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Mid-span deflection of beam, 10 kN load

Short answer

Using the deflection formula for a simply supported beam with a point load, the mid-span (3.5 m) deflection is 17.76 mm downward.

Find deflection of the beam shown below at middle of the span. Given that: Modulus of Elasticity of the material is 20000 MPa. Section of the beam is 150 mm×250 mm150\ \text{mm}\times250\ \text{mm}. Point Load W = 10 kN. [Figure: simply supported beam with a 10 kN load W at 4 m from the left support and 3 m from the right support (span 7 m).]

Given: simply supported beam, span L=7L = 7 m = 7000 mm, load W=10W = 10 kN at a=4a = 4 m from the left support (b=3b = 3 m from the right). E=20000 N/mm2E = 20000\ \text{N/mm}^2, section 150×250150\times250 mm.

              W = 10 kN
                 v
 A ^-------------|-------^ B
   |<----4 m---->|<-3 m->|
   mid-span at 3.5 m

Moment of inertia

I=150×250312=1.953×108 mm4I = \frac{150\times250^3}{12} = 1.953\times10^8\ \text{mm}^4 EI=20000×1.953×108=3.906×1012 N mm2EI = 20000\times1.953\times10^8 = 3.906\times10^{12}\ \text{N mm}^2

Deflection at mid-span

The mid-span point x=3500x = 3500 mm lies in the left segment (x<a=4000x < a = 4000 mm). For a point load, in the segment x≤ax \le a:

y=W b x6EIL(L2−b2−x2)y = \frac{W\,b\,x}{6EIL}\left(L^2 - b^2 - x^2\right) y=10×103×3000×3500 (70002−30002−35002)6×3.906×1012×7000y = \frac{10\times10^3\times3000\times3500\,(7000^2 - 3000^2 - 3500^2)}{6\times3.906\times10^{12}\times7000} y=1.05×1011×2.775×1071.6406×1017=17.76 mmy = \frac{1.05\times10^{11}\times2.775\times10^7}{1.6406\times10^{17}} = 17.76\ \text{mm}

Answer: deflection at mid-span =17.76= 17.76 mm downward.

✓ Answer checked by Claude Opus 5.5 (high effort)

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