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A cantilever beam of span 6 m is subjected to transverse shear force of 5 kN at free end. Beam is having rectangular cross-section of dimension 40 mm x 50 mm. Determine the magnitude of shear stress (kPa) on an element a lying at a distance of 20 mm from neutral axis. (6 pts)

Answer :

The magnitude of shear stress (kPa) on an element a lying at a distance of 20 mm from neutral axis. (6 pts) are given, p = 2kn e = 210gpa distance from constant cease = 20mm = 0.2m.

What is the cantilever formula?

The cantilever beam equations (deflection) w = load. l = member length. e = young's modulus. i = the beam's moment of inertia.

  • theta= 2000 2*210*10^ 9 * (0.006)^ four 12 [(0.1)^ 2 -(0.08)^ 2 ] =0.1587 rad
  • ø at q i.e. x = 100 - 20 = 80mm = 0.08m from loose cease.
  • theta = p/(2ei) * (l ^ 2 - x ^ 2) phase of facet a. i = (a ^ four)/12 for square.

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