Shear force and bending moment diagrams

[Stacyg]

Hi. The question is:
Draw the shear force and bending moment diagrams for the simply supported beam shown below. State the values of the greatest shear force and bending moment, and the bending moment at D.
The diagram is attached as is my attempt at the answer. I hvae not stated the values of greatest shear force, bending moment or shown the bending moment at D as I am not sure how to do this.
Also with my Bending moment diagram I am not sure if I have started it right.
Any help would be great thanks.

 

[uradnky]

It's been awhile since I've done this but hopefully I can help a little.

I can't read your SFD too well, but to solve for the bending moment at D..

Take moments from point A

Ma = 0 kn . mm
Mb = 0 + (Length x Width on your SFD)

The bending moment at D will be Mc + (Bound area from C-D)

Labeling the points on your SFD should help you out with this.

 

[Mene]

Hi there,

Shear force and bending moment diagrams are graphical representations of the internal forces and moments acting on a beam. They are important tools used in structural analysis and design.

To draw the shear force and bending moment diagrams for the simply supported beam shown, we first need to determine the reactions at the supports. Since the beam is simply supported, the reactions at each support will be equal and opposite.

Let's label the reactions as RA and RB. By taking moments about point A, we can determine that RA = 10 kN and RB = 20 kN. These reactions will be shown on our diagrams as point loads acting upwards at A and B.

Next, we can draw the shear force diagram by starting at the left end of the beam and moving to the right. At point A, the shear force is equal to RA = 10 kN, so we draw a horizontal line at this value. As we move towards the middle of the beam, the shear force decreases linearly until we reach point B, where it becomes zero. This is because there are no external forces acting on the beam between A and B.

At point B, the shear force suddenly increases to RB = 20 kN and remains constant until the end of the beam. This is because the reaction at B is acting upwards to counteract the downward force of the beam's own weight.

Next, we can draw the bending moment diagram by starting at the left end of the beam and moving to the right. At point A, the bending moment is zero. As we move towards the middle of the beam, the bending moment increases parabolically until we reach point B, where it is equal to the product of the reaction RB and the distance from B to the end of the beam.

At point B, the bending moment suddenly decreases to zero and then remains constant until the end of the beam. This is because the reaction at B creates an equal and opposite moment to counteract the bending moment caused by the weight of the beam.

The greatest shear force occurs at point B, where it is equal to RB = 20 kN. The greatest bending moment also occurs at point B, where it is equal to RB multiplied by the length of the beam.

To find the bending moment at point D, we can use the equation M = Wx, where W is the force acting on the beam and x is the distance from the point of interest. In this case, the force acting on the