For the 14 m beam with 240 kN at 3 m and 160 kN at 9.5 m from support 0, what are the reaction at support 0 and the bending moments under the loads?

R0 = 240 kN; M2 = 720 kNm; M3 = 720 kNm

R0 = 160 kN; M2 = 480 kNm; M3 = 720 kNm

R0 = 240 kN; M2 = 720 kNm; M3 = 1080 kNm

R0 = 200 kN; M2 = 600 kNm; M3 = 900 kNm

How is the conjugate beam set up for this simply supported real beam?

Simply supported with the same span, loaded with the M/EI diagram

Cantilever fixed at support 0, loaded with the real loads

Simply supported, loaded with the shear force diagram

Beam with internal hinges at both supports, loaded with M/EI

After finding theta0 = 3863.57/EI, how is the slope at point 2 obtained?

Subtract the area of triangle 0-2 (1080) from theta0, i.e. conjugate shear at point 2

Add the area of triangle 0-2 (1080) to theta0

Take the moment of the 1080 load about point 2

In y2 = 3863.57x3 - 1080x1, what does the lever arm of 1 m belong to?

The distance from point 2 to the centroid of triangle 0-2 (3 - 2 m)

The distance from point 2 to point 3

The distance from support 0 to the centroid of triangle 0-2

Asked 2/24
2070 Bhadra · 10 marks
2066 Kartik · 10 marks

Conjugate beam: slopes and deflections under loads

Short answer

Using conjugate beam method, M/EI areas give conjugate reactions as slopes. Slopes: left support 0.0115 rad clockwise, right support 0.0105 rad anticlockwise; under 240 kN load 0.00828 rad clockwise, under 160 kN load 0.00564 rad anticlockwise. Deflections: 31.28 mm under 240 kN, 39.86 mm under 160 kN.

Using conjugate beam method, calculate slopes at the supports and at the points beneath the loads for the given simply supported beam and also calculate the deflections of the points beneath the loads. Take $EI = 3.36\times10^{11}\ \text{kNmm}^2$ . [Figure: simply supported beam 0-1; point 2 carries 240 kN at 3 m from support 0; point 3 carries 160 kN at 6.5 m from point 2 and 4.5 m from support 1 (span 14 m).]

Given data

Simply supported beam, span $L = 14$ m (support 0 at the left, support 1 at the right). Load $P_2 = 240$ kN at point 2 (3 m from support 0); $P_3 = 160$ kN at point 3 (9.5 m from support 0, 4.5 m from support 1). $EI = 3.36\times10^{11}$ kNmm $^2 = 3.36\times10^{5}$ kNm $^2$ .

Step 1: Reactions and bending moments

R_1 = \frac{240\times3 + 160\times9.5}{14} = 160\ \text{kN},\qquad R_0 = 400 - 160 = 240\ \text{kN}

M_2 = 240\times3 = 720\ \text{kNm},\qquad M_3 = 160\times4.5 = 720\ \text{kNm}

Step 2: Conjugate beam (simply supported, loaded with $M/EI$ )

The $M/EI$ diagram is a trapezoid with peak $720/EI$ between points 2 and 3.

       720/EI ____________
            /|            |\
          /  |            |  \
  0 ----/----2------------3-----\---- 1
        3 m       6.5 m       4.5 m

Loads on the conjugate beam (in units of $1/EI$ ):

Triangle 0-2: $\tfrac12\times3\times720 = 1080$ , centroid 2.0 m from 0.
Rectangle 2-3: $6.5\times720 = 4680$ , centroid at 6.25 m from 0.
Triangle 3-1: $\tfrac12\times4.5\times720 = 1620$ , centroid 11 m from 0.

Step 3: Conjugate reactions = slopes at supports

\theta_0 = R'_0 = \frac{1080\times12 + 4680\times7.75 + 1620\times3}{14} = \frac{54090}{14} = \frac{3863.57}{EI}

\theta_1 = R'_1 = 1080 + 4680 + 1620 - 3863.57 = \frac{3516.43}{EI}

Step 4: Slopes at points 2 and 3 (conjugate shear)

\theta_2 = 3863.57 - 1080 = \frac{2783.57}{EI}

\theta_3 = 3863.57 - 1080 - 4680 = -\frac{1896.43}{EI}

(The sign changes between 2 and 3, so the maximum deflection lies between the loads.)

Step 5: Deflections (conjugate moment)

y_2 = 3863.57\times3 - 1080\times1 = \frac{10510.7}{EI}

y_3 = 3863.57\times9.5 - 1080\times7.5 - 4680\times3.25 = \frac{13393.9}{EI}

Step 6: Numerical values with $EI = 3.36\times10^5$ kNm $^2$

Point	Slope (rad)	Deflection (mm)
0 (support)	0.01150 (clockwise)	0
2 (240 kN)	0.00828 (clockwise)	31.28
3 (160 kN)	0.00564 (anticlockwise)	39.86
1 (support)	0.01047 (anticlockwise)	0

Answer: slopes at supports: $0.0115$ rad and $0.01047$ rad; slopes under loads: $0.00828$ rad (point 2) and $0.00564$ rad (point 3, opposite sense); deflections: $31.28$ mm under the 240 kN load and $39.86$ mm under the 160 kN load.

✓ Answer checked by Claude Opus 5.5 (high effort)

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