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The Geometry of Bending: A Central Point Load

This lesson builds intuition for the bending moment diagram of a simply supported beam using equilibrium and symmetry.

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The Simply Supported Beam
Beam Setup10 kN4 m
L=4 m,W=10 kNL = 4\text{ m}, \quad W = 10\text{ kN}
Imagine a beam spanning four meters, held by two supports at the ends. When we place a ten kilonewton load exactly in the middle, how does the beam 'feel' this pressure?
Step-by-step solver
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(a) Define variables

Identify the beam length and the central point load applied at the midpoint.

L=4 m,W=10 kN at x=2 mL = 4\text{ m}, \quad W = 10\text{ kN at } x = 2\text{ m}
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(b) Find reactions

Use vertical equilibrium to determine that the total load is shared equally between the supports due to symmetry.

RA=RB=102=5 kNR_A = R_B = \frac{10}{2} = 5\text{ kN}
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(c) Express moment

Calculate the bending moment M(x) as the reaction force multiplied by the distance x from the support.

M(x)=5xfor 0≤x≤2M(x) = 5x \quad \text{for } 0 \le x \le 2
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(d) Determine max moment

Evaluate the moment at the central point where the shear force changes sign.

M(2)=5×2=10 kN⋅mM(2) = 5 \times 2 = 10\text{ kN}\cdot\text{m}

Original question

A simply supported beam of length 4 m carries a point load of 10 kN at its centre. Determine the maximum bending moment and draw the bending moment diagram.

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