shear exceeds about 60 per cent of the shear capacity, and neither standard requires an interaction check for them. Plate girders that rely on tension field action are different: the tension field and the flanges share the web, and the standards limit the combination. A quick test for any beam is to compare the factored shear with about 0.6 of the shear resistance at the sections of high moment; if it is lower, as it is for every member in the mezzanine, the two checks are independent.

Practical points

  • Check shear in short, heavily loaded beams and at coped ends; for typical floor beams it rarely governs.
  • Give beams adequate bearing length on walls and use bearing plates sized for the concrete or masonry.
  • Where a column bears on a beam, check web yielding and crippling, and consider bearing stiffeners or a heavier web.
  • Avoid stiffeners where a slightly heavier rolled section would do; fabrication time costs more than steel weight.

Common mistakes

  • MistakeUsing only the web clear depth for shear area in a rolled shape.

    InsteadS16 takes \(A_w = dw\), the full depth times web thickness.

  • MistakeChecking web yielding but not crippling.

    InsteadCrippling often governs at beam ends, especially for thin webs.

  • MistakeForgetting that a cope reduces both shear area and flexural strength at the beam end.

    InsteadCheck the coped section and block shear at the connection.

  • MistakeAdding stiffeners by habit.

    InsteadA longer bearing or a heavier section is often cheaper.

Summary and key terms

  • In an I-section the web carries nearly all the shear at an almost uniform stress.
  • S16: \(V_r = \phi A_wF_s\), \(F_s = 0.66F_y\) for \(h/w \le 439\sqrt{k_v/F_y}\).
  • Under concentrated loads, webs are checked for yielding and crippling; crippling often governs at ends.
  • Bearing stiffeners act as short columns; longer bearings or heavier sections are alternatives.
  • Thin plate girder webs develop tension field action between stiffeners after buckling.
  • The mezzanine beam end, W460x60 on a 150 mm plate, has a crippling resistance of 255 kN against 152 kN.
Shear yielding
Yielding of the web under shear stress near Fy/√3.
Shear buckling
Buckling of a thin web panel under shear.
Web crippling
Local folding of the web just below a concentrated load.
Web yielding
Compressive yielding of the web over the length on which a concentrated force spreads.
Bearing stiffener
A plate welded to the web under a concentrated load to act with the web as a column.
Cope
A cutout at the end of a beam that removes part of a flange and web to clear another member.
Tension field action
Post-buckling shear resistance of a stiffened web through diagonal tension.
Plate girder
A girder built up from plates welded together.

Practice problems

Work each problem on paper first, then reveal the answer to check.

  1. Find \(V_r\) for a W410x39 (\(d = 399\) mm, \(w = 6.4\) mm).

    Show answer

    \(h/w = 59.6 > 54.6\), so the shear stress is reduced for buckling; the simple value \(0.90(399)(6.4)(0.66)(345) = 523\) kN is an upper bound.

  2. Why does crippling depend on the web thickness squared?

    Show answer

    It is a plate buckling problem, and plate buckling strength grows with the square of thickness.

  3. How long a bearing plate would make end web yielding of the W460x60 equal to 300 kN?

    Show answer

    \(300\,000 = 0.80(8.0)(N + 53.2)(345)\), so \(N = 83\) mm; crippling would still govern at 255 kN.

  4. In tension field action, what plays the role of the truss verticals?

    Show answer

    The transverse stiffeners, which carry compression.

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