10) Horizontal beam supported on both ends - square profil 100 × 100 × 5, length is 800 mm.It is bent by the force in the middle of the beam by 0.2 mm. To achieve approximately half of the value of deflection,you will use the beam:
To determine the beam that would achieve approximately half of the given deflection, we need to calculate the deflection of a beam with different dimensions and select the one that meets the desired criteria.
To calculate the deflection of a beam, we can use the formula for deflection in a simply supported square beam:
δ = (F * L^3) / (48 * E * I)
Where:
- δ is the deflection
- F is the applied force
- L is the length of the beam
- E is the modulus of elasticity of the material
- I is the moment of inertia of the beam cross-section
Given:
- Length of the beam (L) = 800 mm = 0.8 m
- Deflection of the current beam (δ) = 0.2 mm = 0.0002 m
We want to find a beam that gives half of the given deflection, so:
Desired deflection (δ_desired) = 0.0002 m / 2 = 0.0001 m
Now, we can rearrange the deflection formula to solve for the moment of inertia (I):
I = (F * L^3) / (48 * E * δ)
To find the appropriate beam dimensions, we can assume the same material and modulus of elasticity as the current beam since it is not mentioned.
Since we already know the deflection, we can find the moment of inertia for the current beam using its dimensions:
I_current = (F * L^3) / (48 * E * δ_current)
Now, we can set up a ratio and solve for the dimensions of the new beam:
(I_new / I_current) = (δ_new / δ_current)^3
(I_new / I_current) = (0.0001 m / 0.0002 m)^3 = 0.5^3 = 0.125
Therefore, the moment of inertia of the new beam should be approximately 0.125 times the moment of inertia of the current beam.
To achieve this, you would need to select a square beam with larger dimensions to increase the moment of inertia. The exact dimensions would depend on the available options and specific requirements.