What is the consistent-deformation (compatibility) condition used to find R_D?

The net vertical deflection at D is zero: Delta_D0 + f_DD R_D = 0

The slope at A is zero: theta_A0 + f_AA R_D = 0

The sum of vertical forces is zero: R_D = 20 kN

The deflection at D under the loads equals f_DD, so R_D = 1

What does f_DD = 216/EI represent?

The vertical deflection at D of the primary cantilever due to a unit upward load at D, found as the integral of m^2/EI

The deflection at D due to the real loads on the primary structure

The reaction at D produced by a unit deflection

The fixed-end moment at A divided by EI

How is the final BM obtained and what is its value at A?

M = M0 + R_D m; at A it is -140 + 11.713 x 6 = -69.72 kNm

M = M0 only; at A it is -140 kNm

M = M0 - R_D; at A it is -151.7 kNm

M = M0 x R_D; at A it is -1640 kNm

2079 Baishakh · 10 marks

Consistent deformation: frame BMD with 20 kN

Short answer

Release the roller at D, taking its vertical reaction as redundant. Compute primary moment M0 and unit-load moment m, integrate to get ΔD0 = –2530/EI and fDD = 216/EI. Compatibility gives RD = 11.71 kN upward. Final moments: MA = 69.72 kNm (outer face tension), peak sagging 35.14 kNm at C.

Use the consistent deformation method to draw the BM diagram of the given frame. [Figure: frame with column AB (A fixed at the bottom, B at top, height 4 m) carrying 10 kN/m UDL along its height, beam BCD with BC = CD = 3 m, 20 kN downward at C, EI constant, D supported at the right end.]

Reading of the figure: A is fixed at the base of the column AB (4 m), the beam BCD has $BC = CD = 3$ m, the column carries a horizontal UDL of 10 kN/m (towards the right), a 20 kN load acts down at C, and D is on a roller (vertical support). EI is constant.

The frame is indeterminate to the first degree ( $r = 3 + 1 = 4$ ). Take the vertical reaction $R_D$ at D as the redundant.

        20 kN
          |
  B ------C------ D  (roller)
10 ->|     v     o
kN/m |
  A (fixed)

Step 1: Primary structure (D released), loads only

It is a cantilever fixed at A. Downward deflection of D under the loads is $\Delta_D$ .

Section	$M_0$ (kNm)	$m$ (unit load at D)
D	0	0
C	0	3.00
B (beam)	-60.00	6.00
B (column)	-60.00	6.00
mid AB	-80.00	6.00
A	-140.00	6.00

( $M_0$ is the moment of the loads on the cantilever (hogging, negative) and $m$ is the moment due to a unit upward force at D (sagging, positive), so their product is negative.)

For example, at A: $M_0 = -(10 \times 4 \times 2 + 20 \times 3) = -140$ kNm and $m = +6$ kNm per kN.

Step 2: Deformations by the unit-load method

\begin{aligned} \Delta_{D0} &= \int \frac{M_0\, m}{EI}\,dx = \frac{-2530}{EI}\ \ (\text{i.e. }2530/EI\ \text{downward}) \\ f_{DD} &= \int \frac{m^2}{EI}\,dx = \frac{216}{EI} \end{aligned}

(Column AB: $6^2 \times 4 = 144$ ; beam BC: $\frac{3}{3}(36 + 18 + 9) = 63$ ; beam CD: $\frac{3}{3}(9) = 9$ ; total 216.)

Step 3: Consistent deformation (deflection at D is zero)

\Delta_{D0} + f_{DD} R_D = 0 \ \Rightarrow\ R_D = \frac{2530}{216} = 11.713\ \text{kN}\ (\uparrow)

Step 4: Other reactions (statics)

$V_A = 20 - 11.713 = 8.287$ kN ( $\uparrow$ )
$H_A = 10 \times 4 = 40$ kN ( $\leftarrow$ )
$M_A$ from $\Sigma M_A = 0$ : $M_A = 40 \times 2 + 20 \times 3 - 11.713 \times 6 = 69.72$ kNm (anticlockwise reaction).

Step 5: Bending moments $M = M_0 + R_D\, m$

Point	BM (kNm) (+ve = tension on inner/under face)	Tension side
A	-69.72	outer (left) face of column
Mid AB	-9.72	outer face
B	10.28	inner (right) face of column / underside of beam
C	35.14	underside
D	0

The BM in the column is $M(y) = -69.72 + 40y - 5y^2$ (y from A), which changes sign at $y = 2.57$ m from A. In the beam, BM rises linearly from 10.28 kNm at B to 35.14 kNm under the 20 kN load, then falls linearly to zero at D.

Answer: $R_D = 11.71$ kN up; $M_A = 69.72$ kNm; peak sagging BM 35.14 kNm at C.

✓ Answer checked by Claude Opus 5.5 (high effort)

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