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

L = unsupported beam length = 400 mm
      total length is 500 mm, supports 50 mm from ends
b = beam width = 100 mm
h = beam height = 24.7 mm
E = Modulus of elasticity = 9.03x1010 Pa = 9.03x104 N/mm
2
I = Moment of Inertia = b*h3/12 = 1.26x105 mm4
E
I = (9.03x104*1.26x105) = 1.13x1010 N mm2
Z = Section Modulus =
I/(h/2) = 1.02x104 mm3
W = central load = 0.5 Kg = 4.9 N/mm2

x = point coordinate relative to support
Y(x) = deflection = (Wx/48E
I)(3L2 - 4x2)
S(x) = extreme fiber stress = -Wx/2Z

Calculate central deflection and stress: x = L/2
Ycent =  Ymax = Y(L/2) = WL3/48E
I= (4.90*4003)/(48*1.13x1010) = 580 nm
                            
See Beam Test Data 2 in good agreement

Scent = Smax = -WL/4Z= -(4.90*400)/(4*1.02x104) = .048 N/mm2 = 5x103 Pa

The Schott web site lists bending strengths for bonded Zerodur pieces in the range of 25 to 50 MPa; an older Schott publication listed Zerodur bending strengths of 70 to 120 MPa depending on surface finish.

Tilt = T = dY/dx = d/dx[(W/48EI)(3xL2-4x3)] = [W/16EI](L2 - 4x2
                with max at supports, minimum at center 
Tmax = T(0) = WL2/16E
I = (4.90*4002)/(16*1.13x1010)= 4.3 microradians
Tcent = T(L/2) = 0

See Beam Test Data 1, in good agreement.