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Maximum Shear Stress Example

Medium DifficultyFE Mechanics of Materials

Find the maximum shear stress in a simply supported beam carrying a uniform distributed load and a center point load. The beam has an I-beam cross-section.

Simply supported beam with UDL 5 kN/m and center point load 10 kN, with I-beam cross-section.

Concept

The shear stress formula at a horizontal cut in a beam is: τ = \frac{V Q}{I b} where V = shear force at the section, Q = first moment of area above (or below) the cut about the neutral axis, I = moment of inertia of the full section, b = width at the cut. For an I-beam, maximum shear stress occurs at the neutral axis (in the web).

Problem

A simply supported beam of length L carries a uniform distributed load w over its entire span and a concentrated point load P at midspan. The beam has an I-beam cross-section. Find: Support reactions at A and B Maximum shear force in the beam Maximum shear stress in the beam

Given

L = 10 m w = 5 kN/m (UDL, downward) P = 10 kN (point load at midspan, downward) I-beam cross-section (from diagram): h = 250 mm, b = 200 mm, t_{f} = 25 mm, h_{w} = 200 mm, t_{w} = 40 mm

Support reactions (symmetry)

By symmetry, each support carries half the total load. Total load = w L + P . R_{A} = R_{B} = \frac{w L + P}{2} R_{A} = R_{B} = \frac{50 + 10}{2} = 30 kN ↑

Maximum shear force

The shear force is largest at the supports. Just to the right of A, V = R_{A} . Moving toward midspan, the UDL reduces the shear; the point load at midspan causes a step change. So the maximum magnitude occurs at the supports. ∣ V ∣_{m a x} = 30 kN At the supports (before UDL reduces the shear)

Moment of inertia (I)

Same I-beam as the bending stress example. From the cross-section dimensions (web + two flanges with parallel-axis theorem), the moment of inertia about the neutral axis is: I = 153.75 \times 1 0^{6} mm^{4} = 1.5375 \times 1 0^{- 4} m^{4}

First moment of area (Q) at the neutral axis

For maximum shear stress at the NA, take the area above the neutral axis. This is the top flange plus the part of the web above the NA. Q = \sum A_{i} \overset{y}{ˉ}_{i}, where \overset{y}{ˉ}_{i} is the distance from the NA to each sub-area centroid. Q_{flange} = A_{f} \cdot \overset{y}{ˉ}_{f} = (200) (25) (112.5) = 562500 mm^{3} (Top flange: area 200\times25 mm², centroid 112.5 mm above NA.) Q_{web above NA} = A_{w} \cdot \overset{y}{ˉ}_{w} = (40) (100) (50) = 200000 mm^{3} (Web above NA: 40\times100 mm², centroid 50 mm above NA.) Q = Q_{flange} + Q_{web} = 762500 mm^{3} = 7.625 \times 1 0^{- 4} m^{3}

Maximum shear stress

At the neutral axis, width b = t_{w} = 40 mm = 0.04 m . Apply the shear formula with V_{m a x} = 30 kN, and convert units to N and m for τ in Pa (then MPa). τ_{m a x} = \frac{V _{m a x} Q}{I b} τ_{m a x} = \frac{( 30 \times 1 0 ^{3} ) ( 7.625 \times 1 0 ^{- 4} )}{( 1.5375 \times 1 0 ^{- 4} ) ( 0.04 )} = 3.72 MPa

Final Answer

(1) Support reactions at A and B R_{A} = R_{B} = \frac{50 + 10}{2} = 30 kN ↑ (2) Maximum shear force in the beam ∣ V ∣_{m a x} = 30 kN (3) Maximum shear stress in the beam τ_{m a x} = \frac{( 30 \times 1 0 ^{3} ) ( 7.625 \times 1 0 ^{- 4} )}{( 1.5375 \times 1 0 ^{- 4} ) ( 0.04 )} = 3.72 MPa At the neutral axis, in the web of the I-beam.

Key Formulas

τ = \frac{V Q}{I b} (shear stress at a cut; b = width at cut)

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