This section is from the book "Cyclopedia Of Architecture, Carpentry, And Building", by James C. et al. Also available from Amazon: Cyclopedia Of Architecture, Carpentry And Building.

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

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.

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.

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).

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.

Fig. 16.

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