# Stress

## Stress is force applied on cross-sectional area

Stress is the ratio between *applied force* and *cross-sectional area* where the applied force is acting.

### Normal Stress

**Normal stress** can be expressed as

σ = F_{N}/ A (1)

where

σ = normal stress (N/m^{2}, Pa, psi)

F_{N}= applied force perpendicular to the area -Normal force(N, lb)

A = cross-sectional area (m^{2}, in^{2})

### Shear Stress

**Shear stress** can be expressed as

τ = F_{V}/ A (2)

where

τ= shear stress (N/m^{2}, Pa, psi)

F_{V}= applied force in plane of the area -Shear force(N, lb)

### Example - Normal Stress in a Column

A *10000 N* force is acting in the direction of a British Universal Column UB 152 x 89 x 16 with cross sectional area *20.3 cm ^{2}.* The normal stress in the column can be calculated as

*σ = ( 10000 N) / ((20.3 cm^{2}) (0.0001 m^{2}/cm^{2}) *

* = 4926108 Pa (N/m ^{2})*

* = 4.9 MPa *

The Yield strength - the amount of stress that a material can undergo before moving from elastic deformation into plastic deformation - is typical 250 MPa for steel.

### Example - Shear Stress in a Beam with Point Load

For a beam with single point load supported on both ends - the shear force *F _{v}* (or

*V*in the figure above) is equal in magnitude to support force

*R*.

_{1}or R_{2}Reaction forces can be calculated due to moment equilibrium around support 1

*F L / 2 = R _{2} L (4)*

*R _{2} = F / 2 (5)*

*R _{1} = R_{2} = F / 2 (6)*

For a *10000 N* point load perpendicular on a beam similar to the example above - supported at both ends - the magnitude of the reaction and shear forces can be calculated as

*R _{1} = R_{2} *

* = V _{1} *

* = V _{2} *

* = (10000 N) / 2*

* = 5000 N *

* = 5 kN *

The shear stress can be calculated as

*τ = (5000 N) / ((20.3 cm^{2}) (0.0001 m^{2}/cm^{2}) *

* = 2463054 Pa*

* = 2.5 MPa *