Meriam, Kraige & Bolton — Equilibrium of Rigid Bodies: Problem 4_58

⚡ Mecademy AIENG정역학 · ch4  Problem Statement Determine the force in member of the regular pyramid with square base. The height of the pyramid is and the side length of the square base is . A vertical downward force of is applied at vertex , and a horizontal force is applied at joint parallel to the side . The joints , and are supported as shown. Problem 4/58 (a) Analysis of the regular pyramid truss 1. Formula: For a space truss in equilibrium, we apply the method of joints and global equilibrium of the structure. 2. Substitution and Geometric Setup: Set up a coordinate system with the origin at the center of the base . Apex Joint Joint Joint Joint Applied forces: At vertex : At joint : (parallel to ) Reactions at provide necessary support. From global moment equilibrium about the and axes through the center, we find the vertical reactions: ? Actually, using as reference for moments: . BD 6a5a2L ALDEB B,CE F= ∑ 0,M= ∑0 (0,0,0) A=(0,0,6a) B=(2.5a,−2.5a,0) C=(2.5a,2.5a,0) D=(−2.5a,2.5a,0) E=(−2.5a,−2.5a,0) AF = A −2Lk DF = D LiEB B,C,E xy M =∑ x 0⇒−2L(0)+R (2.5a)− Cz R (2.5a)− Bz R (2.5a)= Ez 0ER = Cz L,R = Bz 0,R = Ez L 3. Calculation: Step 1: Joint and leg forces. At joint , the only member with a -component is . Since there is no -load at , we have: At joint , by symmetry and from equilibrium equations in and : The horizontal projection of the force in member on joint has components: Since is compression, it pushes joint in the direction from to , which is . Step 2: Joint e