Skip to main content

Overhang beam free end deflection by virtual work

Short answer

Applying the virtual work method with a unit load at D: the vertical deflection at the free end D is 40/EI (kN·m³) downward.

Calculate the vertical deflection of free end D of the beam loaded as shown in fig-2 by using virtual work method. Take EI as constant throughout. [Figure: beam ABCD, A hinge, B 2 m from A with 20 kN downward, C roller 2 m from B, free end D 2 m from C with 10 kN downward.]

Given data

Beam ABCD: A hinge, B at 2 m (20 kN down), C roller at 4 m, D free end at 6 m (10 kN down). EIEI constant.

    A     B     C     D
    ^-----+-----^-----+
        20 kN         10 kN
      2 m   2 m   2 m

Real loading

∑MC=0\sum M_C = 0: RA×4+10×2=20×2⇒RA=5R_A\times4 + 10\times2 = 20\times2 \Rightarrow R_A = 5 kN (up); RC=25R_C = 25 kN (up).

Bending moments (sagging +, xx from A): AB: M=5xM = 5x (10 kNm at B); BC: M=5x−20(x−2)=40−15xM = 5x - 20(x-2) = 40 - 15x, so MC=−20M_C = -20 kNm; CD: M=−10sM = -10s (ss from D), giving MC=−20M_C = -20 kNm.

Virtual load: unit downward load at D

Reactions: RA=−0.5R_A = -0.5, RC=1.5R_C = 1.5. Moments: AB: m=−0.5xm = -0.5x (hogging), BC: m=−0.5xm = -0.5x (to C: −2-2), CD: m=−sm = -s.

Integrals

PortionMMmm∫Mm dx\int Mm\,dx
AB (0-2)5x5x−0.5x-0.5x−2.5×83=−6.667-2.5\times\frac{8}{3} = -6.667
BC (2-4)5x−20(x−2)5x - 20(x-2)−0.5x-0.5x+20.000+20.000 (see below)
CD (0-2, from D)−10s-10s−s-s10×83=26.66710\times\frac{8}{3} = 26.667

Evaluating the BC integral: ∫24(−0.5x)(40−15x) dx=∫24(−20x+7.5x2)dx=−10(16−4)+2.5(64−8)=−120+140=20\int_2^4 (-0.5x)(40-15x)\,dx = \int_2^4(-20x + 7.5x^2)dx = -10(16-4) + 2.5(64-8) = -120 + 140 = 20.

ΔD=−6.667+20+26.667EI=40EI\Delta_D = \frac{-6.667 + 20 + 26.667}{EI} = \frac{40}{EI}

Answer: vertical deflection of the free end D =40/EI= 40/EI (kNm3^3/EI), downward.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Theory of Structures I chapter 3 questions →