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ENGR5004 / Week 3
Old spec Statics Session 6 ProblemTutorialOld spec ENGR2162:212 min

ENGR216 (old spec) Statics session 6 worked example (Problem, slides 13-15) official answers

A simply supported beam AB of length LL (pin at A, roller at B) carries the downward distributed load

w(x)=w0sin⁡πxLw(x) = w_0 \sin\frac{\pi x}{L}

with xx measured from A.

Simply supported beam AB of length L (pin at A, roller at B) under a half-sine downward load w = w0 sin(pi x / L), zero at both ends and w0 at midspan.
Simply supported beam AB of length L (pin at A, roller at B) under a half-sine downward load w = w0 sin(pi x / L), zero at both ends and w0 at midspan.
Formulas you may need
  • Resultant of a distributed load P=∫0Lw dxP = \int_0^L w\,dx at its centroid (on the formula sheet)
  • Load, shear and moment: dVdx=−w(x)\dfrac{dV}{dx} = -w(x), dMdx=V(x)\dfrac{dM}{dx} = V(x), V(x)−VA=−∫xAxw dxV(x) - V_A = -\int_{x_A}^{x} w\,dx, M(x)−MA=∫xAxV dxM(x) - M_A = \int_{x_A}^{x} V\,dx (on the formula sheet)
  • Maximum bending stress σmax=∣M∣cI\sigma_{max} = \dfrac{|M|c}{I}; square of side aa: I=a4/12I = a^4/12, c=a/2c = a/2 (on the formula sheet)
  1. (a)
    Determine the equations of the shear load V(x)V(x) and the bending moment M(x)M(x).
  2. (b)
    Plot V(x)V(x) and M(x)M(x) along the beam axis.
  3. (c)
    Assuming the cross section is square with side length aa, determine the position along the beam where the maximum normal stress occurs and the value of that stress.