ben-ouezdou-step by step shear and moment

7/25/2019 Ben-Ouezdou-Step by Step Shear and Moment

1/181

A STEP BY ATEP SHEAR AND MOMENT DIGRAM

Example 1- Draw the shear and bending moment diagrams of a beam by writing the equations.

Solution.

Step 1: Get support reactions (Ay and Dy)

+ MD=0

Ay (12)15 (6) (9) + 15050 (3) = 0.

Ay = 67.5 kN.

+ Fy =0.

Ay15 (6)50 + Dy = 0.

Dy = 72.5 kN.

Step 2: Define the regions of loading change (or support change). Choosing to start from left to

right or the opposite direction make it easier for the solution but either ways will give the same

end results.

Region 1: 0 x < 6 m.between A and B.

Region 2: 6 x < 9 m.between B and C.

Region 3: 9 x < 12 m.between C and D.

Step 3 to 5:For each region, draw the Free Body Diagram of a section inside the region (not at

the border), i.e. x between the limits of the region a not at the edge of the region.

Then find the shear (V) and the moment (M) as a function of the abscissa x.

Take some points to find V and M, as to allow the drawing of the V and M diagram.

Step 3:Region 1: 0 x < 6 m.

Shear:

+ Fy = 0.

67.515 (x)V = 0

V =15 x + 67.5 linear

@ x = 0 V = 67.5 kN.

@ x = 6 m V =22.5 kN.

Ay Dy

S

67.5 kN

x V

M

15 kN/m


2/182

6 m

150 kN.m

As the shear passes from a > 0 value to a negative value, it goes through 0. Thus we have to

determine the abscissa, where the V = 0. At this x value, the moment will be maximum (or

minimum).

Mmaxwhen V= 0.

V =15 x + 67.5 = 0. @ x0= 4.5 m.

Moment

+ MS= 0

67.5 (x)15 (x) (x/2)M = 0.

M =7.5 x2 + 67.5 x Parabolic

@ x = 0 M = 0.

@ x = 6 m M = 135 kN. m.

@ x = 4.5 m M = 151.9 kN.m. This is Mmaxfor this region only.

Step 4:Region 2: 6 x < 9 m.

Shear:

+ Fy = 0.

67.515 (6)V = 0

V = - 22.5 kN.m Constant.

Moment

+ MS= 0

67.5 (x)15 (6) (x3) + 150M = 0.

M =22.5 x + 420. linear

@ x = 6 m M = 285 kN.m.

@ x = 9 m M = 217.5 kN. m.

S

67.5 kN

xV

M

15 kN/m

S

67.5 kN

x V

M

15 kN/m


3/183

6 m

150 kN.m

50 kN

Step 5:Region 3: 9 x 12 m.

Shear:

+ Fy = 0.

C67.515 (6)50V = 0

V =72.5 kN.m Constant.

Moment

+ MS= 0

67.5 (x)15 (6) (x3) + 15050 (x9)M = 0.

M = 72.5 x+ 870. linear

@ x = 9 m M = 217.5 kN.m.

@ x = 12 m M = 0.

Note that at both ends (when there is no applied couple) and with the condition of either a hinge

or a roller, the moment is zero (0).

Step 6:Shear and moment bending diagram.

Draw the shear forces and the bending moment diagram according to the results found above,

under the figure of the beam.

3 m S

67.5 kN

x

V

M

15 kN/m


4/184

.

Figure 3

4.5 m

+

+

+

+

-


5/185

Example 2a- Draw the shear and bending moment diagrams of a the cantilever beam by writing

the equations.

Method 1: Start from the right (free end).

By starting from the right there is no need to look for the reactions et the left support, unless

you want to confirm the left end results. So we start straight forward by step 2:

Step 2: Define the regions. The regions change each time the load changes. Here we have 2

regions:

Region 1: 0 x < 0.8 m C B

Region 2: 0.8 m x 2 m B A

Step 3 to step 4:

For each step:

i. Make the cut between C and B;

ii. Draw the FBD between 0 and the cut

iii. Find the shear by taking + Fy = 0.

iv. Find the moment by taking + MS= 0

A B C


6/186

0.8 mX0.8

0.8 mX0.8

Step 3: Region 1: 0 x < 0.8 m C B

Shear.

+ Fy = 0.

V600 = 0.

V = 600 N Constant.Moment

+ MS= 0

- M600 (x)100 = 0

M = - 600 (x) - 100. Linear.

