Reactions Of Supports Continued 030017

Fig. 10.

R1 x 10 - 1,000 X 9 - 2,000 X 4 - 3,000 X 2 = 0. The first equation reduces to

10 R2 = 1,000+12,000 + 24,000 = 37,000; or R2= 3,700 pounds. The second equation reduces to

10 R1 = 9,000+8,000+6,000 = 23,000; or R1= 2,300 pounds.

The sum of the loads is 6,000 pounds and the sum of the reactions is the same; hence the computation is correct.

2. Fig. 10 represents a beam supported at B and D (that is, it has overhanmncr ends) and sustaining three loads as shown. "We wish to determine the reactions due to the loads.

Let R1 and R2 denote the reactions as shown; then the moment equations are: For origin at B,

-2,100x2+0+3,600x6-R2X14+1,600x18 = 0. For origin at D,

-2,100x16 + x14 - 3,600x8 + 0 + 1,600x4 = 0. The first equation reduces to

14 R2 = - 4,200+21,600 + 28,800 = 46,200; or R2 = 3,300 pounds. The second equation reduces to

14 R1 = 33,600+28,800-6,400 = 56,000; or R1= 4,000 pounds.

The sum of the loads equals 7,300 pounds and the sum of the reactions is the same; hence the computation checks.

3. What are the total reactions in example 1 if the beam weighs 400 pounds?

(1.) Since we already know the reactions due to the loads (2,300 and 3,700 pounds at the left and right ends respectively (see illustration 1 above), we need only to compute the reactions due to the weight of the beam and add. Evidently the reactions due to the weight equal 200 pounds each; hence the left reaction = 2,300 + 200 = 2,500 pounds, and the right " =3,700+200=3,900 " .

(2.) Or, we might compute the reactions due to the loads and weight of the beam together and directly. In figuring the moment due to the weight of the beam, we imagine the weight as concentrated at the middle of the beam; then its moments with respect to the left and right supports are (400 X 5) and-(400 X 5) respectively. The moment equations for origins at A and E are like those of illustration 1 except that they contain one more term, the moment due to the weight; thus they are respectively:

1,000 X 1 + 2,000 X 6 + 3,000 X 8 - R2 X 10 + 400 X 5=0,

R1 X 10 - 1,000 X 9 - 2,000 X 4 - 3,000 X 2 - 400 X 5=0. The first one reduces to

10 R2= 39,000, or R = 3,900 pounds; and the second to

10 R1 = 25,000, or R = 2,500 pounds.

4. What are the total reactions in example 2 if the beam weighs 42 pounds per foot ?

As in example 3, we might compute the reactions due to the weight and then add them to the corresponding reactions due to the loads (already found in example 2), but we shall determine the total reactions due to load and weight directly.

The beam being 20 feet long, its weight is 42x20, or 840 pounds. Since the middle of the beam is 8 feet from the right and 6 feet from the left support, the moments of the weight with respect to the left and right supports are respectively:

840x8 = 6,720, and - 840x6 = -5,040 foot-pounds. The moment equations for all the forces applied to the beam for origins at B and D are like those in example 2, with an additional term, the moment of the weight; they are respectively: - 2,100x2+0+3,600x6 - R2 x 14+1,600x18 + 6,720 = 0, - 2,100xl6 + R1xl4 - 3,600x8 + 0+1,600x4 - 5,040 = 0.

The first equation reduces to

14 R2=52,920, or R2=3,780 pounds, and the second to

14 R1= 61,040, or R1= 4,360 pounds.

The sum of the loads and weight of beam is 8,140 pounds; and since the sum of the reactions is the same, the computation checks.

Examples For Practice

1. AB (Fig. 11) represents a simple beam supported at its ends. Compute the reactions, neglecting the weight of the beam.

Ans.

Right reaction

=

1,443.75 pounds.

Left reaction

=

1,556.25 pounds.

Reactions Of Supports Continued 030018

Fig. 11.

2. Solve example 1 taking into account the weight of the beam, which suppose to be 400 pounds.

Ans.

Right reaction

=

1,643.75 pounds.

Left reaction

=

1,756.25 pounds.

3. Fig. 12 represents a simple beam weighing 800 pounds supported at A and B, and sustaining three loads as shown. What are the reactions ?

Ans.

Right reaction

=

2,014.28 pounds.

| Left reaction

=

4,785.72 pounds.

Reactions Of Supports Continued 030019

Fig. 12.

4. Suppose that in example 3 the beam also sustains a uniformly distributed load (as a floor) over its entire length, of 500 pounds per foot. Compute the reactions due to all the loads and the weight of the beam.

Ans.

Right reaction

=

4,871.43 pounds.

Lett reaction

=

11,928.57 pounds.