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Maximum Bending Stress Example

Medium DifficultyFE Mechanics of Materials

Find the maximum bending 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 flexure formula gives the bending stress at any point in a beam: σ = \frac{M y}{I} = \frac{M c}{I} where M = bending moment, c = distance from neutral axis to extreme fiber, I = moment of inertia. Maximum stress occurs where M is largest and y = c .

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 bending moment Maximum bending 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 bending moment

For a simply supported beam, the maximum moment occurs at midspan. The UDL and point load both contribute: M_{m a x} = \frac{w L ^{2}}{8} + \frac{P L}{4} M_{m a x} = \frac{( 5 ) ( 1 0 ^{2} )}{8} + \frac{( 10 ) ( 10 )}{4} = 62.5 + 25 = 87.5 kN \cdot m

Distance from neutral axis to extreme fiber (c)

The neutral axis is at mid-height of the symmetric I-section. So c is half the total depth: c = \frac{h}{2} = \frac{250}{2} = 125 mm = 0.125 m

Moment of inertia (I)

For the I-section, find I about the horizontal centroidal axis by adding the web inertia and two flanges (using the parallel-axis theorem for each flange). All dimensions in mm; convert to m⁴ at the end. I_{web} = \frac{t _{w} h _{w}^{3}}{12} = \frac{( 40 ) ( 200 ) ^{3}}{12} = 26.67 \times 1 0^{6} mm^{4} I_{flange} = \frac{b t _{f}^{3}}{12} + A d^{2} = \frac{( 200 ) ( 25 ) ^{3}}{12} + (200) (25) (112.5)^{2} \approx 63.54 \times 1 0^{6} mm^{4} (Flange centroid is d = 125 - 12.5 = 112.5 mm from the NA.) I = I_{web} + 2 I_{flange} = 26.67 \times 1 0^{6} + 2 (63.54 \times 1 0^{6}) = 153.75 \times 1 0^{6} mm^{4} = 1.5375 \times 1 0^{- 4} m^{4}

Maximum bending stress

Apply the flexure formula at the section of maximum moment. Maximum stress occurs at the extreme fibers where y = c . σ_{m a x} = \frac{M _{m a x} c}{I} σ_{m a x} = \frac{( 87.5 \times 1 0 ^{3} ) ( 0.125 )}{1.5375 \times 1 0 ^{- 4}} = 71.1 MPa

Final Answer

(1) Support reactions at A and B R_{A} = R_{B} = \frac{50 + 10}{2} = 30 kN ↑ (2) Maximum bending moment M_{m a x} = \frac{( 5 ) ( 1 0 ^{2} )}{8} + \frac{( 10 ) ( 10 )}{4} = 62.5 + 25 = 87.5 kN \cdot m (3) Maximum bending stress in the beam σ_{m a x} = \frac{( 87.5 \times 1 0 ^{3} ) ( 0.125 )}{1.5375 \times 1 0 ^{- 4}} = 71.1 MPa At midspan, top and bottom fibers of the I-beam.

Key Formulas

σ_{m a x} = \frac{M c}{I} or σ_{m a x} = \frac{M}{S} where S = I / c is the section modulus.

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