Week 8 - Assignment 7 Shear Stress.docx

# Week 8 - Assignment 7 Shear Stress.docx - 18 M 0=0 5000 N...

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18 M 0 = 0 5000 N × 50 mm F × 15 mm = 0 F = 16666.6 N τ = shear force crosssectionareaof key →τ = F A τ = 16666.6 N 20 mm× 5 mm = 166.66 N mm 2 = 166.66 MPa 19 The length and width of adhesive tape segments taken for the test of shear strength are 5 inches and 1 inch respectively. The force, F , necessary to cause the adhesive layer to slide off the table for different lengths ‘ a ’ attached are listed in the table below S. No Length, ‘ a Force, F , 1 0.5 0.2 2 1 0.4 3 1.5 0.6 4 2 0.75 5 2.25 0.82 6 2.5 0.875

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The shear stress acting on the tape for a particular length of adhesiveness to the table areaof adhesivenessattached 0.5 × 1 ¿ = 0.4 lb ¿ 2 × 1 psoi 1 lb ¿ 2 = 0.4 psi τ = pullforce ¿ the table ¿ = F a×b →τ = 0.2 lb ¿ The shear stress in tape for different length is listed below S. No Length, ‘ a ’ (inches) Shear stress, τ , (psi) 1 0.5 0.4 2 1 0.4 3 1.5 0.4 4 2 0.375 5 2.25 0.364 6 2.5 0.35 The graphs of pull force versus adhesive length, a , The graph of shear stress versus adhesive length, a
The adhesive tape breaks at a length of 2.8 inches on the table before it sheared off the table.

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