Your Question:

Q.1 A simply supported beam ABCDEF as shown in Figure Q1 is subject to three sets of loads, including two point loads at A and C respectively, and a uniformly distributed load (UDL) on segment DEF

(a) Find the reaction forces at supports B and E.

(6 marks)

(d) Sketch the deflected shape of the beam

Step By Step Answers with Explanation

ΣM_B = 0

where:

R_B = 80 + 50 - 12 * 4 = 104 kN

Moments about B:

(b) Drawing the shear force diagram

To draw the shear force diagram, we need to consider the vertical forces acting on the beam at different sections.

Section 3 (C to D)

In addition to the point load at C, the uniformly distributed load also acts on this section. The uniformly distributed load is equal to 12 kN/m, and it acts over a distance of 4 m. Therefore, the total downward force acting on this section is equal to 50 kN + 12 kN/m * 4 m = 108 kN. The reaction force at E also acts on this section, but it is an upward force. Therefore, the shear force in this section is equal to 108 kN - 68 kN = 40 kN.

The shear force diagram is shown below:

The point with zero shear force is located at a distance of 4 m from support C.

Section 2 (B to C)

In addition to the moment due to the reaction force at B, the moment due to the point load at C also acts on this section. The moment due to the point load at C is equal to 50 kN * (x - 4 m). Therefore, the bending moment in this section is equal to 104 kN * x - 50 kN * (x - 4 m).