Three-point bending¶

Simply supported beam with central load¶

What is the maximum deflection of a simple supported beam with central load, if there is an uncertainty of only 1% for all input parameter? The input parameter are load P, beam length L, beam width W, beam height H and young’s modulus E.

$P &= 5\, kN \pm 1\% L &= 2\, m \pm 1\% W &= 50\, mm \pm 1\% H &= 100\, mm \pm 1\% E &= 30000\, N/mm^2 \pm 1\%$

Three point bending test [wikipedia.org]

$w(x) &= {\begin{cases}-{\frac {Px(4x^{2}-3L^{2})}{48EI}},&{\mbox{for }}0\leq x\leq {\tfrac {L}{2}}\\{\frac {P(x-L)(L^{2}-8Lx+4x^{2})}{48EI}},&{\mbox{for }}{\tfrac {L}{2}}<x\leq L\end{cases}} w &= w_{L/2} = \tfrac {PL^{3}}{48EI} A &= WH I &= WH^3$

Code¶

import phuzzy as ph
number_of_alpha_levels = 31

# load P
P0 = 5000.  # N
dP = 0.01 * P0  # N
P = ph.Triangle(alpha0=[P0 - dP, P0 + dP], alpha1=[P0], name="P", number_of_alpha_levels=number_of_alpha_levels)

# dimensions L, W, H
W0 = 50  # mm
H0 = 100  # mm
L0 = 2000  # mm

dW = 0.01 * W0  # mm
dH = 0.01 * H0  # mm
dL = 0.01 * L0  # mm

L = ph.Triangle(alpha0=[L0 - dL, L0 + dL], alpha1=[L0], name="L", number_of_alpha_levels=number_of_alpha_levels)
W = ph.Triangle(alpha0=[W0 - dW, W0 + dW], alpha1=[W0], name="W", number_of_alpha_levels=number_of_alpha_levels)
H = ph.Triangle(alpha0=[H0 - dH, H0 + dH], alpha1=[H0], name="H", number_of_alpha_levels=number_of_alpha_levels)

# material

E0 = 30000.  # N/mm2
dE = 0.1 * E0  # N/mm2
E = ph.TruncNorm(alpha0=[E0 - dE, E0 + dE], alpha1=[E0], name="E", number_of_alpha_levels=number_of_alpha_levels)

I0 = W0 * H0 ** 3 / 12.
w0 = P0 * L0 ** 3 / (48 * E0 * I0)

print("I0 = {:.4g} mm^4".format(I0))
# I0 = 4.167e+06 mm^4
print("w0 = {:.4g} mm".format(w0))
# w0 = 6.667 mm

I = W * H** 3 / 12.
I.name = "I"
w = P * L ** 3 / (48 * E * I)
w.name = r"P L^3 / (48 EI)"

print("I = {} mm^4".format(I))
# I = FuzzyNumber(W*H^3/12.0:[[4002483.375, 4335850.041666667], [4166666.6666666665, 4166666.6666666665]]) mm^4

print("w = {} mm".format(w))
# w = FuzzyNumber(P*L^3/E*48*W*H^3/12.0:[[5.594629603627992, 8.024370049019725], [6.666666666666667, 6.666666666666667]]) mm

w_mean = w.mean()
dw_l = w_mean - w.min()
dw_r = w.max() - w_mean
print("w = {:.4g} mm (- {:.4g}|+ {:.4g})".format(w_mean, dw_l, dw_r))
# w = 6.703 mm (- 1.109|+ 1.321)
print("w = {:.4g} mm [{:.4g},{:.4g}]".format(w_mean, w.min(), w.max()))
# w = 6.703 mm [5.595,8.024]

Parameter¶

used Parameter

Results¶

results area moment of inertia I and deflections w

Source code ssb.py