Question

The bar shown has a cross-sectional area of 0.001 m² and a modulus of elasticity of 100 GPa. It is subjected to a uniformly distributed axial force q= 50 kN/m pointed to the left. An external axial force F= 20 kN, pointed to the left, is applied at the middle of the bar, x=L/2. a. What is the axial force P in the bar as a function of x? b. What is the bar's total change in length? ttttttttta x = 1/2 x = L = 0.5 m = 1 m Note: All forces are axial, even though the diagram puts q above the bar.

Answer 1

Given bar of Area. A=0.000m² = 1000 mm? Modulny of elasticity E-loopa = 100 kN/mma. q=SOEN/m F= 20KN spomnos for part AB. AB. Consider at x distance Consider et from at sea section xx free end. PEE A. Ebro paqx. ww Ef=0 & P= qx. kn P-50 EN where xisin nete for past BC Efx=0 - P=ftqx P=20kn +502 kN P-(50x +20) FW Part(b) Now ban's total change in length a p dx 0.5 m xxtos, P=soa -for for Oosim Xelm P> 504+20 "Gox) dx (50x+20) dx = Foxlóf) [erry + 16" (some +208]o's - Cox10°) poetry + 10" (25{1*-0.54) +20(1-os] = 6.25*10*8m + 78.75X105 AL= 35x10 m AL= 35x16² mm TAL=0.35mm ] Shortning.. since bas is subjected to - Compressive forcer