% Students Correctly Answered:
59.5%
Solution
Answer: B) 52 kip-ft & 108 psi
1. Calculate the self-weight of the RC beam:
Uniform self-weight (\(W_{sw}\))= Width x thickness x RC:
\[= \frac{6 \ in}{12 \ in/ft} \times \frac{24 \ in}{12 \ in/ft} \times \frac{150 \ pcf}{1000} = 0.150\ kip/ft\\\]
2. Calculate Reactions:
Sum of Moment around Point b = 0
\[-F_{ya} \times 20 + 0.15 \times 20 \times 20/2 -10\times 5 -0.15\times5\times 5/2 =0 \\\]
\[F_{ya} = -1.094\ kip\quad (downward)\\\]
Sum of Moment around point a = 0
\[-F_{yb} \times 20 +10 \times (5+20) + 0.15 \times (5+20) \times (5+20)/2 = 0 \\\]
\[F_{yb} = 14.85\ kip\\\]
Verify the sum of all vertical forces = 0
\[F_{ya} + F_{yb} = 0.15 \times 25 + 10 = 13.75\ kip \\\]
\[F_{ya} + F_{yb} = -1.094 + 14.85 = 13.75\ kip\ - OK\]
Sum of Moment at point b considering right side section
\[10\times5 + 0.15 \times 5 \times \frac{5}{2} = 51.875\ kip.ft\\\]
\(M_{max}\ \text{at point b or at mid span of the beam}\\\)
\[M_{mid\ span}\ = \frac{M_{b}}{2}-\frac{W_{sw} L^2}{8}=\frac{51.9}{2}-\frac{0.15 \times 20^2}{8} = 18.45\ kip.ft \\\]
\[Hence\ M_{b} = M_{max} = 51.88\ kip.ft\\\]
\[\sigma_{max} = M_{max} \times Y_{max} / I \\\]
\[I = \frac{B W^3}{12} = \frac{6 \times 24 ^3}{12} = 6912\ in^3\]
\[Y_{max }= \frac{24}{2} = 12\ in\]
\[\sigma_{max} = 51.88\ kip.ft \times 1000 \ lb/kip \times 12 \ in/ft \times \frac{12 \ in}{6912 \ in^3} = 1080\ psi\\\]
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