I can't find the formula for beam with fixed end (instead of simply supported at ends).
What is the formula for fixed ends?
And given there is fixed end formula, can you apply it sideways to the weak axis of a wide flange? Or do you somehow use some kinds of components (since it's sideways and there is also gravity load)? Or does the formula for deflection only depend on the load, Young's modulus and moment of inertia and you can apply this at any angle of the beam??
Let's compute.
Let's use the formula for simply supported with uniform load.
Deflection = 5/384 (w L^4 / EI)
given: a 6" CHB wall has 2.73 kN/m2 x 1.5 meter = 4kN/m (half wall contribution to the top I-beam support of 3 meter high wall).
(In lateral movement computations, half of wall goes to top support and bottom support (or floor/diaphragm)).
Let's take lateral acceleration to be 0.25 g or about 1 kN/m. This is used as load for the weak axis of the I-beam, right?
Length of I-beam between column supports = 6 meters
Young's modulus = 200 GPa (for A36 steel wide flange).
Moment of Inertia = 0.0000040666 m^4
Or let's convert all of them from metric to imperial units so the numbers can be more familiar to you.
uniform load = 1 kN/m = 0.0057 Kips/Inch
l = 6 meters = 236.22 inches
Young's modulus = 200GPa = 29000 ksi
Moment of Inertia for W8x21 wide flange weak axis = 0.0000040666 m^4 = 9.77 in^2
Deflection = 5/384 * (w L^4 / EI) = 0.817 inch.
When designing masonry buildings ACI 530 states an out of plane deflection limit of 0.007*h.
Since wall height is 3 meters or 118 inches. And 0.007 x 118 " = 0.826 inch.
So the 6 meter length I-beam weak axis deflection is within the wall deflection limit. But then, what is the formula for fixed ends with uniform load and can you apply this sideways to the I-beam weak axis?
Many thanks!