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## Calculate tilt of Zerodur beam supported at ends, centrally loaded, given:

L = beam length = 500 mm
wd = beam width = 100 mm
ht = beam height = 24.7 mm
E = Modulus of elasticity = 9.03x1010 Pa = 9.03x104 N/mm2

From Machinery’s Handbook, case 2:
I = Moment of Inertia = wd*ht3/12 = 1.256x105 mm4
EI = (9.03x103*1.256x105) = 1.134x1010 N mm2
Z = Section Modulus = I/(ht/2) = 1.017x104 mm3
W = Load = 0.5 Kg = 4.804 N/mm2
X = point coordinate relative to support
Y(x) = Deflection = [Wx/48EI](3L2-4x2)
S(x) = extreme fiber stress = -Wx/2Z (positive indicates tension)

For testing beam was supported 50 mm from ends, giving
L= unsupprted length = 400 mm
W = applied load = 0.5 kg = 4.904 N/mm2

Calculate central deflection and stress:
Ycent = WL3/48EI= (4.904*.4003)/(48*1.134x1010) = 0.58x10-3 mm = 580 nm

See Beam Test Data 2 in excellent agreement

Scent = Smax = -WL/4Z= -(4.904*400)/(4*1.017x104) = .048 N/mm2 = .05x105 Pa

I have yet to establish yield stress for Zerodur but this is about 3 orders less than values listed for metals in the AIP Physics Vade Mecum, section 1.06.D.

One could calculate tilt from difference in deflection for X and X + increment, but better to take the derivative, since Tilt = T = dY/dX = d/dX[W/48EI](3XL2-4X3) = [W/16EI](L2 - 4X2) with max at ends (X=0) and minimum at center (X=L/2)

Tmax = T(0) = WL2/16EI = (4.904*4002)/(16*1.134x1010)= 4.3 microradians
Tcent = T(L/2) = 0

See Beam Test Data 1, in good agreement.

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