4_2 - Strength of Materials Prof. M. S. Sivakumar Problem...

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Strength of Materials Prof. M. S. Sivakumar Indian Institute of Technology Madras Problem 1: Computation of Reactions Problem 2: Computation of Reactions Problem 3: Computation of Reactions Problem 4: Computation of forces and moments Problem 5: Bending Moment and Shear force Problem 6: Bending Moment Diagram Problem 7: Bending Moment and Shear force Problem 8: Bending Moment and Shear force Problem 9: Bending Moment and Shear force Problem 10: Bending Moment and Shear force Problem 11: Beams of Composite Cross Section
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Strength of Materials Prof. M. S. Sivakumar Indian Institute of Technology Madras Problem 1: Computation of Reactions Find the reactions at the supports for a simple beam as shown in the diagram. Weight of the beam is negligible. Figure: Concepts involved Static Equilibrium equations Procedure Step 1: Draw the free body diagram for the beam. Step 2: Apply equilibrium equations In X direction F X = 0 R AX = 0 In Y Direction FY = 0
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Strength of Materials Prof. M. S. Sivakumar Indian Institute of Technology Madras R AY +R BY – 100 –160 = 0 R AY +R BY = 260 Moment about Z axis (Taking moment about axis pasing through A) M Z = 0 We get, M A = 0 0 + 250 N.m + 100*0.3 N.m + 120*0.4 N.m - R BY *0.5 N.m = 0 R BY = 656 N (Upward) Substituting in Eq 5.1 we get MB = 0 R AY * 0.5 + 250 - 100 * 0.2 – 120 * 0.1 = 0 R AY = -436 (downwards) TOP
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Strength of Materials Prof. M. S. Sivakumar Indian Institute of Technology Madras Problem 2: Computation of Reactions Find the reactions for the partially loaded beam with a uniformly varying load shown in Figure. Neglect the weight of the beam Figure (a) (b) Procedure: The reactions and applied loads are shown in figure (b). A crude outline of the beam is also shown to indicate that the configuration of the member is not important for finding out the reactions. The resultant force P acting though the centroid of the distributed forces is found out. Once a free body diagram is prepared, the solution is found out by applying the equations of static equilibrium. F x = 0 R Ax = 0 M A = 0 Anticlockwise +12 X 2 - R By X 6 = 0
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Strength of Materials Prof. M. S. Sivakumar Indian Institute of Technology Madras R By = 4 kN downwards M B = 12 * (6 - 2) - R A y * 6 => R A y = 8 kN Check F y = 0 8 + 12 − 4 = 0 ok! TOP
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Strength of Materials Prof. M. S. Sivakumar Indian Institute of Technology Madras Problem 3: Computation of Reactions Determine the reactions at A and B for the beam shown due to the applied force. Figure Solution At A, the reaction components is x and y directions are R Ax and R Ay . The reaction R B acting at B and inclined force F can be resolved into two components along x and y directions. This will simplify the problem. Calculation: F y = 12, F x = 9; (By resolving the applied force) M A = 0 12 X 3 – R By X 9 = 0 R By = 4 kN = R B x M B = 0 12 X 6 - R Ay X 9 = 0 R Ay = 8kN F x = 0 R Ax – 9–4 = 0 R Ax = 13kN
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Strength of Materials Prof. M. S. Sivakumar Indian Institute of Technology Madras So, () 22 A B R = 8 + 13 = 233 kN R = 4 + 4 = 4 2 kN Check: F y = 0 + 8 – 12 + 4= 0 ok!
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4_2 - Strength of Materials Prof. M. S. Sivakumar Problem...

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