The 1-meter aluminum (E=70 GPa) beam of hollow square cross-section (20x20 mm with 2 mm thickness) shown is under distrubuted loading according to w(x)=100sin(TTX) N/m. The maximum shear force V is N. Type your choice in the blank: 26.2

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The 1-meter aluminum (E=70 GPa) beam of hollow square cross-section (20x20 mm with 2 mm thickness) shown is under distrubuted loading according to w(x)=100sin(TTX) N/m. The maximum shear force V is N. Type your choice in the blank: 26.2 27.5 28.9 30.3 31.8 The maximum bending moment Mis The maximum normal stress due to bending ox-Mc/l is The maximum strain Ex=0x/E is The maximum slope angle is The maximum deflection is C RECOMMENDATION: Spend no more than 25 minutes for this question. (0= 100sin(x) N.m. Type your choice in the blank: 9.2 9.6 10.1 10.6 11.2 MPa. Type your choice in the blank: 12.9 13.5 14.2 14.9 15.6 x10-4 m/m assuming elastic deformation. Type your choice in the blank: 1.75 1.84 1.93 2.03 2.13 0. Type your choice in the blank: 0.28 0.30 0.34 0.37 0.39 mm. Type your choice in the blank: 1.40 1.54 1.69 1.86 1.96 Ty=+bh³ Ty=b³h 4-bh³ 1-meter aluminum beam above has a hollow square cross section of 20x20 mm with 2 mm wall thickness Jc-bh(b²+ h²) ΕΙ d'y --(x) E-V(x) E-M(x) dx [si sin ar dr Jo cos ar dr = 1 a - cos ax + C a sin ax + C Integrating M(x) will provide a new equation (with an integral constant C₁) as dv dx This second equation is related to deflection angle as ~tan 0 = dv/dx Integrating this equation will provide a third equation (with a new integral constant C₂) as ΕΙ Elv This third equation is related to deflection amount, v The integral constants C₁ and C₂ are obtained using boundary conditions.

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