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Figure P3.18(not drawn to scale 3.19 A shear force V20 kN acts on the thin cross section shown in Figure P3.19. The cross section has a uniform thickness of 10 mm. Determine the equation of shear flow along the center- lines and sketch it. 100 mm 25 mm 25 mm Figure P3.19(not drawn to scale)

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25 rom 2-3 d r To indv vay da da tdy 1惢(A.ds) (51) (%) v25 xID T 18 333 33 4134 583-33)1+83-33 21 1225916 s4 m ˇ 3:0. (here ilake shean flow as, v) 1 224516 bx(-20 X10) χ 100メ10) 6257 12295414 sq xlo 3 65 L-toxios)メ(625x10) xts) -132) し12.245416.sq xio) ND: -13.12 N/mm ←13-71 13 2 1

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