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 calculate the required specifications of the beam to achieve approximately half of the deflection, we can use the formula for the deflection of a uniformly loaded beam:
δ = (F * L^3) / (48 * E * I)
Where:
δ = deflection
F = force applied to the center of the beam
L = length of the beam
E = Young's modulus of the material
I = moment of inertia of the cross-sectional area
Given that the beam is a square profile with dimensions 100 × 100 × 5 mm, we can calculate the moment of inertia using the formula:
I = (b * h^3) / 12
Where:
b = width of the beam
h = height of the beam
Let's calculate the required specifications for the beam:
Given:
L = 800 mm
δ = 0.2 mm (desired deflection)
F = unknown (to achieve approximately half of the deflection)
E = unknown (specific to the material)
First, let's calculate the moment of inertia:
I = (100 * 100^3) / 12
I = 833333.33 mm^4
Now, rearranging the formula for deflection, we can solve for F:
F = (δ * 48 * E * I) / L^3
Since we want to achieve approximately half of the deflection, we can substitute δ/2 for δ:
F = ((δ/2) * 48 * E * I) / L^3
Plugging in the known values:
F = ((0.2/2) * 48 * E * 833333.33) / (800^3)
Simplifying the equation:
F = (0.1 * 48 * E * 833333.33) / (800^3)
F = (3999999.984 * E) / 512000000
To find the appropriate specifications for the beam, we would need to know the desired force F and the Young's modulus E of the material being used.