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Frame vertical deflection at C

Short answer

Using virtual work with EI = 6000 kN·m²: the vertical deflection at point C is 270/(EI) = 0.045 m or 45 mm downward.

Determine the vertical deflection of point C of the frame shown in fig-1. E=200 kN/mm2E = 200\ \text{kN/mm}^2, I=30×106 mm4I = 30\times10^6\ \text{mm}^4. [Figure: frame with fixed support A at the base of the column AB (4 m high, 2I) and horizontal member BC (3 m, I); 10 kN downward load at the free end C.]

Given data

Frame: fixed at A (base of column AB), column AB: 4 m, 2I2I; horizontal member BC: 3 m, II; 10 kN downward at the free end C. E=200E = 200 kN/mm2=2×108^2 = 2\times10^8 kN/m2^2, I=30×106I = 30\times10^6 mm4=3×10−5^4 = 3\times10^{-5} m4^4.

EI=2×108×3×10−5=6000 kNm2EI = 2\times10^8\times3\times10^{-5} = 6000\ \text{kNm}^2

Virtual work (unit vertical load at C)

Distances ss from C along CB, then ss from B down BA.

Memberss (m)MM (kNm)mm∫Mm ds\int Mm\,dsstiffnesscontribution
CB0-310s10sss10×9=9010\times9 = 90EIEI90
BA0-43030330×3×4=36030\times3\times4 = 3602EI2EI180

Hence

ΔCv=90+180EI=2706000=0.045 m\Delta_{Cv} = \frac{90 + 180}{EI} = \frac{270}{6000} = 0.045\ \text{m}

Answer: vertical deflection at C =45= 45 mm downward.

✓ Answer checked by Claude Opus 5.5 (high effort)

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