Ans.

V1

=

V2'

-2,100 pounds,

V2"

=

V3

V4

=

v20

V6

V7

V8

=

+

1,900,

V8

=

V9

V10

V11

=

V12

V13

V14

V15

=

V16

-1,700,

V16

=

V12

V18

=

V19

=

V20

=

+ 1,600.

2. Solve the preceding example, taking into account the weight of the beam, 42 pounds per foot. (The right and left reactions are 3,780 and 4,360 pounds respectively; see example 4, Art. 33.)

Ans.

V0

=

-

2,100

lbs.

V7

=

+

1,966

lbs.

V14

=

-

1,928

lbs.

V1

=

-

2,142

V8

=

+

1,924

V15

=

-

1,970

V2

=

-

2,184

V8

=

-

1,676

V16'

=

-

2,012

V2"

=

+

2,176

V9

=

-

1,718

V16'

=

+

1,768

V3

=

+

2,134

V10

=

-

1,760

V17

=

+

1,726

V4

=

+

2,092

V11

=

-

1,802

V18

=

+

1,684

V5

=

+

2,050

V12

=

-

1,844

VI9

=

+

1,642

V6

=

+

2,008

V13

=

-

1,886

V20

=

+

1,600

3. Compute the values of the shear at sections one foot apart in the beam of Fig. 11, neglecting the weight. (The right and left reactions are 1,444 and 1,556 pounds respectively; see example 1, Art. 33.)

Ans.

V0" =V1=V2'=+ 1,556 pounds,

V2" =V3 ,=V =V 5=V6'=+956,

V6" =V7=+56,

V7 " =V8=V9=V10=V11 =V12 =V13 ' = - 4447

V13"=V14=V15=V16'= -1,444.

4. Compute the vertical shear at sections one foot apart in the beam of Fig. 12, taking into account the weight of the beam, 800 pounds, and a distributed load of 500 pounds per foot. (The right and left reactions are 4,870 and 11,930 pounds respectively; see examples 3 and 4, Art. 33.)

Ans.

V0

=

0

V7

=

+

6,150

lbs.

v15

=

+

830

lbs.

V1'

=

-

540

lbs.

V 8

=

+

5,610

V16

=

+

290

V1'

=

-

2,540

v8"

=

+

4,610

V17

=

-

250

V2

=

-

3,080

v9

=

+

4,070

vI7"

=

-

3,250

V3

=

-

3,620

v10

=

+

3,530

V18

=

-

3,790

V4

=

-

4,160

V11

=

+

2,990

V19

=

-

4,330

V5

=

-

4,700

v12

=

+

2,450

V 20

=

-

4,870

V'6

=

-

5,240

v13

=

+

1,910

V 20

=

0

V6"

=

+

6,690

V14

=

+

1,370

38. Shear Diagrams. The way in which the external shear varies from section to section in a beam can be well represented by means of a diagram called a shear diagram. To construct such a diagram for any loaded beam,

1. Lay off a line equal (by some scale) to the length of the beam, and mark the positions of the supports and the loads. (This is called a "base-line.")

2. Draw a line such that the distance of any point of it from the base equals (by some scale) the shear at the corresponding section of the beam, and so that the line is above the base where the shear is positive, and below it where negative. (This is called a shear line, and the distance from a point of it to the base is called the "ordinate" from the base to the shear line at that point.)

We shall explain these diagrams further by means of illustrative examples.

Examples. 1. It is required to construct the shear diagram for the beam represented in Fig. 13, a (a copy of Fig. 9).

Lay off A'E' (Fig. 13, b) to represent the beam, and mark the positions of the loads B', C and D'. In example 1, Art. 37, we computed the values of the shear at sections one foot apart; hence we lay off ordinates at points on A'E' one foot apart, to represent those shears.

Use a scale of 4,000 pounds to one inch. Since the shear for any section in AB is 2,300 pounds, we draw a line ab parallel to the base 0.575 inch (2,300 4,000) therefrom; this is the shear line for the portion AB. Since the shear for any section in BC equals 1,300 pounds, we draw a line b'c parallel to the base and

External Shear And Bending Moment Part 2 030020

Fig. 13.

