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Forces in all the members are calculated with all details as shown.
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The length of each member in the truss is L feet. Note: On your FBD draw...
Logout Course Q2 Weight: 20.00 % For the symmetrical truss, use the method of joints and calculate: Note: On your FBD draw all internal forces in the positive direction (tension). F1 lb F, Ib F, lb 6 0° F2 lb 4 ft 6ft B 6ft G 6ft 1 h ft h ft Fs lb COLLAPSE IMAGES F1 80 lb F2 240 lb F3 405 lb 0 80 deg h 10 ft Best Score: 2.00/5 Current Score: 2.00/5 The force in...
For the truss, use the method of sections Note: On your FBD draw all internal forces in the positive direction (tension. FkN FKN PKN A 0.51 m F Im FKNM COLLAPSE IMAGES F = 53 KN P = 98 N h 3m 1 = 5m 0.35 deg Best Score: 4.00 / 5 F = 53 kN P = 98 kN h = 3m /=5 m = 35 deg Best Score: 4.00/5 Current Score: 0.00 / 5 Calculate A. (the x...
やrcurPoint や rourPoint For The Given Truss, C/X G For the sy metrical t /X e The Length or Each M × e The Weight or The Tru e Civil Engineering que -) С https://www.msufourpoint.com/flassessment/1718/214/Homework%2010 FourPoint Course - 4 LogoutMICHIGAN STATE UNIVERSITY Q.1 For the symmetrical truss, use the method of joints and calculate Note: On your FBD draw all internal forces in the positive direction (tension) Q.2 Q.3 F1 lb F, lb Q.4 Q.5 F2 lb 6 ft B...
The truss is supported by wheels at A and a hinge at B and subjected to 2 equal vertical forces P lb at C and G, and a horizontal force P lb at D. Draw the reactions in the positive coordinate directions. P lb y L ft P lb D X A Lft L ft P lb COLLAPSE IMAGES P 175 lb L5 ft The x component of the reaction at A, Ax. ENTER 3 tries remaining 1 point(s) possible...
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Determine the force in each member of the Howe roof truss shown. Take P1 = P2 = P3 = 600 lb. (Round the final answers to their nearest whole number.) P2. P1 P3 300 lb 300 lb 6 ft 6 ft E 8f-8f8f8ft l b. (Compression) b. (Tension) The force in member AB (FAB) is The force in member AC(Fad is The force in member BC (Fed is O. The force in member CE(Fce is The...
For the beam shown, draw the reactions at supports A and B in the positive direction, and also draw the shear and bending moment in the positive direction on your FBD. w lb/ft aft L ft aft COLLAPSE IMAGES w = 480 lb/ft L = 13 ft 9 = 2 ft Calculate the support reaction Ay lb ENTER 3 tries remaining. 1 point(s) possible Calculate the support reaction Ax lb ENTER 3 tries remaining. 1 point(3) possible Calculate the support...
For the beam shown, draw the reactions at supports A and B in the positive direction, and also draw the shear and bending moment in the positive direction on your FBD. w lb/ft aft L ft aft COLLAPSE IMAGES w = 480 lb/ft L = 13 ft 9 = 2 ft Calculate the support reaction Ay lb ENTER 3 tries remaining. 1 point(s) possible Calculate the support reaction Ax lb ENTER 3 tries remaining. 1 point(3) possible Calculate the support...
The hook and spring assembly shown in the figure are equilibrium, calculate: F. lb k lb/ft А (1) Falb В 02 2 ft F3 lb COLLAPSE IMAGES F1 = 110 lb F3 = 52 lb 01 - 21 deg 02 - 35 deg k = 1506 lb/ F2 = 98 lb The magnitude of the resultant force. Ib ENTER 3 tries remaining. 1 points) possible The orientation counter-clockwise from the positive x axis of the spring. deg ENTER 3 tries...
Consider the diagram below: Note: On your FBD draw all internal forces in the positive direction (tension). RN QN E SN PN TNB а т т а т т а т т т а т COLLAPSE IMAGES P = 113N Q = 312 N R = 129 N S = 82 N T = 81 N = 45 deg a = 2 m Determine the force in member EB N ENTER 3 tries remaining. 1 point(s) possible
For the beam shown, draw the reactions at supports A and B in the positive direction, and also draw the shear and bending moment in the positive direction on your FBD. Calculate w lb/ft PID Mb.ft с Lft B a ft bft COLLAPSE IMAGES w = 140 lb/ft P= 200 lb L = 7 ft a = 4 ft b = 4 ft M = 80 lb.ft The vertical reaction at A. (Ay) ID ENTER 3 tries remaining. 1 point(3)...