Practice Question: Maximum Deflection on a Steel Box Section

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Solution

Answer: B) 0.45 inch

Refer to PE Civil Reference Handbook Section 4.1.7 for beam deflections.

By the superposition principle, deflection can be calculated for each load separately and added to get the total deflection on the beam.

Both the point load in the middle of the beam and the uniform load will result in a maximum deflection at the beam center.

\[L=18ft=18\times 12 = 216\ in\]

\[E= 29000\ ksi\]

\[W= 4\ kip/ft\\\]

\[B_{out} = 6 \ in\]\[T_{out} = 20 \ in\]

\[B_{in} = 6-2 \times 3/8 = 5.25 \ in\]
\[T_{in} = 20-2 \times 3/8 = 19.25\ in\]I
\[I= B_{out} T^3_{out} / 12- B_{in} T^3_{in} 12\]

\[= 6 \times 20^3 / 12-5.25 \times 19.25^3 /12 = 880\ inch^4\\\]

Uniform Load Maximum Deflection:

\[\Delta_{max_1}=\frac{5wl^4}{384EI} = \frac{5 \times 4 \times 1000/12 \times 216^4 }{384\times 29000 \times 880 } =0.37\ inch.\\\]

 

Point Load Maximum Deflection:

 

\[\Delta_{max_2}=\frac{Pl^3}{48EI} = \frac{10 \times 1000\times 216^3 }{48\times 29000 \times 880 } =0.08\ inch.\\\]

\[\Delta_{max}\ =\ \Delta_{max1}+\Delta_{max2}= 0.37+0.08 = 0.45\ inch\\\]
 

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