The horizontal beam shown in the figure has fixed supports at F and L. Internal hinges are provided at locations G and J. The beam supports vertically downward concentrated loads as well as uniformly distributed loads, as shown in the figure. (Figure not to scale) Span F-G = 4 m carries a UDL of 5 kN/m. A 20 kN point load acts right at hinge G. A 30 kN point load acts at H, the midpoint of the hinge-to-hinge span G-J (G-H = H-J = 2 m each, hinge at J). Span J-K = K-L = 2 m each, with span K-L carrying a UDL of 4 kN/m. The upward vertical reaction (in kN) at the support L is
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Split the beam at the hinges into three pieces; the middle piece G-J is simply supported by the shear forces the hinges transmit, so solve it first, then use that to load piece J-L and find V_L.
Step 1: Set up the compound beam. The beam runs F-G-H-J-K-L with FG = 4 m, GH = HJ = JK = KL = 2 m each, total length 12 m. F and L are fixed supports. G and J are internal hinges, so the bending moment is zero at G and at J, and the beam behaves as three rigid pieces pinned together at those hinges: piece 1 = F-G, piece 2 = G-J (the suspended middle span), piece 3 = J-L.
Step 2: Solve the middle piece G-J on its own. Piece 2 spans 4 m (G to J) and carries only the 30 kN point load, applied at H, which sits exactly at its midspan (2 m from G and 2 m from J). Calling the unknown vertical hinge forces \(V_{G2}\) and \(V_{J2}\) that the neighbouring pieces must feed into piece 2: \[ V_{G2}+V_{J2}=30 \] Taking moments about G: \[ V_{J2}(4) = 30(2) \implies V_{J2} = 15\ kN \] so \(V_{G2}=30-15=15\ kN\) too, which also makes sense by symmetry since the load sits at midspan. By Newton's third law, piece 2 pushes back down with 15 kN on each of piece 1 (at G) and piece 3 (at J).
Step 3: Load up piece 3 (J to L). Piece 3 spans J-L = 4 m (J to K to L). It carries the 4 kN/m UDL over the 2 m stretch K-L, total \(4\times2=8\ kN\), plus the 15 kN downward push transmitted at its left end J from piece 2 in Step 2. There is no other load on this piece, and it is fixed only at L. Summing vertical forces on piece 3: \[ V_L = 15+8 = 23\ kN \]
Step 4: Sanity-check using the whole beam. For piece 1 (F to G): loads are the 5 kN/m UDL over 4 m (=20 kN), the 20 kN point load applied right at hinge G, and the 15 kN push-back from piece 2 at G, giving \(V_F = 20+20+15=55\ kN\). Total applied load on the whole beam \(=20+20+30+8=78\ kN\), and \(V_F+V_L=55+23=78\ kN\), matching exactly and confirming the result.
Final Answer: The upward vertical reaction at support L is 23 kN. \[ \boxed{V_L = 23\ kN} \]
