A 5 kN/m 5 m ▬▬▬▬ C ↓ EZZJ B 7.5 m 2.5 m The simply-supported beam shown has a total length of 10 m with a pin support at A and roller support at C. The beam is subject to a distributed load of 5 kN/m over its entire length in addition to a concentrated load of 40 kN located 2.5 m from B. a) Draw the free body diagram (FBD) of the beam showing all external forces (including reactions). b) Solve for the reactions at A and B and state their directions. c) If we make a cut at location C (midpoint of the beam), show the positive orientations of all internal forces at C on both halves of the beam.

In the system shown in the figure, parts AB, BC and CD are of equal length and 4 m. Clockwise from point AA moment of 10 kNm is applied in the opposite direction. The distributed load of 3 kN/m at BC is upward. Beam DA load of 5 kN was applied downwards from the point. The profile of the beam has a 100 mm edge as seen on the right side.was formed by subtracting a circle with a diameter of 80 mm from the center of a regular hexagon of length. Accordingly; A)Calculate the reaction forces at B and C B)Draw shear force and bending moment diagrams C)Calculate the maximum bending and show it on the bending moment diagram. D)Calculate the sectional moment of inertia of the profile E)Calculate the maximum stress in the beam NOTE:For option C, you need to obtain a parobolic expression when drawing the bending moment diagram. You need to use derivation.

Problem 1 Complete 5.117 from the course text. 5/117 Determine the reactions at A and B for the beam subjected to the distributed and concentrated loads. 2 kN/m Hint: w | w = w₁ + kx² 0.6 m A 1.4 m 6 kN/m Problem 5/117 Tapvideo B 0.4 m 0.6 m 2 4 kN 0.5 m *. First, turn the distributed load into a point force applied at the area centroid. Next, arefully draw the FBD and apply equilibrium equations.

A simply supported beam supports the applied loads as shown. If E = 200 GPa and I = 0.0015 m4. Using double integration method, Determine the maximum delection along span AB only.