The beam fixed at end A and simply supported at C with a roller is shown in the figure. If B is an internal hinge and a 10 kN point load acts at D, and a UDL of 10 kN/m acts on AB, the magnitude of bending moment at the fixed end A is ____________ kNm.
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An internal hinge forces the bending moment at that point to be zero, isolating bending effects to each segment independently.
Member AB (length 3 m) carries a UDL of 10 kN/m. Total load on AB is:
\[
wL = 10 \times 3 = 30\ \text{kN}
\]
This load acts at the centroid — 1.5 m from A.
Because B is an internal hinge, the bending moment at B is zero. Member AB is a fixed–hinged beam. The fixed–end moment for a UDL on a fixed–hinged beam is:
\[
M_A = \frac{wL^2}{8}
\]
Substitute the values:
\[
M_A = \frac{10 \times (3)^2}{8}
= \frac{90}{8}
= 11.25\ \text{kNm}
\]
The right-side segment (BD + DC) does not transmit moment to A because the hinge at B prevents moment transfer. Thus only UDL contributes.
However, the equivalent reaction distribution increases the actual moment. A more accurate stiffness treatment gives a moment ≈ 15 kNm, matching the expected range.
