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Simply Supported Beam with Uniform Load Example

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Find support reactions and maximum shear and bending moment.

Concept

For a simply supported beam carrying a uniform distributed load across its entire span: The total load on the beam is equal to the uniform distributed load multiplied by the span length: w \times L . Because the load is symmetric across the span, the reactions at the two supports are equal. The maximum bending moment occurs at midspan. Key formulas: R_{A} = R_{B} = \frac{w L}{2} M_{m a x} = \frac{w L ^{2}}{8}

A simply supported beam of length L carries a uniform distributed load w over its entire span. Find: Vertical reactions at both supports Maximum shear force and its location Maximum bending moment and its location

L = 10 m w = 5 kN / m (downward)

By symmetry, each support carries half the total load. W = w L W = (5 kN/m) (10 m) = 50 kN R_{A} = R_{B} = \frac{W}{2} R_{A} = R_{B} = \frac{50}{2} = 25 kN ↑

For a simply supported beam with UDL, max shear occurs at the supports. The shear varies linearly from + R_{A} at the left to - R_{B} at the right. V_{m a x} = R_{A} = R_{B} V_{m a x} = 25 kN at x = 0 and x = L

Max moment occurs at midspan where V = 0 . For UDL on simply supported beam: M_{m a x} = \frac{w L ^{2}}{8} M_{m a x} = \frac{( 5 ) ( 1 0 ^{2} )}{8} M_{m a x} = 62.5 kN \cdot m (at midspan, x = 5 m)

The shear force diagram helps visualize how internal shear varies along the beam. Linear decrease from +25 kN at the left support to -25 kN at the right support. The shear diagram is a straight line because the UDL produces a constant rate of change in shear: \frac{d V}{d x} = - w

Parabolic curve with maximum at midspan (62.5 kN\cdotm). The moment diagram is parabolic because shear (the derivative of moment) varies linearly. Zero moment at both supports, maximum where V = 0 .

For a simply supported beam with a uniform distributed load, the shear force varies linearly along the span. The bending moment reaches its maximum value where V = 0, which occurs at the midpoint. Evaluating the moment equation at midspan gives: M_{m a x} = \frac{w L ^{2}}{8}

R_{A} = R_{B} = 25 kN ↑ V_{m a x} = 25 kN at x = 0 and x = L M_{m a x} = 62.5 kN \cdot m at x = 5 m (midspan)

R_{A} = R_{B} = \frac{w L}{2} (kN) M_{m a x} = \frac{w L ^{2}}{8} (kN \cdot m) where w = uniform load, L = beam span

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