A timber column 15cm x 10cm in section is 3m long with both ends
hinged. Determine the safe load, the column can carry with a factor of
safety as 10. Tame E for timber as 1.055 x 102 t/m2
To determine the safe load that the timber column can carry, we'll need to use the formula for Euler's column buckling load:
P = (π^2 * E * I) / (K * L^2)
Where:
P = Safe load the column can carry
E = Modulus of Elasticity for timber (given as 1.055 x 10^2 t/m^2)
I = Moment of Inertia of the column cross-section
K = Effective Length Factor, depends on the end conditions of the column
L = Length of the column
First, let's calculate the Moment of Inertia (I) for the timber column. The Moment of Inertia for a rectangular cross-section can be calculated using the following formula:
I = (b * h^3) / 12
where:
b = breadth of the rectangle (15 cm in our case)
h = height of the rectangle (10 cm in our case)
I = (15 cm * 10 cm^3) / 12
I = 12,500 cm^4
Next, we need to determine the Effective Length Factor (K) for the hinged column ends. For both ends hinged, the effective length factor is 2.
Now, we can substitute the given values into the formula:
P = (π^2 * E * I) / (K * L^2)
P = (3.14^2 * 1.055 x 10^2 t/m^2 * 12,500 cm^4) / (2 * 300 cm)^2
Now, we need to convert the units to match each other. Let's convert cm to meters and t/m^2 to N/m^2:
P = (3.14^2 * 1.055 x 10^2 N/m^2 * 12,500 x 10^-8 m^4) / (2 * 3 m)^2
P = (3.14^2 * 1.055 x 10^2 N/m^2 * 12.5 x 10^-4 m^4) / (18 m^2)
P = (9.865 x 10^3 N/m^2 * 12.5 x 10^-4 m^4) / 18
P ≈ 6.867 N
Lastly, we need to apply the factor of safety. Since the factor of safety is given as 10, we multiply the calculated load by the factor of safety:
Safe Load = P * Factor of Safety
Safe Load = 6.867 N * 10
Safe Load ≈ 68.67 N
Therefore, the safe load that the timber column can carry with a factor of safety of 10 is approximately 68.67 Newtons.