Taking moments about A for the real loading, what gives E_y = 47.5 kN?

E_y x 4 = 10 x 3 + 80 x 2, with the 80 kN UDL resultant acting 2 m from A

E_y x 4 = 80 x 2 only, since the 10 kN load is horizontal

Why does member DE contribute nothing to the integral of Mm?

Its real moment M is zero because no load acts beyond the section on the E side

It is shorter than the other members

Beam CD has stiffness 2EI. How is this treated in the sum for the horizontal deflection of E?

The integral of Mm over CD is divided by 2EI

All members are assumed to have EI

CD is ignored because it carries a UDL

The integral over CD is multiplied by 2

2074 Bhadra · 10 marks

Frame horizontal deflection at E

Short answer

Horizontal deflection at E is 921.7/EI to the right. Virtual work with a unit horizontal load at E gives ∫M m/EI = 921.7/EI kNm³.

Determine horizontal deflection at E of the frame shown in figure below. [Figure: frame; left column A-B-C with A (hinge) at the bottom, B 3 m above A, C 3 m above B; 10 kN horizontal load at B; top beam CD of 4 m (2EI) with 20 kN/m UDL; right column D-E of 4 m (EI) with E roller; left column EI; width 4 m.]

Given data

Frame: A(0,0) hinge, B(0,3), C(0,6), D(4,6), E(4,2) roller. Left column AC: $EI$ (3 m + 3 m); beam CD: $2EI$ , 4 m, UDL 20 kN/m; right column DE: $EI$ , 4 m. 10 kN horizontal at B (to the right). Required: horizontal deflection of E.

Real loading: reactions

\sum M_A = 0:\; E_y\times4 = 10\times3 + 80\times2 \Rightarrow E_y = 47.5\ \text{kN},\quad A_y = 80 - 47.5 = 32.5\ \text{kN},\quad A_x = 10\ \text{kN (left)}

(The 10 kN load gives 30 kNm clockwise about A, the UDL resultant 80 kN at $x = 2$ m gives 160 kNm.)

Virtual load: unit horizontal load at E (to the right)

$A_x = 1$ (left); moments about A: $E_y\times4 = 1\times2$ , so $E_y = 0.5$ (up), $A_y = 0.5$ (down).

Moments and integrals ( $s$ from the start of each member, A-B-C-D-E)

Member	$s$ (m)	stiffness	$M$ (kNm)	$m$	$\int Mm\,ds$ / stiffness
AB	0-3	$EI$	$10s$	$s$	90.00
BC	0-3	$EI$	30	$s + 3$	405.00
CD	0-4	$2EI$	$30 + 32.5s - 10s^2$	$6 - s/2$	426.67
DE	0-4	$EI$	0	$4 - s$	0

\Delta_{Eh} = \frac{90 + 405 + 426.67}{EI} = \frac{921.67}{EI}

Answer: horizontal deflection of E $= 921.7/EI$ (kNm $^3$ /EI), to the right.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Frame horizontal deflection at E

Given data

Real loading: reactions

Virtual load: unit horizontal load at E (to the right)

Moments and integrals (sss from the start of each member, A-B-C-D-E)

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