Question 17.5: Determine the bending ligament efficiency in the header in E...
Determine the bending ligament efficiency in the header in Example 17.3.
p = 4in.
T_{0} = 0.125in. d_{0} = 1.625in.
b_{0} = 4 − 1.625 = 2.375in.
T_{1} = 0.375in. d_{1} = 1.5in.
b_{1} = 4 − 1.5 = 2.5in.
T_{2} = 0.25in. d_{2} = 1.375in.
b_{2} = 4 − 1.375 = 2.625in.
\Sigma A X = 2.375 × 0.125(0.0625 + 0.375 + 0.25) + 2.5 × 0.375(0.1875 + 0.25) + 2.625 × 0.25(0.125)
\Sigma A X = 0.6963
\Sigma A = 2.375 × 0.125 + 2.5 × 0.375 + 2.625 × 0.25 = 1.8906
\bar{X}=\frac{0.6963}{1.8906} =0.3683 in.
I=\frac{1}{12} [(2.375)(0.125)^{3}+(2.5)(0.375)^{3}+(2.625)(0.25)^{3}] + (2.375)(0.125) \times (0.625+0.375+0.25-0.3683)^{2} +(2.5)(0.375) (0.1875+0.25 -0.3683)^{2} +(2.625)(0.25)(0.3683-0.125)^{2}I = 0.0884
c = (the larger of 0.3683 or 0.75 − 0.3683)
= 0.3817
From Eq. (17.14), the equivalent diameter is
D_{ E }=p-\frac{6 I}{t^{2} c} (17.14)
D_{ E }=4-\frac{(6)(0.0884)}{(0.75)^{2}(0.3817)} = 4 − 2.47 = 1.53in.
The ligament efficiency for bending stress is calculated from Eq. (17.15) as
e_{ b }=\frac{P-D_{ E }}{p} (17.15)
e_{ b }=\frac{4-1.53}{4}=0.617