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ILD: shear at section 2-2, propped cantilever

Short answer

Use redundant RB = x^2(3L-x)/(2L^3); shear is -RB for a load left of the section and 1-RB for a load right. Ordinates at 2 m intervals are 0, -0.0859, -0.3125/0.6875, 0.3672, 0, with a unit jump at the section.

Draw influence line diagram for the shear at section 2-2 of the propped cantilever beam shown in figure below. Find the ordinates at 2 m interval. [Figure: beam AB, A fixed, B roller, span 8 m, section 2-2 at 4 m from A.]

Sign convention: sagging moment (+), hogging moment (-); shear positive when the left part tends to go up; reaction positive upward. Ordinates are for a unit load moving along the beam.

Method

Propped cantilever AB, L=8L = 8 m, A fixed, B roller, section 2-2 at 4 m from A. Redundant: RB=x2(3L−x)2L3R_B = \dfrac{x^2(3L-x)}{2L^3}. Shear at the section (left part): load left of the section, V=−RBV = -R_B; load right, V=1−RBV = 1 - R_B.

Sample: x=6x = 6: RB=36×182×512=0.633R_B = \dfrac{36\times18}{2\times512} = 0.633, V=0.367V = 0.367. At the section: RB(4)=16×201024=0.3125R_B(4) = \dfrac{16\times20}{1024} = 0.3125, giving V=−0.3125V = -0.3125 just left and +0.6875+0.6875 just right.

Ordinates (2 m interval)

x from A (m)02468
V2−2V_{2-2}0-0.0859-0.3125 / 0.68750.36720

ILD shape

                        |***
                        |   ******
                        |         ******
                        |               ******
*************-----------|---------------------***
             *********  |
                      **|


^                                               ^

Jump of 1.0 at the section (from −0.3125-0.3125 to +0.6875+0.6875); the ILD is zero at A and B.

✓ Answer checked by Claude Opus 5.5 (high effort)

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