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](3L24x2)
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)
= 0.58x103 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](3XL24X3)
= [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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