Beam: Shear and Moment Diagrams

A beam is supported as shown in the figure; it is intended to resist concentrated vertical load F

1

,

located L

1

from the left end, and distributed vertical load F

d2

, which acts over a length L

2

from the right

end of the beam. The beam is sectioned into units with endpoints A, B, C, and D. The assignment in

this problem is to draw the Shear and Moment diagrams for this beam. The X direction is along the

beam.

F

1

F

d2

L

L

1

L

2

A

B

C

D

The first step is to draw a free-body diagram (FBD) of the overall beam to calculate the reaction

forces at A and D. To determine the reaction forces, the distributed vertical load may be replaced with a

concentrated vertical load F

2

, which acts at a length L

2

/2 from the right end of the beam.

F

1

F

2

L

L

1

L /2

2

A

x

A

y

D

y

We write three equations of static equilibrium:

2/0

0

0

2211

21

LLFLFLDM

FFDAF

AF

yAz

yyy

xx

The reaction force solutions are:

yy

x

y

DFFA

A

L

LLFLF

D

21

2211

0

2/

To determine the shear forces in each subsection of the beam, free-body diagrams of each subsection are

drawn, representing the forces on the left and right sides. For this problem, the shear forces are:

xFFAxv

FAxv

Axv

dyCD

yBC

yAB

21

1

The shear forces in the first two subsections AB and BC are constant. The shear forces in CD

starts from the BC value and decreases linearly with slope F

d2

, ending at D

y

. The end reaction +D

y

then closes the shear diagram back to zero. Note v

CD

above assumes that x starts at zero; hence the axes

must be shifted horizontally to fit in with the overall shear diagram.

In order to determine the moment diagram given the shear diagram, one integrates the shear

functions in each subsection. The difference between the moment subsection endpoint values in any

case equals the area under the shear curve in that subsection.

2/

2

21

1

xFFAmxm

xFAmxm

xAmxm

dyCCD

yBBC

yAAB

Again in the moment equations, each x range starts at zero at the left of the subsection; hence the

axes must be shifted horizontally to fit in with the overall moment diagram. Note, due to the beam

support conditions in this problem, the moment must be zero at the left and right ends of the beam. In

the moment equations, the constants of integration are:

211

1

0

LLLFAmm

LAm

m

yBC

yB

A

User sets:

F

1

and F

d2

(N), plus L

1

and L

2

(m):

2/0

2/0

000,20

000,100

2

1

2

1

LL

LL

F

F

d

Computer sets:

L = 10 m

Visualize:

Beam with loading conditions, Shear and moment diagrams

Numerical Display:

Shear forces v

A

, v

B

, v

C

, and v

D

, moment magnitudes m

A

, m

B

, m

C

, and m

D

, plus

reaction forces A

x

, A

y

, and D

y

.

User Feels:

Any of the above shear forces, moment magnitudes, or reaction forces ( Y force

only with Joystick user feels appropriate force or moment magnitude as they move along shear/moment

diagram).

Examples:

There are two distinct shear/moment diagrams, depending on whether 0

1

FA

y

or

0

1

FA

y

; the above shear and moment equations apply equally to both cases, but the resulting

diagrams have different characteristics.

Example 1: 0

1

FA

y

When the user enters F

1

= 5,000, L

1

= 2.5, F

d2

= 1,000, and L

2

= 3 (N and m), the results are:

v

A

= 4,200, v

B

= -800, v

C

= -800, and v

D

= -3,800 (N);

m

A

= 0, m

B

= 10,500, m

C

= 6,900, and m

D

= 0 (Nm);

A

x

= 0, A

y

= 4,200, and D

y

= 3,800

The associated shear/moment diagrams ( N, Nm) for Example 1 are:

0

1

2

3

4

5

6

7

8

9

10

-2000

0

2000

4000

X

Shear

0

1

2

3

4

5

6

7

8

9

10

0

5000

10000

15000

X

Moment

Example 2: 0

1

FA

y

When the user enters F

1

= 1,000, L

1

= 2, F

d2

= 2,000, and L

2

= 4.5 (N and m), the results are:

v

A

= 2,825, v

B

= 1,825, v

C

= 1,825, and v

D

= -7,175 (N);

m

A

= 0, m

B

= 5,650, m

C

= 12,038, and m

D

= 0 (Nm);

A

x

= 0, A

y

= 2,825, and D

y

= 7,175

The associated shear/moment diagrams ( N, Nm) for Example 2 are:

0

1

2

3

4

5

6

7

8

9

10

-6000

-4000

-2000

0

2000

X

Shear

0

1

2

3

4

5

6

7

8

9

10

0

5000

10000

15000

X

Moment

In Example 1, the constant shear

1

FA

y

in BC is negative, while in Example 2, the constant

shear

1

FA

y

in BC is positive. Thus, in Example 1, the linearly-changing moment in BC has negative

slope, while in Example 2, this slope is positive. Further, though the linearly-changing shear in CD has

negative slope F

d2

in both examples, the moment parabola in CD is strictly decreasing to zero in

Example 1, while this parabola increases before decreasing to zero moment in Example 2. Note further

that at point B, there is a discontinuity in shear, which translates to a change in moment slope at point B,

but there is no discontinuity in shear at point C, hence moment slopes match at point C; this is true in

both examples.

Comprehension Assignment:

Once you get the feel for this simulation, run the program several times to collect and plot data:

for a constant applied distributed load F

d2

= 1000 (N/m) and L

2

= 5 m, and a fixed value of L

1

, vary

applied load F

1

over its allowable range and determine the resulting maximum shear v

MAX

, maximum

moment m

MAX

, plus reaction forces A

x

, A

y

, and B

x

. Plot v

MAX

, m

MAX

, A

x

, A

y

, and B

x

vs. F

1

. Repeat these

plots for various values of L

1

over its allowable range. Discuss the trends you see do the results make

sense physically?

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