## Description

### Unformatted Attachment Preview

Purchase answer to see full attachment

## Explanation & Answer

Please view explanation and answer below.Here is the answer

CIVE 302

Season & Year

Professor Name

Statement of Objective

Lab 4. Torsion of Member with Circular Cross-Section: Principal Strains and Stresses

Thus far our only focus has been on axial deformations which cause elongation or compression

of a member along their axis of application. Other types of deformation exist, though, such as

shear deformations which translate or slide the face on which they act. To help visualize this

behavior, in this lab a solid steel shaft with circular cross-section is loaded under pure torsion.

Torsion is caused when a moment is applied in-plane with the cross-section of the member,

twisting it and causing it to deform in shear. A visualization of this is provided below in Figure

4-1.

Procedure

Recall that the measurement devices used thus far only measure elongation or contraction along

their primary axis. Meaning that strain gages cannot directly measure these values. Looking at

more detail at Figure 4.1, it is clear that the shear strain, γ , is not a linear quantity but an angle of

distortion. We thus need additional techniques to be able to somehow determine shear strain

from linear-based measurement equipment.

For a bar in pure torsion, each section along the longitudinal axis rotates strictly in a plane; that

is, all points within the member follow a strictly a circular arc in the plane and do not translate in

either direction along the member axis. Note that this is not the case for non-circular crosssections, as torsion develops warping deformations which displace points axially in addition to

rotationally. This behavior is shown below in Figure 4-2, where it can be seen that point B at the

outermost fiber of the cross-section travels along the circumference of the circle to B’ at an

applied torsion of T₀.

Torsional loading at a section causes shear stresses that are zero at the center of the member and

increase linearly with radius r to a maximum at the surface of the cylinder at r = R. Shear stress

is found from the section torsion T, radius r, and polar moment of inertia J as

For a circular section, the polar moment of inertia is equal to the sum of moments of inertia

about the x and y axes as

2

Shear strain is found by dividing the shear stress by the shear modulus G, where the shear

modulus is found for a given material from material testing as the slope of the shear stress versus

shear strain plot. Thus shear strain is found from

Note the similarities in this equation to those seen in the previous labs. To calculate axial

stresses, we had a driving axial load P that was divided by the cross-sectional area – a geometric

property of the member. Then, to calculate strains, we divided this stress by the modulus of

elasticity – a material property of the section – and concluded that the deformation is a function

of the driving load P, the geometrical resistance of the member, and the material resistance of the

member. In the case of torsional strain, we see a similar pattern: a torque T drives the

deformation of the member, a geometri...