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# 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 = FN / A                           (1)

where

σ = normal stress (N/m2, Pa, psi)

FN = applied force perpendicular to the area - Normal force (N, lb)

A = cross-sectional area (m2, in2)

### Shear Stress

Shear stress can be expressed as

τ = FV / A                           (2)

where

τ = shear stress (N/m2, Pa, psi)

FV = 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 cm2. The normal stress in the column can be calculated as

σ = (10000 N) / ((20.3 cm2) (0.0001 m2/cm2

= 4926108 Pa (N/m2)

= 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 Fv (or V in the figure above) is equal in magnitude to support force R1 or R2.

Reaction forces can be calculated due to moment equilibrium around support 1

F L / 2 = R2 L                              (4)

R2 = F / 2                              (5)

R1 = R2 = 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

R1 = R2

= V1

= V2

= (10000 N) / 2

= 5000 N

= 5 kN

The shear stress can be calculated as

τ = (5000 N) / ((20.3 cm2) (0.0001 m2/cm2)

= 2463054 Pa

= 2.5 MPa

## Related Topics

• Mechanics - Forces, acceleration, displacement, vectors, motion, momentum, energy of objects and more.
• Statics - Loads - forces and torque, beams and columns.

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