4-e P₁ P₁₂ D 2m B 1.5 m Take P₁ = 45 kN and P2 = 30 KN. You found that BC=0KN and CD=45kN(c). Find the force of members BD and AD. Use the method of your choice (sections or joints).
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- Correct Answer is written below. Detailed and complete fbd only please. I will upvote, thank you. 1: The assembly shown is composed of a rigid plank ABC, supported by hinge at A, spring at B and cable at C.The cable is attached to a frictionless pulley at D and rigidly supported at E. The cable is made of steel with E = 200,000MPa and cross-sectional area of 500 mm2. The details of pulley at D is shown. The pulley is supported by a pin, passingthough the pulley and attached to both cheeks. Note that E is directly above B.Given: H = 3 m; L1 = 2 m; L2 = 4 m; w = 12 kN/m; x:y = 3:4Spring Parameters:Wire diameter = 30 mmMean Radius = 90 mmNumber of turns = 12Modulus of Rigidity = 80 GPaAllowable stresses:Allowable shear stress of Pin at D = 85 MPaAllowable normal stress of cheek at D = 90MPaAllowable bearing stress of cheek at D = 110MPa1. Calculate the reaction of spring Band tension in cable at C.2. Calculate the vertical displacementat C and the required diameter ofpin at D.3.…4m A 72 kN C E B D F 144 kN 3 m 3 m 3 m Figure Q16 Fill in the multiple blanks below. To find the reactions the starting point is to take moments at a suitable node location. Since node unknowns it is the ideal location to first take moments. By taking moments in a clockwise orientation we find a moment of there is an additional moment of 288 kNm from the load at C. From combining all moments together, we can then find the vertical reaction at F which is RFy= place. For best practice, it is a good approach to take moments at has two kNm due to the force load at node B and KN to 1 decimal in order to the find the vertical reaction RAY- Finally, we can sum forces in the horizontal direction to find the reaction RAX = -72 kN, assuming the reaction at A acts left-to-right. After which we can then sum forces in the vertical direction to verify the sum of RAY plus Rgy is the same as the total downwards force which should be KN.I am not entierly sure where I when wrong here but I ended up getting 38.188lbs as an incorrect answer. I am pretty confident in my meathod I just belive I made an algibraic error. Could you solve this for me? thank you so much.
- Include a free body diagram please.- For the frame of the figure:a) Draw up the free-body diagram of the three elements that compose it, for the BCD element, set theequilibrium equations (summation of forces along the longitudinal axis "t" and the perpendicular "n", as well as thesummation of moments at point C). Note: it is not required to evaluate the reactions or the charges on the elements.b) Make a cut at the midpoint between C and D and “CALCUE” the internal loads numerically.Use answer from first part without spring to build onto next part, thanks.
- Previously we addressed the problem of a pinned-joint truss in the shape of a regular pentagon that is supported on rollers and loaded by a single vertical force F. We want to determine the loads in the members and the displacements of the joints. 4 F You can assume the following: 3 Since this is an equilibrium problem, it can be tackled as a virtual work problem. Also, the loads are constant, so their virtual work can be described as a change in energy. Thus the system is a conservative system. Instead of approaching the problem by writing the equilibrium equations, solve it using the minimization of energy. = 6 -E4(A)², where A = the change in length of the Note that the energy for each truss member is member, E is the modulus of elasticity for all members and A is the cross sectional area of all members. All outside members are of length L, and the interior members are determined by the geometry to be 1.61804L. Also, note that the system is symmetrical about the vertical centerline.…Solve fast with suitable diagramanswer all parts

