Explanation Given information: The structure is given in the Figure. The young’s modulus E and area A is constant. Apply the sign conventions for calculating reactions, forces, and moments using the three equations of equilibrium as shown below. For summation of forces along x -direction is equal to zero ( ∑ F x = 0 ) , consider the forces acting towards right side as positive ( → + ) and the forces acting towards left side as negative ( ← − ) . For summation of forces along y -direction is equal to zero ( ∑ F y = 0 ) , consider the upward force as positive ( ↑ + ) and the downward force as negative ( ↓ − ) . For summation of moment about a point is equal to zero ( ∑ M at a point = 0 ) , consider the clockwise moment as negative and the counter clockwise moment as positive. Calculation: Refer the given structure. The structure is symmetric with respect to the vertical s axis passing through the hinge E . Divide the magnitudes of forces and moments of the given loading by 2 to obtain the half loading. Sketch the half loading for the given structure as shown in Figure 1. Draw the reflection of half loading about the specified axis s . Sketch the reflection of half loading as shown in Figure 2. Add the half loading and reflection of half loading to find the symmetric component. Sketch the symmetric loading component as shown in Figure 3. Subtract the symmetric loading component from the given loading to obtain the antisymmetric component. Sketch the antisymmetric loading component as shown in Figure 4. Find the member forces due to symmetric loading component: Sketch the substructure with symmetric boundary conditions as shown in Figure 5. Find the reactions and member end forces of substructure using equilibrium equations and to the left of s axis. The member end forces to the right of s axis are obtained by the reflection. Summation of forces along y -direction is equal to 0. + ↑ ∑ F y = 0 − ( 30 ) ( 8 ) + A y = 0 A y = 240 kN ↑ Summation of moments about E is equal to 0. ∑ M E = 0 − A x × 12 − A y × 5 + 30 × 8 × 4 = 0 − A x × 12 − 240 × 5 + 30 × 8 × 4 = 0 − 12 A x = 240 A x = 20 kN → Summation of forces along x -direction is equal to 0. + → ∑ F x = 0 20 − E x = 0 E x = 20 kN ← Consider frame AD , Summation of forces along x -direction is equal to 0

Find the member end forces of the frame using structural symmetry.

Structural Analysis, SI Edition

6th Edition

ISBN: 9780357030981

Author: Aslam Kassimali

Publisher: Cengage Learning US

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