# deformation

## Shearing Deformation

**Shearing Deformation**

Shearing forces cause shearing deformation. An element subject to shear does not change in length but undergoes a change in shape.

The change in angle at the corner of an original rectangular element is called the shear strain and is expressed as

The ratio of the shear stress τ and the shear strain γ is called the *modulus of elasticity in shear* or modulus of rigidity and is denoted as G, in MPa.

The relationship between the shearing deformation and the applied shearing force is

where V is the shearing force acting over an area A_{s}.

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## Solution to Problem 217 Axial Deformation

## Solution to Problem 216 Axial Deformation

**Problem 216**

As shown in Fig. P-216, two aluminum rods AB and BC, hinged to rigid supports, are pinned together at B to carry a vertical load P = 6000 lb. If each rod has a cross-sectional area of 0.60 in.^{2} and E = 10 × 10^{6} psi, compute the elongation of each rod and the horizontal and vertical displacements of point B. Assume α = 30° and θ = 30°.

## Solution to Problem 215 Axial Deformation

**Problem 215**

A uniform concrete slab of total weight W is to be attached, as shown in Fig. P-215, to two rods whose lower ends are on the same level. Determine the ratio of the areas of the rods so that the slab will remain level.

**Solution 215**

## Solution to Problem 214 Axial Deformation

**Problem 214**

The rigid bars AB and CD shown in Fig. P-214 are supported by pins at A and C and the two rods. Determine the maximum force P that can be applied as shown if its vertical movement is limited to 5 mm. Neglect the weights of all members.

## Solution to Problem 213 Axial Deformation

**Problem 213**

The rigid bar AB, attached to two vertical rods as shown in Fig. P-213, is horizontal before the load P is applied. Determine the vertical movement of P if its magnitude is 50 kN.

## Solution to Problem 208 Axial Deformation

**Problem 208**

A steel tire, 10 mm thick, 80 mm wide, and 1500.0 mm inside diameter, is heated and shrunk onto a steel wheel 1500.5 mm in diameter. If the coefficient of static friction is 0.30, what torque is required to twist the tire relative to the wheel? Neglect the deformation of the wheel. Use E = 200 GPa.

## Solution to Problem 206 Axial Deformation

**Problem 206**

A steel rod having a cross-sectional area of 300 mm^{2} and a length of 150 m is suspended vertically from one end. It supports a tensile load of 20 kN at the lower end. If the unit mass of steel is 7850 kg/m^{3} and E = 200 × 10^{3} MN/m^{2}, find the total elongation of the rod.

## Solution to Problem 205 Axial Deformation

**Problem 205**

A uniform bar of length L, cross-sectional area A, and unit mass ρ is suspended vertically from one end. Show that its total elongation is δ = ρgL^{2}/2E. If the total mass of the bar is M, show also that δ = MgL/2AE.

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