ILDs for reactions, SF and BM, 20 m beam

Data: simply supported beam $AB$, $L = 20$ m. Section $C$ at $a = 5$ m from $A$, $b = 15$ m. Let a unit load be at a distance $x$ from $A$.

ILD for reaction $R_A$

$R_A = \dfrac{20-x}{20}$: a straight line from 1 at $A$ to 0 at $B$.

ILD for reaction $R_B$

$R_B = \dfrac{x}{20}$: a straight line from 0 at $A$ to 1 at $B$.

ILD for shear force at C

• Load to the left of $C$ ($x<5$): from the right part, $V_C = -R_B = -\dfrac{x}{20}$. At $C$ the ordinate is $-\dfrac{5}{20} = -0.25$.
• Load to the right of $C$ ($x>5$): $V_C = R_A = \dfrac{20-x}{20}$. At $C$ the ordinate is $+\dfrac{15}{20} = +0.75$.

So the ILD is two parallel lines: from 0 at $A$ to $-0.25$ just left of $C$, a jump of 1 at $C$ to $+0.75$, and then down to 0 at $B$.

ILD for bending moment at C

• Load left of $C$: $M_C = R_B\times 15 = \dfrac{15x}{20}$.
• Load right of $C$: $M_C = R_A\times 5 = \dfrac{5(20-x)}{20}$.
• Peak at $C$: $\dfrac{ab}{L} = \dfrac{5\times 15}{20} = 3.75$ m.

 R_A:  1 \
          \_____ 0 at B
 SF_C: 0 \-0.25 | +0.75 \ 0
 BM_C: 0 /\ 3.75 (at C) \ 0

The positive area of the SF ILD is $\tfrac12\times 15\times 0.75 = 5.625$ m and the negative area is $\tfrac12\times 5\times 0.25 = 0.625$ m. The area of the BM ILD is $\tfrac12\times 20\times 3.75 = 37.5$ m².