3.37 kN 32.63 kN | 15 kN M 7 kN/m B |C D E 4 m 4 m 4 m 3 m 21 A, 4, A, A,-15 A,-15 O 23.3 kN-m O 26.1 kN-m O 19.7 kN-m
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A: Replacing support at A and with the tension in the rope and in member BC, below given is the Figure.…
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Q: A A) RA= 1 50 KN 1m 1.5m B) RA= 45 KN/m 40 KN/m 6.00 m KN, R₂ = 3.00 m KN, R₂ = 20 275 KN 1m 4.00 m…
A: find the reactions on the supports.
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A:
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A: Given:The deformation per unit length,
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A:
Q: Q1: MO =S 400 lb F1 = 400 1 F2 = -600 k F3 = 500 sin 45 i-500 cos 45 k 3" 7" r2=? 600 lb 45 M-…
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Q: 3 kN/m В 8 m 4 m D 10 m 5 kN H. = 5 kN A A M, = 66 kNm a Va= 24 kN
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Q: m₂ = 0.70 m₂ = 0.40 (a) 37° Stickman provides F Find F when: a) to break free. @ 20.kg b) to move…
A: (a) for break free ; Fapplied=f=μsmgcosθ=0.7×20×9.81×cos37°=109.7 N
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Q: G2 A 16" 13" 3" Answers: NA = 277.3 lb NB: 277.3 lb
A: Draw the free-body diagram of the system.
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A: According to the principal of engineering mechanics Reaction and moment at support is equal to zero
Q: 6 KN 5 kN 4 kN 4 kN 5 kN, K 3 kN 4 m 2 kN HA A G |B C |D F 2 m 2 m 2 m T 2 m T 2 m T 2 m
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Q: 12 kN 12 kN C VE VG 1.8 m Bo D 4 panels @ 2.4 m ' 9.6 m
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Q: Im 2m 3m Im 3m Question 4 10 kN 20 kN UDL = 20 kN/mD E A B 2.5m 2m 2m 3m RBY ReY 16
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Q: V₁ = 3 m/s D=450 mi 2₁:5 mm 3m 3m 2354.5 V₂ = 2.5 m/s D₂ = 300 mm V3 = ? = 200 mm
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A: The given is as shown below,
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A:
Q: 12 KN 2.5 KN/M 介介企 A 12 m B 4m
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A: Apply the equilibrium condition of force in the vertical direction. Apply the equilibrium condition…
Q: F1 = 5 kN F2 = 13 kN 0 = 45° B = 60° C F3 = 4 kN A B 4 m 8 m F4 = 22 kN 6m
A: given:
Q: 1.0 m 2.0 m 05 m 05 m 20 kN 4 KN 10 kN 4.0 m
A: Given Data: Force (F) = 10 kN
Q: 20 kN D 20 kN 20 kN 3 m 10 kN 3 m 3 m A 4 m 4 m E 4 m 4 m B,
A:
Q: З KN 8 kN A 25 kN m 2 m 2 m 2 m 5 kN
A: Note: As per bartleby guidelines for more than one question asked only one question to be solved…
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A: A beam is defined as a structural member subjected to transverse shear load during its…
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Q: 400 lb F1 = 400 F2 = -600 k F3 = 500 sin 45 i-s00 cos 45k 7t Y600 lb 45 M- rxF,. M,- r2x , M,- rgxf,…
A: "Since you have asked multiple questions, we will solve the first question for you. If you want any…
Q: F = 800 N F = 800 N F = 800 N 60° 60° 60L xr FR FR 0 = 90° (b) (c) F2 (a) Solution: The magnitude of…
A:
Q: Identify the zero-force members in each truss
A: There are two rule to identify zero force in the truss members:- (a) Let's consider a joint which is…
Q: Q1: MO=? 400 lb F1 = 400 j F2 = -600 k %3D F3 = 500 sin 45 i -500 cos 45 k 3 7° r2=? 600 lb r3=? y…
A: Given data: The first force is F1=400j^ The second force is F2=-600k^ The third force is…
Q: E D 1 m A F F 1m 1 m
A:
Q: F = 375 N F2 = 500 N 5 3 В. 0.5 m 8 m 6 m - 5 m- 370 F3 = 160 N
A: The given system of forces is as shown below,
Q: FAH = 0 FAI 5. 4 Ax = 0.9 kN FAB Ay = 5.1 kN %3D O A. FAI = 10.125 kN (T) O B. FAI = 6.375 kN (C)…
A: Force in y-direction FAlsinθ sinθ=PH=45
Q: 2.6 kN 2.4 kN-m B A 1.0 m 1.0 m 1.0 m
A: To draw the shear force or bending moment diagram, one should first determine the reaction at the…
Q: A Mansard roof truss is loaded as shown. Determine the force in members DF, DG, and EG.
