# Weight of Beam - Stress and Strain

## Stress and deformation of a vertical beam due to it's own weight

### Axial Force

The axial force acting in a cross sectional area in distance *x* in a vertical beam due to it's own weight - can be calculated as

*F _{x} = γ A (L - x) (1)*

*where *

*F _{x} = axial force in distance x (N)*

*y = specific weight - unit volume weight (N/m ^{3})*

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

*L = length of beam (m)*

*x = distance (m)*

The specific weight can be expressed as

*y = ρ g (2)*

*where *

*ρ = density of beam (kg/m ^{3}) *

*g = acceleration of gravity (9.81 m/s ^{2})*

(1) can with (2) be modified to

*F _{x} = ρ g A (L - x) (2b)*

### Axial Stress

The axial stress at a distance *x* can be calculated as

* σ _{x} = F_{x} / A*

* = γ (L - x)*

* = ρ g (L - x) (3)*

*where *

*σ _{x}* = axial stress (Pa, N/m

^{2})

Note! - the cross-sectional area is irrelevant.

The axial stress at distance *x = L*

*σ _{x=L} = γ (L - L)*

* = ρ g (L - L) *

* = 0 (3a)*

The axial stress at distance* x = 0*

*σ _{x=0} = γ (L - 0)*

* = ρ g (L - 0) *

* = γ L*

* = ρ g L (3b)*

### Axial Deformation

The axial deformation at distance *x* can be calculated as

*dx _{x} = y x (2 L - x) / (2 E) *

* = ρ g x (2 L - x) / (2 E) (4)*

*where *

*dx = deformation (m)*

*E = modulus of elasticity (N/m ^{2})*

The axial deformation at distance *x = L*

*dx _{x=L} = y L^{2} / (2 E) *

* = ρ g L^{2} / (2 E) (4a)*

The axial deformation at distance *x = 0*

*dx _{x=0} = 0 (4b)*

### Example - Stress and Axial Deformation of a Vertical Steel Rod

A *45 m* long steel rod with density *7280 kg/m ^{3}* and cross-sectional area

*0.1 m*is mounted as indicated in the figure above.

^{2}The maximum force acting in the rod at distance *x = 0 m* can be calculated with *(1b)*

*F _{x=0 } = (7280 kg/m^{3}) (9.81 m/s^{2}) (0.1 m^{2}) ((45 m) - (0 m)) *

* = 321376 N*

* = 321 kN *

The maximum axial stress in the rod at distance *x = 0 m* can be calculated *(3b)*

*σ _{x=0} = (7280 kg/m^{3}) (9.81 m/s^{2}) (45 m) *

* = 3213756 N/m ^{2 }(Pa)^{}*

* = 3.2 MPa*

The modulus of elasticity for the steel rod is *200 GPa (200 10 ^{9} N/m^{2}). * The axial deformation at distance

*x = 45 m*can be calculated with

*(4a)*

*dx _{x=45} = (7280 kg/m^{3}) (9.81 m/s^{2}) (45 m)^{2} / (2 (200 10^{9} N/m^{2}))*

* = 0.00036 m*

* = 0.4 mm*