Question

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

• A. 23
• B. 33
• C. 43
• D. 58

Hint:

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.

Answer 1

The Correct Option is A

Step 1: Write the bending moment as a function of x.
Measure $x$ from F. For a beam fixed at F, the bending moment at any section can be written $M(x) = -M_F + V_F x - (\text{moment of all downward loads to the left of } x)$, where $M_F$ and $V_F$ are the unknown fixed-end reaction moment and vertical reaction at F. Since internal hinges carry zero bending moment, this function must equal zero at $x=4$ (hinge G) and at $x=8$ (hinge J).

Step 2: Apply the zero-moment condition at hinge G ($x=4$).
Loads to the left of $x=4$: the 5 kN/m UDL over F-G totals 20 kN acting at its centroid, 2 m from F, giving a moment about the section of $20(4-2)=40$; the 20 kN point load sits exactly at $x=4$, so its lever arm about the section is zero and it drops out. So: \[ 0=-M_F+4V_F-40 \implies M_F=4V_F-40 \]

Step 3: Apply the zero-moment condition at hinge J ($x=8$).
Loads to the left of $x=8$: the 20 kN UDL resultant now has lever arm $8-2=6$, giving $20(6)=120$; the 20 kN point load at $x=4$ has lever arm $8-4=4$, giving $20(4)=80$; the 30 kN point load at H ($x=6$) has lever arm $8-6=2$, giving $30(2)=60$. Total moment of loads $=120+80+60=260$. So: \[ 0=-M_F+8V_F-260 \implies M_F=8V_F-260 \]

Step 4: Solve the two hinge equations together.
Since both expressions equal $M_F$: \[ 4V_F-40=8V_F-260 \] \[ 220=4V_F \] \[ V_F=55\ kN \]

Step 5: Close the loop with overall vertical equilibrium.
Total downward load on the whole beam $=20+20+30+8=78\ kN$. Since $V_F+V_L=78$: \[ V_L=78-55=23\ kN \]

Final Answer:
Solving the hinge conditions directly as simultaneous equations gives the same reaction found by the free-body method. \[ \boxed{V_L=23\ kN} \]