Question

0 Q c) 28125000 1 Question 4 2 the shown column subjected to the shown forces Answer the following questions 100 O F c) 42187500 150 c) 204.44444 F, [kN] F; [kN] [kN] 300 500 400 c) 44.444444 5) moment of inertia about x-axis Ix (mm) a) 83333333 b) 42187500 6) moment of inertia about y-axis ly a) 83333333 b) 28125000 7) Normal stress at point A[Mpa) a) 186.66667 b) 44.4444444 8) Normal stress at point B [Mpa) a) 62 222222 b) 186.666667 9) Normal stress at point o [Mpal .) 35.555556 10) Transverse shear stress at point A[Mpa) a) 44444444 b) 11) Transverse shear stress at point B[Mpa) a) 400000 b) 4.44444444 12) Transverse shear stress at point o [Mpa) a) 106.66667 b) c) -8.8888889 12 c) 35000 c) 400000 10 11 ) moment in x direction Mx [kN.mm) al -35000 b) 65000 132) moment in y direction My [kN.mm) a) 20000 -20000 153) Normal force in z direction [kN] a)-800 b) 800 174) Cross section arca A [mm] b) 10000 c) 80000 co 16 c) 200 c) 4.4444444 18 al 22500 c) 90000 19

Answer 1

1.) Moment in x-direction:-

= (-F₂*100) + (F₁*50) + (F₃*0)

= -500*100 + 300*50

= -35000 kN-mm

2.)

Moment in y-direction:-

= (-F₂*100) + (F₁*100)

= -500*100 + 300*100

= -20000 kN-mm

3.)

Normal force in z-direction:-

= -F₁ - F₂

= -500 -300

= -800 kN.

4.)

Cross section area:-

= L* B

= 150 * 150 mm²

= 22500 mm²

As per HOMEWORKLIB RULES, I have solved first four parts of the question