ILDs for BM and SF at 10 m, 25 m beam

Data: simply supported beam $AB$, $L = 25$ m. Section $C$ at $a = 10$ m from $A$, $b = 15$ m. Unit load at $x$ from $A$.

Reactions

$R_A = \dfrac{25-x}{25}$, $R_B = \dfrac{x}{25}$.

ILD for bending moment at C

• Load left of $C$ ($x\le 10$): take the right part: $M_C = R_B\times 15 = 0.6x$
• Load right of $C$ ($x\ge 10$): take the left part: $M_C = R_A\times 10 = 0.4\,(25-x)$

Peak under $C$: $\dfrac{ab}{L} = \dfrac{10\times 15}{25} = 6$ m. The ILD is a triangle: 0 at $A$, $+6$ at $C$, 0 at $B$.

  A o----------C-------------------o B
               /\ 6.0
  ILD M_C:  0 /  \ 0

ILD for shear force at C

• Load left of $C$ ($x<10$): $V_C = -R_B = -\dfrac{x}{25}$; at $C^-$: $-0.4$
• Load right of $C$ ($x>10$): $V_C = R_A = \dfrac{25-x}{25}$; at $C^+$: $+0.6$

  ILD V_C:   0 \ -0.4 | +0.6 \ 0     (two parallel lines, jump of 1 at C)

Areas: $M_C$ ILD $=\tfrac12\times 25\times 6 = 75$ m²; $V_C$ ILD: positive $\tfrac12\times 15\times 0.6 = 4.5$ m, negative $\tfrac12\times 10\times 0.4 = 2$ m.

For a UDL $w$ over the whole span, $M_C = 75w$ and $V_C = 4.5w - 2w = 2.5w$; for a single load $P$ at $x$, $M_C = P\,y$ and $V_C = P\,y$ with $y$ read from the ILD.