4.17. Determine the components of stress from the results obtained in (a) v=rsin 0, ve = 2r cos 0 (b) VT = cos 0, 1/4 = 0 (c) v = V₁ = 0 (d) v = (1 - 4) cos 0, Ve= - - (1 + 4/4) sin 0 - B
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- Question 2) arm of Figure B from point x z plane, a horizontal angle of θ = 44 ° angled F = 26 kN size and a force C at the point z in the direction of M = 18 kn.m acts a moment in size. The lengths of the arm are also given as L 1 = 1.6 m and L 2 = 1.3 m . It is desired to determine the stress state of point A on the aa section taken from the arm. The radius r of the section is r=0.029 m and the shear modulus of the sleeve material is also G = 79 Gpa . According to this; Question 2-C) Find the normal stress caused by the normal force at point A ( Write your result in MPa .) Question 2-D) Find the normal stress caused by the bending moment at point A. (Your result MPa in the size of your font.)Question 2) arm of Figure B from point x z plane, a horizontal angle of θ = 44 ° angled F = 26 kN size and a force C at the point z in the direction of M = 18 kn.m acts a moment in size. The lengths of the arm are also given as L 1 = 1.6 m and L 2 = 1.3 m . It is desired to determine the stress state of point A on the aa section taken from the arm. The radius r of the section is r=0.029 m and the shear modulus of the sleeve material is also G = 79 Gpa . According to this; Question 2-A) Find the shear stress at point A due to the shear force . ( Write your result in MPa .) Question 2-B) Find the shear stress due to the torsional moment at point A. ( Write your result in MPa .) Question 2-C) Find the normal stress caused by the normal force at point A ( Write your result in MPa .)Question 2) arm of Figure B from point x z plane, a horizontal angle of θ = 38 ° angled F = 29 kN force sized and C from the point z direction M = 17 kn.m acts a moment in size. The lengths of the arm are also given as L 1 = 1.5 m and L 2 = 1.1 m . It is desired to determine the stress state of point A on the aa section taken from the arm. The radius r of the section is r=0.015 m and the shear modulus of the arm material is also G = 78 Gpa . According to this; Question 2-A) Find the shear stress at point A due to the shear force . ( Write your result in MPa .) Question 2-B) Find the shear stress due to the torsional moment at point A. ( Write your result in MPa .) Question 2-C) Find the normal stress caused by the normal force at point A ( Write your result in MPa .)
- F2 Q1) Axial displacement of point C in the system shown on the left is 0.01 cm. Find the maximum elongation of the bar and the maximum normal stress. A F1 A. GIVEN: F2 = 100 kN, (1 = 240 cm, €2= 160 cm, bi = 5 cm, bz=10 cm, h=5 cm, E=2.107 N/em², a=20 cm Note: neglect stress concentration. h bị b2 A-A sectionThe distribution of stress in an isotropic aluminium machine component is given (in MPa) as: σ₂ = y +2z² - 6 σ₁ = x+z-6 oy σ₂ = 3x+y-13 6: T =3z² -11 xy (i) (ii) =x²-14 Tyz Txz = y² XZ x, y and z are coordinates of a point within the machine component. By taking Young's modulus, E = 70 GPa, Poisson ratio, v= 0.3 and yield stress, Y = 5 MPa, do the following for a point P located at (4, 1, 2): a) Provide the stress and strain tensors. b) Determine all the principal stresses and principal strains. c) Determine if the machine component will fail based on the failure criteria below: Tresca criterion Von Mises criterionb2 h G B M. Q4) For the given system; a) Draw internal force diagrams and determine the most critical section. b) Show the stress distribution on this critical section and find the maximum normal stress. GIVEN: (1 = 50 cm, {2= 30 cm, M, = 5000 Nem, q. = 20 N/cm, bị = 4 cm, b2 = 2 cm, h = 4 cm, t= 1 cm
- Question 2) arm of Figure B from point x z plane, a horizontal angle of θ = 38 ° angled F = 29 kN force sized and C from the point z direction M = 17 kn.m acts a moment in size. The lengths of the arm are also given as L 1 = 1.5 m and L 2 = 1.1 m . It is desired to determine the stress state of point A on the aa section taken from the arm. The radius r of the section is r=0.015 m and the shear modulus of the arm material is also G = 78 Gpa . According to this; Question 2-C) Find the normal stress caused by the normal force at point A ( Write your result in MPa .) Question 2-D) Find the normal stress caused by the bending moment at point A. (Your result MPa in the size of your font.)State of stress at the wall and the roof of the following circular opening bafore the excavation is (a] = |0 4. Find the induced stress components at point a. R W with respect to the given x-y coordinate systerm, and expresst them in the matrix form. b. Find the total stresses at point W, and give them in matrix form. c. Repeat (a) and (b) for the roof. d. Find the stresses oy, Ciy, and tyy at point A which is located at a radial distance twice the radius of the tunnel.Determine the resultant internal loadings acting on the cross section at B of the pipe shown. End A is subjected to a vertical force P and a horizontal force Q. Use P = 90 N and Q = 70 N. a. Determine the internal axial force (in N) in section B. b. Determine the internal shear force (in N) in section B. c. Determine the internal torque (in N-m) in section B. With FBD please thank you.
- 5/5 find the normal stress at left of point C A 100 C 40 kN 50 kN 40 mm 70 10 kN 20 mm 30 mm 100 mm 300 mm 300 mm 200mm B.1. A circular shaft with a keyway can be approximated by the section shown in Fig. r=b Ay r r=2acose Figure 1: Circular shaft with a keyway. The keyway is represented by the boundary equation r = b and the shaft is represented by the boundary equation r = 2a cos 0. Show that using a Prandtl stress function of the form ²) (₁ (1-2a cos) X v = K (b² − ²) (1 – : will solve the problem of torsion on this shaft and find the constant K (assume an applied torque T). Compute the shear stress components Txz and Tyz. (Hint: start by converting the stress function to Cartesian coordinates).Consider a long cylindrical solid rod - total length L - mass density p; - radius R1 = R2 = R; suspended from the top and hung freely downward along the (-1)Ž direction; - subjected to gravitational pull from the ground in the (-1)Ź direction. R2 +Z Z = h R1 Z = 0 a) Calculate the longitudinal stress o within the rod as a function of h; b) Repeat the calculation ifR| turns into a cone); = 0 and R2 = R (i.e., when the cylindrical rod c) Based on your answers to a) and b), provide some reasoning as to why a space elevator would not work [hint: think about the rod as the load bearing element of the space elevator].
