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7-46. The beam is subjected to a shear of V = 800 N. Determine the average shear stress developed in the nails along the sides A and B if the nails are spaced s = 100 mm apart. Each nail has a diameter of 2 mm.

7-46. The beam is subjected to a shear of V = 800 N. Determine the average shear stress developed in the nails along the sides A and B if the nails are spaced s = 100 mm apart. Each nail has a diameter of 2 mm.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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why is the area of the nails multiplied by 2?

7-46. The beam is subjected to a shear of V = 800 N. Determine the average shear stress developed in the nails along the sides A and B if the nails are spaced s = 100 mm apart. Each nail has a diameter of 2 mm. y I = 0.015(0.03) (0.25) + 2(0.075)(0.15)(0.03) 0.03(0.25) + 2(0.15)(0.03) (0.25)(0.03³) + (0.25)(0.03) (0.04773 -0.015)² 12 + (2)(1/2)(0.03)(0.15³) + 2(0.03) (0.15)(0.075 -0.04773)² 32.164773(106) mª q= Q=y'A' = 0.03273(0.25)(0.03) = 0.245475(10³) m³ VQ I 800(0.245475) (10-³) 32.164773(10-6) F = qs = 6105.44(0.1) = 610.544 N Tavg 0.04773 m 6105.44 N/m Since each side of the beam resists this shear force then F 2A 610.544 2)(0.002²) 97.2 MPa 100 mm 100 mm 30 mm A 30 mm 250 mm B 150 mm 30 mm FJ = 0.03273m
Transcribed Image Text:7-46. The beam is subjected to a shear of V = 800 N. Determine the average shear stress developed in the nails along the sides A and B if the nails are spaced s = 100 mm apart. Each nail has a diameter of 2 mm. y I = 0.015(0.03) (0.25) + 2(0.075)(0.15)(0.03) 0.03(0.25) + 2(0.15)(0.03) (0.25)(0.03³) + (0.25)(0.03) (0.04773 -0.015)² 12 + (2)(1/2)(0.03)(0.15³) + 2(0.03) (0.15)(0.075 -0.04773)² 32.164773(106) mª q= Q=y'A' = 0.03273(0.25)(0.03) = 0.245475(10³) m³ VQ I 800(0.245475) (10-³) 32.164773(10-6) F = qs = 6105.44(0.1) = 610.544 N Tavg 0.04773 m 6105.44 N/m Since each side of the beam resists this shear force then F 2A 610.544 2)(0.002²) 97.2 MPa 100 mm 100 mm 30 mm A 30 mm 250 mm B 150 mm 30 mm FJ = 0.03273m
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