Stress
Stress is force applied on crosssectional area.
Stress is the ratio between applied force and crosssectional 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 = crosssectional 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
Related Topics

Mechanics
The relationships between forces, acceleration, displacement, vectors, motion, momentum, energy of objects and more. 
Statics
Forces acting on bodies at rest under equilibrium conditions  loads, forces and torque, beams and columns.
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