If F = 69 N, determine the internal normal force, shear force, and moment at points D and E in the two members. (Figure 1) Figure D 1 m 0.75 m B < 1 of 1
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- 3 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 0.00Determine the normal force, shear force, and moment at point C. Take that P = 8 kN and M = 33 kN m. (Figure 1) Figure M B . 1 of 1 -1.5 m 1.5 m 1.5 m-1.5 m-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. A
- A folding tray mechanism is attached to a wall as shown. Find the internal forces and bending moment in the lower support arm at section s-s, located midway between points B and E, when a force of F = 270 N is applied at an angle of 0 = 28°. 2013 Michael Swanbom Ⓒ030 BY NC SA k s -Ś a b G с h₁ h₂ ка E S- B Values for dimensions on the figure are given in the following table. Note the figure may not be to scale. Variable Value 16 cm 19 cm 25 cm 23 cm 17 cm The internal axial load at section s-s is A = The internal shear load at section s-s is V = The internal bending moment at section s-s is M = N. N. C. N-m. 0Below 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. 125mmFind 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.
- (a) Develop shear and bending moment diagrams using the graphical method. (b) Determine functions for shear and moment using the method of sections for the region defined by 0The steel box wrench (E=29,000ksi, v=0.32) is loaded with a 40-lb horizontal force and a 25-lb vertical force. Find the following for an element located at point B on Section a-a: 24 in. 1. The non-vanishing resultant internal moments on section a-a are which of the following? 4 in. a. (160 i- 600 j+960 k)in · Ib b. (-160 i+600 j+960 k ) in · lb c. (-960 i+160 j+600 k)in - lb d. (-960 i -160 ј-600 k)in-lb a e. none of the above 2. The normal and shear force components on section a-a are: a. V, = 65lb;V, = 40lb; N¸ = 25lb b. V, = 0;V, =40lb; N, = 25lb c. V, = 64;V, = 25lb; N, = 40lb d. V, =0;V, =-40lb; N¸ = -25lb e. none of the above 3. The non-vanishing stress components at point B are: 0̟ =1.60; t =-4.89ksi; а. 0, =-4.89; 7 =3.60ksi; 0, = 2.50ksi; T, d. o, = 4.91ksi; t =-4.89ksi; b. с. =-4.89ksi; е. none of the above 4. The principal stresses at point B from most tensile to most compressive are: O1,02,0; =4.16,–0.8,–5.76 b. o,,0,,0, =5.76,0,–4.16 а. O1,02,0; =1.9,-2.45,-6.8 с. d.…Consider the pipe assembly shown in (Figure 1). Neglect the weight of the pipe. Take F₁ = (200i 200j - 250k) N and F2 = (2001+ 500j) N. Figure ▸ ▼ Part A Part B Part C 1.5 m Submit Mr, My, Mz = 525,- 250,0 B Previous Answers Im Determine the x, y, and z components of internal moment acting at a section passing through point B on the left part of the body relative to the cross-section. Express your answers in newton-meters to three significant figures separated by commas. ▸ View Available Hint(s) LIVE ΑΣΦ 41 vec 1 m X Incorrect; Try Again; 4 attempts remaining 1 of 1 ? N.mDetermine the ff: •Internal bending moment at point c •internal shear force at point c •internal normal force at point c Asap please.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 Q3A moment about the z-axis of 192 N-m is applied to a beam with dimensions b = 85 mm and h = 302 mm. If there are no other loads applied to the beam (My, P, Vy and Vz = 0), what is the normal stress at point F? Give your answer in kPa to two decimal places with negative indicating compression and positive indicating tension. b/4 М, h E Ihis F h/3 В M. D b/4SEE MORE QUESTIONS