So we need two points to later draw the moment diagram (limits).

@ x = 0 M = 0.@ x = 0.8 m M = - 580 N.m

Step 4: Region 2: 0.8 x < 2 m B A

Shear.

+ Fy = 0.

V400 ( x - 0.8 ) - 600 = 0.V400 ( x ) + 400 x 0.8 - 600 = 0.

V = 400 (x) + 280. Linear.

So we need two points to later draw the shear diagram (limits).

@ x = 0.8 m V = 600 N.

@ x = 2 m V = 1 080 N

Moment

+ MS= 0

- M400 (x0.8) (0.5) ( x0.8 )600 (x)100 = 0.

.- M - 200 ( x0.8 )2600 x100 = 0.

- M -200 (x22 (x) (0.8) + (0.8)

2)600 x100 = 0.

. M200 x2+ 320 (x)128600 x100.

100 Nm

100 NmSx

VM600 N

S

x

VM600 N

C

C

400 N/m

B

100 NmS

x

VM600 N

C

400 N/m

B


7/187

- M200 x2280 (x)228 = 0

M = - 200 x2280 (x)228. Parabola

So we need two points to later draw the diagram (limits).

@ x = 0.8 m M = - 128 -224 -228 = - 580 N.m.

@ x = 2 m M = - 800560228 = 1 588 N.m

To know if the parabola is concave up or down, lets check the sign of the coefficient before x

2

.

Here the coef is < 0 so the parabola is concave down.

V (N)

600+1080

-1588

- 100

M (N.m

- 580

-


8/188

Method 2: Start from the left (fixed end).


9/189

Example 2b- Draw the shear and bending moment diagrams of a the cantilever beam by

graphical method (using the relationship between Load, shear and Moment).

Method 1: Start from the right (free end).

By starting from the right we do not need to find the reactions, unless if you want to check the

shear and Moment at the left end.

1) Shear:

We us the following sign convention for the shear:

From C to B: To start from the right, we begin with the 600 N as a concentrated force. According

to the sign convention (using the right part), the force is down and thus positive. For the shear,

we ignore the applied couple (100 Nm).

For the concentrated load, the change in shear:

V = F (here it is 600 N).

A B C

+

+

Shear, V

Starting from the rightStarting from the left

A B C

V (N)

600


10/1810

Now we use the following table to know the shape of the shear and moment under different

type of loading.

Table 1: Shape of the shear and Moment according to the loading.

Loading Shear Moment

Concentrated (point) Load

Constant Linear

Uniformly Distributed Load

Linear Parabola

Triangular Loading

Parabola Cubic curve

From this table, we can see that for the concentrated Load, the shear is constant.(Add an

horizontal line in the diagram, up to B, where the load changes).

From B to A.

From table 1, you can see that under the distributed load the shear is linear. For this case the

shear is related to the loading by the following relation:

= = Area of the loading

Area of the loading = wL = 400 (1.2) = 480 N.

The new shear at the A is:

VA=VB+ V = 600 + 480 = 1080 N.

So the rest of the diagram is a linear line from 600 to 1080.

+

A B C

V (

600


11/1811

2) Moment

We us the following sign convention for the moment:

Again, we start from the right. Now the applied

external moment (couple) is considered.

The 600 N force would give nothing at the

beginning.

From the above convention we see that

the 100 N.m applied moment gives a negative moment.

+

A B C

1080 +

V (N)

600

+ +

Moment, M


M (

+

A B C

1080

+

V

600

A1

0.8 m

100


12/1812

At B, the jump in the Moment is the area of the shear.

= = = A1= 600 (0.8) = 480 N.m

At B the moment effect of the 100 N load is turning

the Beam clockwise.

According to the moment sign convention, it gives a

negative moment. Thus we should take(480 N.m)

MB= - 100480 = - 580 N.m

Between C and B, the load is a concentrated 100 N

point load, so according to the Table 1,

the diagram is linear. From100 to580 N.m

To find the Moment value at A, again we look for the moment jump.

= = = A2= 600 (1.2)+ (1080-600) (1.2) = 1008 N.m

Hence the moment at A is: MA= MB-768 = - 5801008 = -1588 N.m

Now, according to table 1, the shape of the line between moments at B and A is a parabola

(since the load is uniformly distributed). But the question is it concave up or down? To answer

this question, we use the following figure (Fig. 2) (figure worked out when reading the shear

from left to right, here from A to B).

