**Quick and Easy way for solving Bending Moment and Shear Force problems**

A quick and easy way to solve complicated problems on bending moment and shear force is now revealed. Today we will explain how can you apply online calculators to solve the problems of shear force and bending moment in an incredibly easy way. Let us take the example given below and use online bending moment calculator to determine the values of bending moment and shear force and then draw the diagrams.

**Procedure**

The given beam has two load cases (i) point load and (ii) uniform load. We will solve this problem by applying principle of superposition.

Open **http://civilengineer.webinfolist.com/mech/bmcalc.htm ** for Bending Moment and Shear Force Calculator and follow the steps given below.

**Step 1.** Select the following online calculators and open in two different tabs of the browser.

**Step 2.** Enter the required values of load and distances as per the given problem in the form for input values and calculate the values of shear force and bending moment at C and D by putting x equal to 4 and 8 respectively and also on both the supports A (x=0) and B (x=10). For using appropriate sign convention please visit our page of instructions for calculator.

**a. Point load of 10 kN acting downward**

Enter the values of beam span, load and the position of load as per the diagram shown and enter the position of section where you want to calculate the values of shear force and bending moment and then click "Calculate". You will get the result including support reactions, shear force (Fx) on both sides of the section (left and right) and bending moment Mx at the section as shown in the figure given below.

The calculator also gives the values of maximum bending moment and its location but for this problem we will not use these values. Note down the values of support reaction due to the load case of 10 kN. Also note down the values of Fx and Mx for different points A, B, C and D just by changing the values of x. Enter all the values in the Results table given below.

**b. Uniform load of 20 kN/m acting downward**

In the similar way enter the required input values in the form and then click "Calculate". Here also we will note down the values of Reaction at A, Reaction at B and then the values of Fx and Mx for different points A, B, C and D just by changing the values of x as shown in the figure given below. We will not use the value of maximum bending moment by this load case.

Enter the values obtained in the Results table given above and apply principle of superposition to get the resultant value.

**Plotting the diagrams for Bending Moment and Shear Force**

Once you get the resultant values plot the diagrams for shear force and bending moment as given below by using appropriate sign conventions. To plot shear force diagram take value of shear force on y-axis and distance on x-axis and then join these points to get shear force diagram (SFD). In case of point loads you will get a straight line parallel to x-axis showing that shear force is constant as shown in SFD from C to D and from D to B. You can also notice a stepped change in the values of shear force wherever there is a point load as shown in the diagram at D. It shows two values of shear force (one on the left side and one on the right side). In case of UDL you will get a sloping line showing an increase or decrease in the values of shear force as shown in the diagram between A and C.

Similarly for plotting bending moment diagram (BMD) the values of bending moment should be taken on y-axis and distance on x-axis. The values of bending moment should be joined by straight line if there are point loads as seen in BMD from C to D and from D to B. They should be joined by parabolic curve if there is UDL between two points as shown in BMD from A to C. To get more information visit nature of curve for bending moment diagram.

**Maximum Bending moment**

The value of maximum bending moment occurs at the point of zero shear force. From the shear force diagram it is clear that point of zero shear force will occur between A and B. By using the properties of similar triangles we can easily get the point of zero S.F.

66/x = 14/(4-x);

hence x = 3.3 m from support A.

Now we can easily find out the value of maximum bending moment by putting x = 3.3 in both the calculators.

By Point load the value of B.M = 6.6 kNm

By Uniform load the value of B.M = 102.3 kNm

Therefore resultant bending moment is 108.9 kNm at x = 3.3 which is maximum for this beam.

The results obtained by this online calculator are same as Solved Example 5.2

We hope this will help you in improving your learning about shear force and bending moment.

You can also try other load combinations by using bending moment calculators or visit our solved examples given below for more practice on shear force and bending moment calculation and plotting of diagram.

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