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CivilStructural AnalysismediumFE ExamCoursework

Mid-Span Deflection of a Simply Supported Beam

A simply supported steel beam spans L=6 mL = 6\ \text{m} and carries a uniformly distributed load of w=10 kN/mw = 10\ \text{kN/m} over its full length.

Section properties: E=200 GPaE = 200\ \text{GPa}, I=200×106 mm4I = 200 \times 10^{6}\ \text{mm}^4.

Find the mid-span deflection.

Given
w10 kN/mUniformly distributed load
L6 mSpan
E200 GPaYoung's modulus
I200000000 mm^4Second moment of area
Hint 1

Identify the load case: simply supported, full-span UDL.

Hint 2

δ = 5wL⁴/(384EI) — note the fourth power on the span.

Hint 3

Convert I from mm⁴ to m⁴ using 10⁻¹².

Worked solution — try the problem first

For a simply supported beam under a full-span UDL:

δmax=5wL4384EI\delta_{max} = \frac{5wL^4}{384EI}

Convert to base SI. The II conversion is the usual trap — mm⁴ to m⁴ is 10−1210^{-12}:

w=10,000 N/m,E=200×109 Pa,I=200×106×10−12=2.0×10−4 m4w = 10{,}000\ \text{N/m}, \quad E = 200\times10^{9}\ \text{Pa}, \quad I = 200\times10^{6}\times10^{-12} = 2.0\times10^{-4}\ \text{m}^4

Numerator: 5wL4=5(10,000)(64)=5(10,000)(1296)=6.48×1075wL^4 = 5(10{,}000)(6^4) = 5(10{,}000)(1296) = 6.48\times10^{7}

Denominator: 384EI=384(200×109)(2.0×10−4)=384×4.0×107=1.536×1010384EI = 384(200\times10^{9})(2.0\times10^{-4}) = 384 \times 4.0\times10^{7} = 1.536\times10^{10}

δ=6.48×1071.536×1010=4.219×10−3 m\delta = \frac{6.48\times10^{7}}{1.536\times10^{10}} = 4.219\times10^{-3}\ \text{m}

δ=4.22 mm\boxed{\delta = 4.22\ \text{mm}}

Serviceability check. A common limit is L/360=6000/360=16.7L/360 = 6000/360 = 16.7 mm. At 4.22 mm this beam is comfortably stiff enough.

Concepts:Beam deflectionServiceabilityUnit consistency

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