# What is the 2000 psi shear strength

## Strength calculations (6): shear09.05.2012, 07:02

When the load is sheared off, stresses occur in a cross-sectional area that are parallel to the applied force.

### The external forces act perpendicular to the rod axis. You are trying to move the two cutting edges parallel to each other. The internal force F lies parallel to the cutting surface, resulting in shear stresses τ (Greek letter tau = shear stresses). Components subject to shearing stress must not be destroyed. Exception: when cutting sheet metal there is a separation of materials. When selecting the stress limit values, it must be checked whether it is a shearing or cutting.

Designations:

F shear, cutting force N
S cross-sectional area mm2
τa Shear stress N / mm2 (τ = tau, Greek letter)
τfrom Shear strength N / mm
τaB max maximum shear strength N / mm2

R.m max maximum tensile strength N / mm2
ν Security number (ν = nü, Greek letter), without unit

The Cross-sectional area S consists of the sum of the shear surfaces that result in fracture surfaces when severed. Stress on Shear

The strength calculations are carried out with the shear strengths τ determined by tests or taken from tablesfrom. For steel, τ also applies approximately0,8 • Rm

Formulas summarized:

Shear stress τa = F: S

Allowable shear stress τa perm = τfrom : ν

Cutting power F = S • τaBmax

Maximum shear strength τaBmax = 0.8 • Rm max

Calculation example
Picture above: With what shear force F is the bolt from S275J2G3 loaded in the double-shear connection? Given: bolt cross-section S = 201 mm2; Shear strength τfrom = 440 N / mm2; Safety number ν = 1.6.

Solution:

F = 2 • S • τazul

τazul = τfrom : ν = 440 = N / mm2 : 1.6 = 275 N / mm2

F = 110 550 N = 110.55 kN

Cutting of materials (Picture below)
To calculate the cutting force F, the maximum shear strength is τgB max to use. If this is not known, the tensile strength can also be calculated approximately: τgB max = 0.8 • Rm max

example: A disk with a diameter d = 24 mm is cut out of sheet steel S275J2 with a thickness s = 4 mm (picture). The tensile strength Rm max this steel is between 410 N / mm2 and 560 N / mm2. Which shear force F has to be applied? Solution:

Shear force F = S • τaBmax

Shear area S = circumference U • thickness s = π • d • s =

S = π • 24 mm • 4 mm = 301.6 mm

F = 301.6 mm2 • 0.8 • 560 N / mm2 =

F = 135 117 N = 135.1 kN