Question

Links BC and DE are both made of steel (E = 200 GPa) and have cross-sectional dimensions Problem 2. 1.27 cm x 0.635 cm. Determine (a) the force in each link when a 2.67 kN force P is applied to the rigid member AF shown, (b) the corresponding deflection of point A. P = 2.67 kN 10.16 cm B 5.08 cm 5.08 cm F 10.16 cm 12.70 cm

Answer 1

(a) 2.67KN AA A Ae 5.08 5.08X2 10.16 +5.08X2 10.16cm Ac x Clo-16+5.08X2) 5.08X2 2Ac Ac 5.08cm Ac 2AD DE Let the force in BC be Pac and DE be PorE. 5.08cm Also we know, A= PL AE Deflected shape. Ac = - Pec hBc -PDE LDE ABe E ADE X E Eve s hows deblection opposite to load) sign Ac = 2 4p ABC= ADE = 1:27 X0.635 = 0,80645 cm2 = 0.80645X16m² 2,PDE LDE Aoe X E PochBc ABc E %3D E = 2000la- - Pec LBc =2 POE LDE 200 X/09 N In3 = 200X106 KN/m2 2 PoE LDE %3D LBC

2 Por X 12.7 10.16 2.5PDE. PBc %3D Take moment about F , ŹMp =0; since Fis a pin. EMF = 2.67X CI0-16+5 08+5 08) – PBc XC5.08+5.08) – PpE(5.08) = 0 54.25-10.16 PBc -5.08 PDE =0 10.16 Par+ 5.0s PDE = 54.25 %3D Poc=2.5 PDE=> 10.16 (2.5 Ppe) t5.08 PoR 54.25 30.48 PoE = 54.25 POE = 1.78 kN Compressive) PBc=4.45KN C Tensile) PBC = 2.5 X1-T8= 4.45 kN

PeclBL (b) Deflection at A, AR = 2Ac ABC E 4.45x (10.16/100) 0.80645X104 X200X106 = 5.606X10mm - 5.606 X105 m Ap = 5.606X 16-2 mm
a) Force in link BC= 4.45 kN

Force in link DE= 1.78 kN

b) Deflection at A= 5.606 ×10^-2 mm