# 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.

### Shear stress (shearing)

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 mm^{2}

τ_{a} Shear stress N / mm^{2} (τ = tau, Greek letter)

τ_{from} Shear strength N / mm^{2 }

τ_{aB max} maximum shear strength N / mm^{2}

R._{m max} maximum tensile strength N / mm^{2}

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 tables_{from}. For steel, τ also applies approximately_{0,8} • R_{m}

**Formulas** summarized:

**Shear stress** τ_{a} = F: S

**Allowable shear stress** τ_{a perm} = τ_{from} : ν

**Cutting power** F = S • τ_{aBmax}

**Maximum shear strength** τ_{aBmax} = 0.8 • R_{m 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 mm^{2}; Shear strength τ_{from} = 440 N / mm^{2}; Safety number ν = 1.6.

Solution:

F = 2 • S • τ_{azul}

τ_{azul} = τ_{from} : ν = 440 = N / mm^{2} : 1.6 = 275 N / mm^{2}

**F = 110 550 N = 110.55 kN****Cutting of materials** (Picture below)

To calculate the cutting force F, the maximum shear strength is τ_{gB max} to use. If this is not known, the tensile strength can also be calculated approximately: τ_{gB max} = 0.8 • R_{m 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 R_{m max} this steel is between 410 N / mm^{2} and 560 N / mm^{2}. Which shear force F has to be applied?

Solution:

Shear force F = S • τ_{aBmax}

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

S = π • 24 mm • 4 mm = 301.6 mm^{2 }

F = 301.6 mm^{2} • 0.8 • 560 N / mm^{2} =

**F = 135 117 N = 135.1 kN**

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