4.3-2 Determine the shear force V and bending moment M just right of the 6 kN load on the simple beam AB shown in the figure. 6.0 kN 2.0 kN/m В A 0.5 m +1.0 m1.0 m→* 2.0 m 4.0 m-
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Determine the shear force v and bending moment M just right of the 6KN load on the simple beam AB shown in the figure
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- Solve the preceding problem (W 250 × 44.8) if the resultant force P equals 110 kN and E = 200 GPa.A beam is loaded and supported as shown, with F1 = 2050 lb, F2 = 2080 lb, M = 6070 ft*lb, and w = 1210 lb/ft. The dimensions on the figure are a = 3.8 ft, b = 6.6 ft,and c = 4.3 ft. Develop the expression that describes the shearing forces V(x) in the beam as a functions of x in the interval 3.8ft<x<10.4ft.Beam cross section is shown in Figure Q3 and given the second moment of area is I= 20.85 x 106 mm*. If a resultant shear force, Vacting at the beam cross section is 40 kN, (a) Determine the shear stress at point M, N, O and P based on the given shear force. (b) Sketch the shear stress distribution over the cross section and label the shear stress values V= 40 kN M 80 mm 40 mm y =100 mm 80 mm 80 mm 30 mm 80 mm Figure Q3
- The simply supported beam is subjected to the force F = 700 N and the uniform distributed load with intensity w = 150 N/m. Draw the shear force and bending moment diagrams (in your homework documentation) and determine the equations for V(r) and M(x). Take a = 0 at point A. 19 F a Values for dimensions on the figure are given in the following table. Note the figure may not be to scale. Variable Value a 5.2 m 2.6 m 3.12 m Support Reactions The reaction at A is N. The reaction at D is N. Shear Force and Bending Moment Equations In section AB: V(x)= N and M(x)= N-m. In section BC: v(x)- N and M(x)= N-m. In section CD: V(x)- N and M(x)= N-m. AFind the shear force and bending moment at points B and D. Note: B lies just to the right of the 150 lbf force and D is just to the right of the bearing at C. The bearing at A is a thrust bearing, while the bearing at C is a journal bearing. 150 lb A Answer: VB = -100 lbf MB = 750 lbf in VD = 75lbf Mp = -750 lbf in I 15 in. B 15 in. C. D 75 lb 10 in.Forces of F1=15 kN in the z direction from the B point, F2=10 kN in the z direction from the C point and F3=10 kN in the y direction act on the arm in the figure. The lengths of the arm are also given as L1=0.15 m and L2=0.16 m. The radius r in the a-a section taken over the arm is r=0.2 m and it is desired to determine the stress state at point A. The shear modulus of the sleeve material is G=65 Gpa. According to this; WRITE YOUR RESULTS IN THE BOXES WITHOUT THE UNITS IN kPa. a) Find the shear stress due to the shear force at point A. Response b) Find the shear stress due to the torsional moment at point A. Response c) Find the total shear stress at point A. Response d) Find the normal stress due to the normal force at point A. Response e) Find the normal stress due to the bending moment at point A. Response f) Find the total normal stress at point A. Response
- 2. The component shown always breaks at point C, which is just to the right of the contact point of the roller at B. Find the internal forces in terms of shear, normal and bending to show why this is. 150 mm -250 mm- 400 N OB 125 mm 500 N 110 mm 80 N•m3 For the beam shown, find the reactions at the supports and plot the shear-force and bending-moment diagrams. V = 9 kN, V2 = 9 kN, V3 = 200 mm, and V4 = 1100 mm. ATAT-V3 Provide values at all key points shown in the given shear-force and bending-moment diagrams. X (mm) B A = B = C = D = E= F= P = Q = E * KN * KN * KN × KN KN x KN ✩ kN.mm *kN.mm D 0.00 Reaction force R₁ (left) = In the shear-force and bending-moment diagrams given, +V 0.00 X (mm) 6.3 kN and reaction force R2 (right) = P 11.7 kN. Q 0.00Below Figure shows the section of an angle purlin. A bending moment of 5 kN.m is applied to the purlin in a plane at an angle of 30 deg to the vertical y axis. If the sense of the bending moment is such that both its components Mx and My produce tension in the positive xy quadrant, calculate the maximum direct stress in the purlin, stating clearly the point at which it acts. * 100 mm E 10mm 30 C D -10mm 57 MPa. 89 MPa. Non Above O 72 MPa. 125mm
- Forces of F1=8 kN in the z direction from the B point, F2=5 kN in the z direction from the C point, and F3=10 kN in the y direction act on the arm in the figure. The lengths of the arm are also given as L1=0.15 m and L2=0.16 m. The radius r in the a-a section taken over the arm is r=0.2 m and it is desired to determine the stress state at point A. slip of sleeve material module is G=75 Gpa. Accordingly, a) Find the shear stress due to the shear force at point A. b) Find the shear stress due to the torsional moment at point A. c) Find the total shear stress at point A. d) Find the normal stress due to the normal force at point A. e) Find the normal stress due to the bending moment at point A. f) Find the total normal stress at point A.Q2. In the extruded profile shown in the figure, the maximum allowable stress in tension is 120 MPa and the maximum allowable stress in compression is 150 MPa. Find the maximum bending moment that can be applied to this profile. 20mm 40mm 20 mm 54mm yc 40 mm A= bh h %3D C. Rectangular area A = %3D Triangular areaConsider a two-component rod fixed at the left end. The first component (on the left, positioned from z=0 to z=L/2) is a solid rod with a shear modulus of G and radius c1. The second component (on the right, positioned from z=L/2 to z=L) is a hollow pipe fixed to component 1 at z=L/2 with the same shear modulus and outer radius c1, but with an inner radius of c2. A torque of 6To is applied at z=L/2, and a torque of -2To is applied at z=L. What value of c2 is required to make the maximum shear stress equivalent in each component? What is the angle of twist at the free end at z=L?