For the overhanging beam loaded as shown, calculate the support reactions at A and B.Figure (described in words): Beam ABC is 10 m long. A (left end) is a hinged support and B is a roller support 8 m from A; BC is a 2 m overhang to the free end C. Lo…

Answer: \(R_B=38.37\) kN (upward); at A, \(H_A=15\) kN (to the left), \(V_A=32.61\) kN (upward), so \(R_A=35.89\) kN at 65.3° to the horizontal (up and to the left).

Load resultants:

• Inclined 30 kN at D: vertical \(30\sin60^\circ=25.98\) kN (down), horizontal \(30\cos60^\circ=15\) kN (to the right).
• UDL: \(10\times3=30\) kN acting at the middle of its span, \(x=4.5\) m.
• UVL on BC: \(\frac12\times2\times15=15\) kN acting at \(\frac23\) of its length from the zero end, \(x=8+\frac43=9.333\) m.
• Couple: 20 kN m anticlockwise at \(x=7\) m.

FBD: hinge A gives \(H_A\) and \(V_A\); roller B gives vertical \(R_B\).

\(\sum F_x=0:\ H_A=15\) kN, acting to the left.

\(\sum M_A=0\) (clockwise positive):

\[25.98(2)+30(4.5)+15(9.333)-20-R_B(8)=0\]

\[51.96+135+140-20=8R_B\Rightarrow R_B=\frac{306.96}{8}=38.37\ \text{kN}\ (\uparrow)\]

\(\sum F_y=0:\ V_A+38.37=25.98+30+15\Rightarrow V_A=32.61\) kN \((\uparrow)\).

\[R_A=\sqrt{15^2+32.61^2}=35.89\ \text{kN},\quad\theta=\tan^{-1}\frac{32.61}{15}=65.3^\circ\]

Check, moments about B (clockwise positive): \(V_A(8)-25.98(6)-30(3.5)+15(1.333)-20=260.9-155.9-105+20-20=0\) ✓