CBSEGrade 12MathematicsIntegrals

Deflection of a Beam?

A 2m long, 0.1m wide beam is deflected by 0.05m under a mass of 50kg at its midpoint. Given that the cross-sectional area of the beam is uniform, derive a mathematical expression for the deflection 'y' at any point 'x' from the load, using the equation of the integral of moment of the beam's section.

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📌 CONCEPT: The deflection of a beam is the downward displacement of its midpoint due to an external load, which can be determined using the equation of the integral of moment of the beam's section.

📐 RULE / FORMULA: The deflection 'y' at any point 'x' from the load can be calculated using the formula: y = (w ∗ x^2) / (2 ∗ E ∗ I), where 'w' is the load per unit length, 'x' is the distance from the point of application of the load, 'E' is the modulus of elasticity of the beam material, and 'I' is the moment of inertia of the beam's cross-sectional area.

💡 WORKED EXAMPLE: Suppose a 2m long, 0.1m wide beam is deflected by 0.05m under a mass of 50kg at its midpoint. Given that the cross-sectional area of the beam is uniform and the modulus of elasticity is 2×10^5 N/m^2, calculate the deflection 'y' at a point 0.5m away from the load. Using the formula, y = (50 ∗ 0.5^2) / (2 ∗ 2×10^5 ∗ 0.1^4), we get y = 0.0125m.

⚠️ COMMON MISTAKE: Students often forget to consider the moment of inertia 'I' of the beam's cross-sectional area, which can lead to incorrect calculations of deflection.

19 Aug 26