Mechanics of Materials – MAE 243 (Section 002) Spring 2008 Dr. Konstantinos A. Sierros

Problem 2.7-2 A bar of circular cross section having two different diameters d and 2d is shown in the figure. The length of each segment of the bar is L/2 and the modulus of elasticity of the material is E. (a) Obtain a formula for the strain energy U of the bar due to the load P. (b) Calculate the strain energy if the load P = 27 kN, the length L = 600 mm, the diameter d = 40 mm, and the material is brass with E = 105 GPa.

Problem 2.7-8 The statically indeterminate structure shown in the figure consists of a horizontal rigid bar AB supported by five equally spaced springs. Springs 1, 2, and 3 have stiffnesses 3k, 1.5k, and k, respectively. When unstressed, the lower ends of all five springs lie along a horizontal line. Bar AB, which has weight W, causes the springs to elongate by an amount δ. (a) Obtain a formula for the total strain energy U of the springs in terms of the downward displacement of the bar. (b) Obtain a formula for the displacement by equating the strain energy of the springs to the work done by the weight W. (c) Determine the forces F1, F2, and F3 in the springs. (d) Evaluate the strain energy U, the displacement δ, and the forces in the springs if W = 600 N and k = 7.5 N/mm.

3.1: Introduction to torsion Torsion is the twisting of a straight bar when it is loaded by moments (or torques) that tend to produce rotation about the longitudinal axis of the bar It is a little bit more complicated behaviour than axial tensile/compressive loading

3.1: Introduction to torsion Torsional loading of a bar Moment of a couple may be represented by a vector (fig 3-2b) using the right-hand rule Moment of a couple may be represented by a curved arrow (fig 3-2C)

3.2: Torsional deformations of a circular bar Consider a prismatic bar twisted by torques T acting at the ends The bar is under pure torsion since every cross-section is subjected to the same torque Φ is the angle of twist Longitudinal line pq become a helical curve pq’

3.2: Shear strains at the outer surface Consider an element which is under pure shear Angle of twist per unit length Shear strain of outer surface General cases of torsion Special case: Pure torsion where L is length of prismatic bar

3.2: Shear strains within the bar Previous equations for shear strains apply not only to solid circular bars but also to circular tubes

3.3: Circular bars of linearly elastic materials Torque T tends to rotate the right-hand end of the bar counterclockwise Therefore the shear stresses τ acting on the surface stress element will have the directions shown in fig 3-6b Max shear stress at the outer surface G is shear modulus of elasticity Shear stress at an interior point of radius ρ

Substituting r = d/2 and Ip = πd4/32 into the torsion formula we get; Determine a relationship between the shear stresses and the torque T The torsion formula Polar moment of inertia Substituting r = d/2 and Ip = πd4/32 into the torsion formula we get;

…and for a bar in pure torsion… 3.3: Angle of twist The angle of twist of a bar of linearly elastic material can be related to the applied torque T …and for a bar in pure torsion… Torsional rigidity …and circular tubes

3.3:Limitations The equations shown in this section are valid for bars of circular cross-section (either solid or hollow) that behave in a linearly elastic manner These equations can not be used for bars of other shapes such as rectangular bars and non-circular bars ***Torsion theory originated from the work of C.A. de Coulomb and further developments were due to Thomas Young Coulomb Young

due Wednesday 20 February 2008 Homework 2 due Wednesday 20 February 2008 is this …plus 1st midterm Friday 22nd February…