q(x)= q0[1-c(x/L)] is the distributed load in the truss system given in the figure. It is a built-in support on both supports. a) Calculate the elastic curve equation from the distributed load. b) Find the support reactions. (q0= 6, c=1, L=1) (fixed on both sides)

the 2nd pic is the answer key

2.69 The bars AB, AC, and AD are pinned together as shown in the figure. Hori- zontal movement of the pin at A is prevented by the rigid horizontal strut AE. Cal- culate the axial force in the strut caused by the 10-kip load. For each steel bar, A = 0.3 in.? and E = 29 × 106 psi. For the aluminum bar, A = 0.6 in.² and E = 10 × 106 psi.

1) The truss (containing 12 pin joints, A, B, C, ... L) shown in the figure is simply supported at points H and L that are 4 meters apart. Angles A between different truss members are either 90° or 45° and the member lengths are either 1 m or m. H a) Compute the support reactions when P = 4 kN b) Draw a free body diagram from which internal forces in AB, EJ and IJ can be obtained. c) From the free body diagram of part b obtain forces in members AB, EJ and IJ (indicate if the internal forces are tension or compression) d) List all zero force members that carry no load. (a) (b) 3 (c) 3 (d) 2 RH = FAB = RL = FEJ = Fj =

2KH Goo Support reaction at point B=.

Determine/derive the Shear and Moment Equations of each segment of the beam (kindly include the FBD). Compute the Reactions and Draw the detailed Shear and Moment Diagram indicating values.

Compound axial member ABC has a uniform diameter of d = 1.2 in. Segment (1) is an aluminum [E₁ = 10,000 ksi] alloy and segment (2) is a copper [E₂ = 17,000 ksi] alloy. The lengths of segments (1) and (2) are L₁ = 80 in. and L2 = 140 in., respectively. Determine the force P required to stretch compound member ABC by a total of 0.10 in. LI L2 A d Aluminum Copper Answer: P = i B kips

Member AB has a uniform rectangular cross section of 10 x 24 mm. For the loading Pshown,a) Find reactions at A and B.b) Determine the normal and shearing stresses at Point K and Point H. α (˚) =30      P (kN)=14

Solve with Free body diagram

BJ in kN JC in kN CK in kN JK in kN CD in kN Calculate the force in member