Practice Question: Lifting a beam

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Solution

Answer: A) 7 kip , 74 kip.ft

The weight of the beam will be distributed equally on the lifting points. 

Summation of vertical forces on the beam = 0 

\[F_{ya}  = F_{yb}  = 10.5/2 = 5.25\ kip\]

\[\sum  F_{ya}  = 0\]

\[F_{ac}   \times  9 / 15 = 5.25\ kip,\]

\[F_{ac} = 8.75\ kip = F_{bc}\  (from\ symmetry).\]

Normal force acting on top of the beam:

\[F_{xa}  = F_{xb}    = 8.75  \times  12 / 15 = 7\ kip\]

 

The moment is based on self-weight uniform load and the normal load acting on the top flange of the beam.

Self-weight moment = PL/4

\[= 10.5 \ kip  \times  24 / 4 = 63.0 \ kip. ft\]

Moment due to normal load at top flange of the beam. Consider half the beam depth as the moment arm:

\[= 7 \ kip \times \frac{35.9 \ in}{12 \ in/ft} \times \frac 12 = 10.47\ kip. ft\]

\[Total\ moment = 63.0+10.47 = 73.5\ kip.ft\]

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