A: Given: The loaded truss is shown below:
Q: 2.5 kN 1.5 kN 3 kN 30 A B. 2m 4m 2m
A: Resolve the two forces along the beam and perpendicular to the beam as shown. Write the x component…
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- The Z-section of Example D-7 is subjected to M = 5 kN · m, as shown. Determine the orientation of the neutral axis and calculate the maximum tensile stress c1and maximum compressive stress ocin the beam. Use the following numerical data: height; = 200 mm, width ft = 90 mm, constant thickness a = 15 mm, and B = 19.2e. Use = 32.6 × 106 mm4 and I2= 2.4 × 10e mm4 from Example D-7Use the graphical method to construct the bending-moment diagram and identify the magnitude of the largest moment (consider both positive and negative). The ground reactions and the shear-force diagram are provided. M = 13 kN-m Ay = 3.46 KN Dy Units: KN = 32.54 kN M 4 m 29.5 kN-m 43.3 kN-m 25.8 kN-m O 22.9 kN-m 31.48 kN-m B Ay 4 m 15 kN C Ay 4 m Ay-15 7 kN/m D 3 m 21 Ay-15 E XUse the graphical method to construct the bending-moment diagram and identify the magnitude of the largest moment (consider both positive and negative). The ground reactions are provided. MB = 143 kN-m Mc = 51 kN-m Ay = 16.17 kN (down) Dy = 16.17 kN (up) 1997 4 m 5 m O 80.9 kN-m O 74.1 kN-m O 94.5 kN-m O 50.4 kN-m O 60.6 kN-m 3 m D X
- 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 0.00usics 1 (Engineering) / My courses / PHYS1100 / Quiz 1 for ALL sectiolMOBetma Fomrward 2020 with Confidence Find the cross product between the vectors C%A" for A" =(2i ~-5 j'+k") and C= (3i-2j-4k) of O a. (22i-5j+11k") tion O b. (22i+11 j+11k") O . (221+11 j-19k") O d. (-22i-11 j-11k) Next page pageConstruct the shear-force and bending-moment diagrams and identify the largest negative bending moment. Assume a-3.9 ft, b =7.8 ft, c-3.9 ft. w-1100 lb/ft, and P-3200 lb. A Answer: Mmin a W B b P TTV C lb-ft D
- Find 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.III. Using Three Moment Equation, determine all reactions at supports also, determine shear and bending moments with complete labels. Use I = 1/12 bh^3 and E = 2.15x10^7 KPa 1381 KN/m 1381 KN/m 690.5 KN/m D B 4.0 m 3.0 m 7.0 m 4.0 m b = 0.25m h = 0.50m 4.0 m b = 0.35m h = 0.70m b = 0.45m h = 0.90my Courses E This course EHide block What are the magnitude (F) and the location measured from end A (d) of the equivalent resultant force of the triangular distributed load in the figure below. 5 m 3 m Wo = 7 kN/m %3D A TT Select one: a. F= 10.5 KN ; d= 6 m b. F= 6 KN ; d= 6.5 m c. F= 21 KN; d= 6 m d. F = 6 KN; d = 1 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. AFind the shear force and bending moment 1.5 m, 3 m, and 4.5 m from the left end of the beam. 9 kN/m 3 m Answer: x = 1.5 m: V = 3 m: V = 0, M = 27 kN • m x = 4.5 m: V = -10.125 kN, M : 10.125 kN, M = 18.625 kN·m %3D %3D 18.5625 kN · m %3D481 A Find the shear force and bending moment at points C and D. 300 lb/ft Answer: Vc = -2.11 kip Mc = -3.72 kip ft VD = 900 lbf Mp = -7.35 kip ft D 12 ft 15 ft C weights Lea... B D 3 ft 6 ft I Centroids -... 6000 lb-ft G4.4.1 Weight...