Question

Problem 3: Knowing that the link BD has a uniform cross-sectional area of 900 mm, determine the internal force Nsp in terms of the load P. Draw FBD diagram for all your calculation. 120, 240 Units: KN for force and mm for length

Answer 1

Draw the overall FBD. CX T D Doe 120 mm & P 120 240 mm mm G tre] =0 Take Elec=0 I I 120 P] + [Dy x 240 ... Dy x 240 = -120P Dy = -0.5P Now, Note BD member. . passes through is a two Resultant of BD. force Dx of By

Also, £FY = 0 7 у = 05P EP Icy = 1.5 p Now, Drant The FBD ity = tos p I cx c of member ABC- = tan (4) 2 0 = 53.13° 320 FBD From A B C D tano = BC 240 tano = Be 120 BC=320 mm EMC B =0 I 2 = 6 7 T20 P = (x x 320

x = 0.375 P - I FBD sino +P = 1.5P FBD = P x0.5 sino FBD = (1.25 P) X0:5 FBD = 0.625 P Compression - Internal normal force Ans. Internal normal NBD = 0.625 force P KN - An swer = Normal stress [ Not asked - GBP ] NBD Area = 0.625 P goo = (6.9444 810-4 P) KN