-580

+

A B C

1080

+

V (

600

A1

-

M (N

-100

+ve Shear

Shear Small to Big

Shear Small to Big Shear Big to Small

Fig. 2: Shape of the Moment Diagram

- ve Shear

Shear Big to Small

A2


13/1813

Here shear is positive, thus we use the right part of

the figure. In this example and by looking at the

shear diagram, we can see that that shear goes

from Big to small (from 1080 to 600),

thus the moment curve is concave down.

Do not forget to put the + orsign inside the V and M diagram

PS: During the Quiz and Final Exam, I will provide you with the sign convention (for V and M in

red boxes) and the Fig. 2. You should know the table 1.

-580

-100

+

A B C

1080

+

V (N

600

A1

-

M N

-15

88

-

A2


14/1814

A B C

Method 2: Start from the left (fixed end).

We are going to take the sample example but this time, we are going to solve it by starting from

left. We are going to see that we reach the same answer.

But, this time we need to find the reactions at first (Ay and MA).

The free body diagram is:

So lets find the reactions:

+ Fy = 0

Ay400 (1.2)600 = 0

Solve for Ay = 1080 N.

Moment MA?

+ M/A= 0

MA400 (1.2) (1.2/2)600 (2)100.

Solve for MA = 1588 N.m

A B C

Ay

MA


15/1815

1) Shear:

We us the following same convention for the shear, but this time we use the left part only:

From A to B: To start from the left, we begin with the

reaction Ay = 1080 N as a concentrated force.

According to the sign convention (using the right part),

the reaction Ay is up and thus positive.

For the shear, we ignore the moment reaction MA.

VA= Ay = 1080 N.

We continue by finding the shear at B, VB.

The jump of the shear at B is the area of the uniformly distributed loading w =400 N/m on the

1.2 m length.

Area A0= 400 (1.2) = 480 N.

According to the sign convention of the shear, this loading

is down and thus opposite to the + and should be -.

So the shear at B is:

VB= 1080480 = 600 N.

+

+

Shear, V

Starting from the right

Starting from the left

A B C

V (N)

1080

A B C

V

1080

600


16/1816

In between A and B, the shear is linear (see table 1)

and thus we complete the line between VAand VB

by a linear (straight) line.

From B to C:

To reach the C, we note that since that there is no

loading the shear stays constant (since jump in shear = 0).

Thus the shear at C is VC= 600 N.

We end up with the drawing of an horizontal line

between B and C for the shear diagram.

2) Moment:

We us the same sign convention for the moment, but taking the left part only (blue):

A B C

V (N)

1080

600+

A B C

V (N)

1080600+

+

+ +

Moment, M



17/1817

From A to B: To start from the left, we begin with the

reaction MA, as a concentrated moment.

As this moment turns counterclockwise,

its value is negative, MA= -1588 N.m

Now to find the value of the moment at B,

we consider the Area A2of the shear from A to B.

this area is the jump of the moment.

= = = A2

A2= 600 (1.2)+ (1080-600) (1.2) = 1008 N.m

Hence the moment at B is: MB= MA+1008 = - 580 N.m

To complete the junction between A and B,

we use again Table 1.

Because the load is uniformly distributed,

The moment diagram would be a Parabola.

But to know if it is concave up or down, we use the figure 2.

0.8 m

M

+

A B C

1080

+

V

600

-1

588

A2

+ve Shear

Shear Small to Big

Shear Small to Big Shear Big to Small

Fig. 2: Shape of the Moment Diagram

- ve Shear

Shear Big to Small


18/18

At first, we notice that the shear is positive,

so we use the right part. Then, because the

shear is from big to small, we use the red curve,

that is concave down.

Now to find the value of the moment at right end,

we look for the Area A1under the shear from B to C.

Area A1= 600 (0.8) = 480 N.m

Thus the jump in the moment M = A1= 480 N.m

The moment at C, MCis then:

MC= - 580 + 480 = -100 N.m

Finally we notice that the junction between B and C is

linear according to table 1.

End of the example

-580

M (N.m

+

A B C

1080

+

V (N)

600

-

1588

A2

-

-580

M N

+

A B C

1080

+

V (N

600

A1

-1588

A2

- -100

-

0.8 m

ben-ouezdou-step by step shear and moment

Documents