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Castigliano: fixed-end moment with 50 kNm at overhang

Short answer

Taking roller reaction RB as redundant, Castigliano’s theorem gives δB=0. Integrating virtual work yields RB = 15 kN upward. The fixed-end moment MA = RB*5 - 50 = 25 kNm clockwise. Reactions: RA = 15 kN downward.

Determine the moment at the fixed support of the following loaded beam using Castigliano's theorem. Take EI constant. [Figure: beam ABC, A fixed, B roller at 5 m from A, overhang BC = 2 m with a 50 kNm moment applied at C.]

Reading of the figure: A fixed; the roller B is 5 m from A; the overhang BC (2 m) carries a couple of 50 kNm at C, taken as clockwise. EI is constant. The beam has one redundant; take the roller reaction RBR_B (upward).

 ||==================o=========  <- 50 kNm (cw) at C
 A (fixed)           B        C
 |<------ 5 m ------>|<- 2 ->|

Bending moments (x from C for the overhang, ss from B for AB)

  • Overhang BC: M=−50M = -50 kNm (constant, hogging), independent of RBR_B.
  • Span AB: M=RB s−50M = R_B\,s - 50,  ∂M∂RB=s\ \dfrac{\partial M}{\partial R_B} = s

Castigliano's theorem (δB=0\delta_B = 0)

∫05(RB s−50) s ds=0 ⇒ RB(5)33−50(5)22=0\int_0^5 (R_B\,s - 50)\,s\,ds = 0 \ \Rightarrow\ \frac{R_B (5)^3}{3} - 50\frac{(5)^2}{2} = 0 41.667 RB=625 ⇒ RB=15 kN (↑)41.667\,R_B = 625 \ \Rightarrow\ R_B = 15\ \text{kN}\ (\uparrow)

Moment at the fixed support

At A (s=5s = 5):

MA=RB(5)−50=15(5)−50=+25 kNmM_A = R_B(5) - 50 = 15(5) - 50 = +25\ \text{kNm}

The reaction moment at A is 25 kNm, which is half the applied couple (the carry-over factor of a propped cantilever is 1/2). It acts in the clockwise sense on the beam. The vertical reaction at A is 1515 kN downward.

Check: ΣV=0\Sigma V = 0: RA+RB=0R_A + R_B = 0 so RA=−15R_A = -15 kN.

Answer: moment at the fixed support MA=25M_A = 25 kNm (clockwise); RB=15R_B = 15 kN upward, RA=15R_A = 15 kN downward. For an anticlockwise couple the results have the same magnitudes with all senses reversed.

✓ Answer checked by Claude Opus 5.5 (high effort)

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