Old spec Statics Session 9 Problem 1TutorialOld spec ENGR2162:210 min

ENGR216 (old spec) Statics session 9 worked example 1 (Problem 1, slides 3-5) official answers

Beam AB is simply supported (pin at A, roller at B) and made of three planks glued together into an I section: a top flange 100 mm wide and 20 mm thick, a web 20 mm thick and 80 mm deep, and a bottom flange 60 mm wide and 20 mm thick (total depth 120 mm). The centroid C is 68.3 mm above the bottom face and the centroidal second moment of area about the neutral axis is 8.63×10−6 m48.63\times10^{-6}\ \mathrm{m^4}. Two downward loads P1=P2=3.0P_1 = P_2 = 3.0 kN act at distances a=0.6a = 0.6 m and a+b=1.0a + b = 1.0 m from A, and the span is 2a+b=1.62a + b = 1.6 m. Each glued joint is 20 mm wide.

Determine the shear stress in each glued joint (joint u, between the top flange and the web, and joint l, between the web and the bottom flange) at section n-n, located between A and P1P_1.

Left: simply supported beam AB with loads P1 and P2 at a and a + b from A (span a + b + a); section n-n between A and P1. Right: I section, top flange 100 x 20 mm, web 20 x 80 mm, bottom flange 60 x 20 mm, centroid C 68.3 mm above the bottom; glued joints u (top) and l (bottom).
Left: simply supported beam AB with loads P1 and P2 at a and a + b from A (span a + b + a); section n-n between A and P1. Right: I section, top flange 100 x 20 mm, web 20 x 80 mm, bottom flange 60 x 20 mm, centroid C 68.3 mm above the bottom; glued joints u (top) and l (bottom).
Formulas you may need
  • Shear stress in a beam τ=VQIt\tau = \dfrac{VQ}{It}, Q=yˉAQ = \bar y A of the area beyond the cut (on the formula sheet)
  • Centroid of a composite area yˉ=∑Aiyˉi∑Ai\bar y = \dfrac{\sum A_i \bar y_i}{\sum A_i} (learn this; only needed to check the given data)