Question

A simply supported beam of length 3 m is loaded as shown in the figure. The magnitude of the shear force (in kN) at the mid-point of the beam is ............ (rounded off to one decimal place).

Hint:

When finding shear force at a section, take all loads and reactions to the left (or right) of the section. Be careful to include applied couples only in moment equilibrium, not shear equilibrium.

Answer 1

Correct Answer: 4.9

Step 1: Identify loading.
- Beam length = 3 m. - At $x = 1 \, m$: applied moment = $10 \, kN\!m$ (anticlockwise) and a $30 \, kN$ point load downward. - From $x = 2 \, m$ to $x = 3 \, m$: UDL = $10 \, kN/m$, total load $= 10 \times 1 = 10 \, kN$ acting at 2.5 m.
Step 2: Take reactions.
Let supports at A (x=0) and B (x=3). Using equilibrium: Vertical force equilibrium: \[ R_A + R_B = 30 + 10 = 40 \, kN \] Moment about A: \[ R_B \times 3 - 30 \times 1 - 10 \times 2.5 - 10 = 0 \] (where -10 comes from the applied $10 \, kN\!m$ anticlockwise at 1 m). \[ 3R_B - 30 - 25 - 10 = 0 \] \[ 3R_B = 65 \Rightarrow R_B = 21.67 \, kN \] \[ R_A = 40 - 21.67 = 18.33 \, kN \]
Step 3: Shear force at mid-point (x = 1.5 m).
Shear just right of 1.5 m = \[ V = R_A - 30 = 18.33 - 30 = -11.67 \, kN \]
Step 4: Magnitude.
\[ |V| = 11.7 \, kN \] Final Answer: \[ \boxed{11.7 \, kN} \]