0.325 inch (1,300 4,000) therefrom; this is the shear line for the portion BC. Since the shear for any section in CD is -700 pounds, we draw a line c'd below the base and 0.175 inch (700 4,000) therefrom; this is the shear line for the portion CD. Since the shear for any section in DE equals -3,700 lbs., we draw a lined'e below the base and 0.925 inch (3,700 4,000) therefrom; this is the shear line for the portion DE. Fig. 13, b, is the required shear diagram.

2. It is required to construct the shear diagram for the beam of Fig. 14, a (a copy of Fig. 9), taking into account the weight of the beam, 400 pounds.

The values of the shear for sections one foot apart were computed in example 3, Art. 37, so we have only to erect ordinates at the various points on a base line A'E' (Fig. 14, b), equal to those values. We shall use the same scale as in the preceding illustration, 4,000 pounds to an inch. Then the lengths of the ordinates corresponding to the values of the shear (see example 3, Art. 37). are respectively:

2,500 4,000

0.625 inch

2,460 4,000

0.615 "

1,460 4,000

0.365 "

etc.

etc.

Laying these ordinates off from the base (upwards or downwards according as they correspond to positive or negative shears), we get ah, b'c, c'd, and d'e as the shear lines.

External Shear And Bending Moment Part 2 030021

Scale 1"= 4000 lbs,

Fig. 14.

3. It is required to construct the shear diagram for the cantilever beam represented in Fig. 15, a, neglecting the weight of the beam.

The value of the shear for any section in AB is - 500 pounds; for any section in BC, -1,500 pounds; and for any section in CD, - 3,500 pounds. Hence the shear lines are ab, bc, c'd. The scale being 5,000 pounds to an inch,

Aa

=

500-

5,000

=

0.1

inch,

B'b'

=

1,500

5,000

=

0.3

"

C'c'

=

3,500

5,000

=

0.7

"

The shear lines are all below the base because all the values of the shear are negative.

4. Suppose that the cantilever of the preceding illustration sustains also a uniform load of 200 pounds per foot (see Fig. 16, a). Construct a shear diagram.

External Shear And Bending Moment Part 2 030022

Fig. 15.

First, we compute the values of the shear at several sections.

Thus

V0

=

-

500

pounds,

v1

=

-

500

-

200

=

- 700,

v2

=

-

500

-

200

X

2

=

-

900,

v2

=

-

500

-

200

X

2

-

1,000

=

-

1,900,

v3

=

-

500

-

1,000

-

200

X

3

=

-

2,100,

v4

=

-

500

-

1,000

-

200

X

4

=

-

2,300,

v5

=

-

500

-

1,000

-

200

X

5

=

-

2,500,

v5

=

-

500

-

1,000

-

200

X

5

-

2,000

=

4,500,

v6

=

-

500

-

1,000

-

2,000

-

200

X

6

=

4,700,

v7

=

-

500

-

1,000

-

2,000

-

200

X

7

=

4,900,

v8

=

-

500

-

1,000

-

2,000

-

200

X

8

=

5,100,

v9

=

-

500

-

1,000

-

2,000

-

200

X

9

=

5,300.

The values, being negative, should be plotted downward. To a scale of 5,000 pounds to the inch they give the shear lines ah, b'c, c'd (Fig. 16, V).

Examples For Practice

1. Construct a shear diagram for the beam represented in Fig. 10, neglecting the weight of the beam (see example 1, Art. 37).

2. Construct the shear diagram for the beam represented in Fig. 11, neglecting the weight of the beam (see example 3, Art. 37).

3. Construct the shear diagram for the beam of Fig. 12 when it sustains, in addition to the loads represented, its own weight, 800 pounds, and a uniform load of 500 pounds per foot (see example 4, Art. 37).

4. Figs. a, cases 1 and 2, Table B (page 55), represent two cantilever beams, the first bearing a concentrated load P at the free end, and the second a uniform load W. Figs. b are the corresponding shear diagrams. Take P and W equal to 1,000 pounds, and satisfy yourself that the diagrams are correct.

5. Figs. a, cases 3 and 4, same table, represent simple beams supported at their ends, the first bearing a concentrated load P at the middle, and the second a uniform load W. Figs. b are the corresponding shear diagrams. Take P and W equal to 1,000 pounds, and satisfy yourself that they are correct.

External Shear And Bending Moment Part 2 030023

Fig. 16.