Question

A beam may have zero shear stress at a section but may not have zero deflection; Hence, bending is primarily caused by bending moment In Torsion loading a stress element in a circular rod is subject to shear state The principal plane and the plane on which the shear stresses are maximum, they make 90 degree angle between them. If the Torque on a steel circular shaft (G=80 GPa) is 13.3 kN-m and the allowable shear stress is 98 MPa, then find the fail-safe design RADIUS of the shaft in mm. Young's Modulus (E) and the Shear Modulus (G) of a material are not related to each other and cannot be found if one or the other is known, even the poisson's ratio is known Moment along the length of the beam is directly proportional to the second order derivative of deflection of the beam. After bending due to the vertical load on the beam the plane section of the beam remains plane Poisson's ratio is the direct ratio of the value of the deformation in two orthogonal directions Two beams with equal width but has different depths. It is obvious that the beam with higher depth will have higher deflection Moment of Inertia of a section of a uniformly designed beam changes as a cubic function along the LENGTH of the beam. If 2 normal stresses and 1 shear stress are known at a point at a certain depth across the beam, these 3 numbers are NOT sufficient to calculate the angle of crack that might propagate in the beam if the local principal stress exceeds the ultimate stress.

Answer 1