Shear Strain and Stress and Design

Shear Stress

Shear Stress is a stress that acts tangential to the surface of a material. $\tau = V/A$

Bearing Stress

Bearing stresses are contact stresses that develop under tensile loads between connections. $\sigma_{b} = \frac{F_{b}}{A_{b}}$

where the bearing area is the projected area of the bearing surface.

Single Shear vs. Double Shear

Single Shear: The axial load (P) is cut across one plane of shear. $V = P$ $\tau_{avg} = V/A$

Shear Strain

Shear Strain is the measure of distortion or change in shape. Measured in degrees or radians and denoted by $\gamma$

Hooke’s Law in Shear: $\tau = G\gamma$

Shear Modulus of Elasticity (Modulus of Rigidity): $G = \frac{E}{2(1 + \nu)}$

Factor of Safety

Strength: the ability of a structure to resist loads

Factor of safety: the ratio of actual strength to required strength; must be above 1.0 to avoid failure

$n = \frac{actual\ strength}{required\ strength}$

Margin of Safety: extra capacity a material has beyond failure. This is expressed as a percentage

$Margin\ of\ Safety = (n - 1) \cdot 100\%$

Allowable Stress

Allowable Stress: maximum stress that a structure can safely withstand under working conditions without failure or permanent deformation.

$allowable\ stress = \frac{yield\ strength}{factor\ of\ safety},\ \sigma_{allow} = \frac{\sigma_{Y}}{n},\ \tau_{allow} = \frac{\tau_{Y}}{n}$

If the ultimate stress is used instead of the yield stress, the equations are the same, but with the perspective of factor of safety.

Allowable Loads

Allowable load: The maximum force a structure can withstand before failure or permanent deformation

$P_{allow} = \sigma_{allow}A = \tau_{allow}A = \sigma_{b}A_{b}$

Design Factors

$Required\ Area = \frac{Load\ to\ be\ Transmitted}{Allowable\ Stress}$

Stiffness: ability of a structure to resist change in shape

Stability: ability of a structure to resist buckling under compressive stresses

Example 1:
Problem 1.8-3: Given the diagram below, find $\tau_{avg}$, $\sigma_{bf}$ between the flange and plates, and $\sigma_{bg}$ between the gusset and the pin.

Average Shear Stress:

1. Find area that is resisting shear:

$A = \pi\frac{d_{p}^{2}}{4} = \pi\frac{{(2\ in)}^{2}}{4} = 3.1415\ {in}^{2}$

1. Solve Shear Stress:

$\tau_{avg} = \frac{V}{2A}$ since the pin is in double shear, and V=P/2

$\tau_{avg} = \frac{(160\ kips/2)}{2 \cdot 3.1415\ {in}^{2}} = 12.75\ ksi$

Bearing Stress between the flange plates and the pin:

1. Find the force of the flange plate:

There are two flange plates for each gusset plate, so $F = P/4 = 40\ kips$

1. Find the bearing area:

$A = d_{p}t_{f} = 2\ in\ \cdot 1\ in = 2\ {in}^{2}$

1. Find Bearing Stress

$\sigma_{bf} = \frac{F}{A} = \frac{40\ kips}{2\ {in}^{2}} = 20\ ksi$

Bearing Stress between the gusset plate and the pin:

1. Find the force of the gusset plate:

The gusset plate provides a force of $P/2 = 80\ kips$

1. Find the bearing area:

$A = d_{p}t_{g} = 2\ in \cdot 1.5\ in = 3\ {in}^{2}$

1. Find bearing Stress:

$\sigma_{bg} = \frac{F}{A} = \frac{80\ kips}{3\ {in}^{2}} = 26.7\ ksi$

Example 2:
Problem 1.10-1

What is the outside diameter when P=33 k, the thickness of the tube is 0.25 in, and the allowable tensile stress is 12,000 psi?

1. Use the allowable stress equation to find the needed cross-sectional area

$P_{allow} = \sigma_{allow}A \Rightarrow A = \frac{P_{allow}}{\sigma_{allow}} = \frac{33\ k}{12\ ksi} = 2.75\ {in}^{2}$

1. Use the area of a circle to find the outer diameter

$A = \frac{\pi}{4}(d_{o}^{2} - d_{i}^{2}) \Rightarrow 2.75\ {in}^{2} = \frac{\pi}{4}(d_{o}^{2} - {(2*0.25in)}^{2} \Rightarrow d_{o} =$ 3.75 in