ILD at mid-span: BM and SF under given loads

Data: simply supported beam $AB$, $L = 20$ m. Loads: 50 kN at 3 m from $A$; 10 kN/m UDL from 5 m to 15 m; 30 kN at 3 m from $B$ (17 m from $A$). Section at mid-span, $x = 10$ m.

ILD for bending moment at mid-span

A triangle with peak $\dfrac{L}{4} = 5$ m at mid-span. Ordinate at distance $x$ from the near support $= x/2$.

  A o--------------------o B
              /\ 5.0
  ILD: 0 /            \ 0

• 50 kN at 3 m: ordinate $= 3/2 = 1.5$ m
• UDL from 5 m to 15 m: ordinates $2.5$ at 5 m, $5$ at 10 m, $2.5$ at 15 m; area $= 2\times\dfrac{(2.5+5)}{2}\times 5 = 37.5$ m²
• 30 kN at 17 m: ordinate $= 3/2 = 1.5$ m

$$M = 50(1.5) + 10(37.5) + 30(1.5) = 75 + 375 + 45 = 495\ \text{kN m}$$

ILD for shear force at mid-span

Ordinate $-\dfrac{x}{20}$ for a load left of the section and $+\dfrac{20-x}{20}$ for a load right of it; the jump at mid-span is 1 (from $-0.5$ to $+0.5$).

• 50 kN at 3 m: ordinate $= -\dfrac{3}{20} = -0.15$
• UDL from 5 to 10 m (left of the section): area $= -\tfrac12(0.25+0.5)\times 5 = -1.875$ m
• UDL from 10 to 15 m (right of the section): area $= +\tfrac12(0.5+0.25)\times 5 = +1.875$ m (net area under the UDL $=0$)
• 30 kN at 17 m: ordinate $= +\dfrac{3}{20} = +0.15$

$$V = 50(-0.15) + 10(-1.875+1.875) + 30(0.15) = -7.5 + 0 + 4.5 = -3.0\ \text{kN}$$

Check: $R_A = \dfrac{50(17)+10(10)(10)+30(3)}{20} = 97$ kN; $V = 97 - 50 - 10\times 5 = -3$ kN and $M = 97\times 10 - 50\times 7 - 10\times 5\times 2.5 = 495$ kN m.

Answer: $M_{mid} = 495$ kN m (sagging); $V_{mid} = -3$ kN.