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Weight of Beams - Stress and Strain

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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

Fx = γ  A (L - x)                                    (1)

where

Fx = axial force in distance x (N)

y = specific weight - unit volume weight (N/m3)

A = cross-sectional area (m2)

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/m3)

g = acceleration of gravity (9.81 m/s2)

(1) can with (2) be modified to 

Fx = ρ g  A (L - x)                            (2b)

Axial Stress

The axial stress at a distance x can be calculated as 

σx = Fx / A

    = γ (L - x)

    =  ρ g (L - x)                              (3)

where

σx = axial stress (Pa, N/m2)

Note! - the cross-sectional area is irrelevant.

The axial stress at distance x = L

σx=L = γ (L - L)

    = ρ g (L - L)

    = 0                                    (3a)

The axial stress at distance x = 0

σx=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

dxx = 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/m2)

The axial deformation at distance x = L

dxx=L = y L2 / (2 E) 

         = ρ g L2 / (2 E)                                (4a)

The axial deformation at distance x = 0

dxx=0 = 0                                    (4b)

Example - Stress and Axial Deformation of a Vertical Steel Rod

A 45 m long steel rod with density 7280 kg/m3 and cross-sectional area 0.1 m2 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)

Fx=0  = (7280 kg/m3) (9.81 m/s2) (0.1 m2) ((45 m) - (0 m))

   = 321376 N

   = 321 kN

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

σx=0 = (7280 kg/m3) (9.81 m/s2) (45 m)

        = 3213756 N/m2(Pa)

        = 3.2 MPa

The modulus of elasticity for the steel rod is 200 GPa (200 109 N/m2).  The axial deformation at distance x = 45 m can be calculated with (4a)

dxx=45 = (7280 kg/m3) (9.81 m/s2) (45 m)2 / (2 (200 109 N/m2))

       = 0.00036 m

       = 0.4 mm

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