Can I have a hot tub on the terrace?

[lärjungen]

The formula to calculate the deflection from a point load placed at the center of a beam is as follows: Q*L^3/48*E*I. Q=the point load, L=the span, E=the modulus of elasticity for the material in question, and I=the moment of inertia of the beam. For a rectangular beam, the moment of inertia can be calculated using the formula b*h^3/12 (b and h refer to the cross-sectional dimensions). For steel beams, it is easiest to look up the value in a table. You can use any units (e.g., meters, mm, kg, kN [kilonewton]) as long as you do it consistently.

Taking the current example, a 5-meter-long beam with dimensions of 45x220 mm subjected to a point load of 100 kg (0.98 kN), the calculation is as follows: First, calculate the moment of inertia which is 0.00003993 m4. E is for C24 timber 11000 MPa [megapascal], i.e., 11000000 kN/m2 for deformation calculations. Plugging in all the values, we have the following calculation: 0.98*5^3/48*11000000*0.00003993=0.0058 m, i.e., 5.8 mm. You can make it easier for yourself by entering the formula into an Excel sheet, so you only need to input the span, point load, and cross-sectional dimensions.

Thanks. Saving this. Might come in handy at work

 

[richardtenggren]

To check your calculations, you can perform a dimensional analysis, then you know that you have got all the units right.

The formulas Justus writes can be found among the basic cases in most formula collections if you want to calculate for other load cases.

If you are calculating for steel or another material, remember that the moment of inertia depends on the geometry of the cross-section and that the modulus of elasticity depends on the material.

 

[justusandersson]

Good addition by @richardtenggren! The two other load cases that are most relevant are 1) uniformly distributed load and 2) arbitrarily placed point load. The formulas are similar but not identical. Essentially, this is 18th-century physics.