Conjugate beam: deflection and slope at E and B

Given data and assumptions

(The figure is partly illegible; this is the reading used.) Beam fixed at A. Segment AB: 5 m, rigidity $I$, ending at a link (internal hinge) at B. From B the beam (rigidity $2I$) continues with a 30 kN load 3 m from B, a roller support D at 5 m from B and a free end E at 9 m from B carrying 50 kN. $I = 6\times10^7$ mm$^4 = 6\times10^{-5}$ m$^4$, $E = 200$ kN/mm$^2 = 2\times10^8$ kN/m$^2$, so $EI = 12\,000$ kNm$^2$.

 A=====|B      30 kN          50 kN
 fixed |link ----v---- D(roller) ------v E
  I 5 m  |    3 m    2 m ^      4 m (2I)

Step 1: Split at the link

The link carries shear but no moment. The part BE rests on the roller D and on the link at B (which is carried by the cantilever AB). Moments about D (BE alone):

$$R_B\times5 = 30\times2 - 50\times4 \;\Rightarrow\; R_B = -28\ \text{kN},\qquad R_D = 80 + 28 = 108\ \text{kN}$$

The support force on BE at B is 28 kN downward, so BE pulls the tip of the cantilever AB with an upward force of 28 kN.

Step 2: Cantilever AB (rigidity $EI$)

Upward tip load 28 kN, length 5 m:

$$y_B = \frac{28\times5^3}{3EI} = \frac{1166.67}{EI}\ \text{(upward)},\qquad \theta_{B,left} = \frac{28\times5^2}{2EI} = \frac{350}{EI}\ \text{(anticlockwise)}$$

Step 3: Segment BE by the conjugate beam

First treat BE as a beam with $y = 0$ at B and D (rigidity $2EI$, $x$ measured from B):

(a) Reactions of BE $R_{B} = -28.00$ kN, $R_{D} = 108.00$ kN (positive = upward; check: $\sum F_y = 80.00$ kN equals the total load 80.00 kN).

(b) Bending moment of BE, as $M/EI$ pieces

(Stiffness is in terms of $EI$; "Area" is $\int M/EI\,dx$ in units of $1/EI$; $\bar x$ is the centroid distance from B.)

(c) Conjugate beam The real segment BE has a support at B (provided by the link), a roller at D and a free end at E, so the conjugate beam has a pin at B, an internal hinge at D and a fixed end at E. It carries the $M/EI$ diagram of part (b) as load (sagging $M/EI$ acts downward, hogging acts upward). Conjugate shear $V'$ = slope (positive = clockwise) and conjugate moment $M'$ = deflection (positive = downward).

(d) Conjugate reaction at B (slope at B) Taking moments of the conjugate beam about the support at $x = 5$ m (conjugate moment is zero there):

$$R'_B\times5 = (-63.00)(3.000) + (-142.00)(0.864) = -311.67$$ $$R'_B = \frac{-62.333}{EI}\ \text{(slope at B, anticlockwise)}$$

At the 30 kN load ($x = 3$ m):

$$V' = R'_B - \sum A = -62.333 - [(-63.00)] = \frac{0.667}{EI}\ \text{(slope, clockwise)}$$ $$M' = R'_B\,x - \sum A(x-\bar x) = -187.000 - [(-63.00)(1.000)] = \frac{-124.000}{EI}\ \text{(deflection, upward)}$$

At D ($x = 5$ m):

$$V' = R'_B - \sum A = -62.333 - [(-63.00) + (-142.00)] = \frac{142.667}{EI}\ \text{(slope, clockwise)}$$ $$M' = R'_B\,x - \sum A(x-\bar x) = -311.667 - [(-63.00)(3.000) + (-142.00)(0.864)] = \frac{0.000}{EI}\ \text{(deflection, zero)}$$

At E (free end) ($x = 9$ m):

$$V' = R'_B - \sum A = -62.333 - [(-63.00) + (-142.00) + (-200.00)] = \frac{342.667}{EI}\ \text{(slope, clockwise)}$$ $$M' = R'_B\,x - \sum A(x-\bar x) = -561.000 - [(-63.00)(7.000) + (-142.00)(4.864) + (-200.00)(2.667)] = \frac{1104.000}{EI}\ \text{(deflection, downward)}$$

Step 4: Add the movement of B

Point B actually moves up by $1166.67/EI$ while D does not move, so BE also rotates as a rigid body clockwise by $1166.67/(5EI) = 233.33/EI$. The rigid-body deflection at distance $x$ from B is $1166.67\,(1 - x/5)/EI$ upward, which becomes negative (downward) beyond D. Adding this to the results of Step 3:

(At E: Step 3 gives $1104/EI$ downward and $342.67/EI$ clockwise; the rigid-body rotation adds $0.8\times1166.67/EI = 933.33/EI$ downward and $233.33/EI$ clockwise: $1104 + 933.33 = 2037.33$ and $342.67 + 233.33 = 576$. At B, right of the link: Step 3 gives $62.33/EI$ anticlockwise, and $62.33 - 233.33 = -171$, i.e. $171/EI$ clockwise.)

Numerical values ($EI = 12\,000$ kNm$^2$)

• Deflection at B $= 97.2$ mm upward.
• Slope at B: left $= 0.02917$ rad (anticlockwise), right $= 0.01425$ rad (clockwise).
• Deflection at E $= 169.8$ mm downward; slope at E $= 0.0480$ rad (clockwise).

Answer: $\Delta_B = 97.2$ mm upward; $\Delta_E = 169.8$ mm downward; $\theta_E = 0.0480$ rad clockwise. At the link the slope jumps from $0.0292$ rad (anticlockwise, left) to $0.0143$ rad (clockwise, right).