Weight of Beams  Stress and Strain
Stress and deformation of vertical beams due to 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 = crosssectional 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 crosssectional 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 crosssectional area 0.1 m^{2} is mounted as indicated in the figure above.
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
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