The bent rod is supported at a. b. and c by smooth journal bearings.

no title provided The bent rod is supported at A, B, and C by smooth journal bearings. Compute the x, y, z components of reaction at the bearing A if the rod is subjected to forces F1 = 800 lb and F2 = 450 lb.F1 lies in the y?z plane. The bearings are in proper alignment and exert only force reactions on the rod.
Compute the x, y, z components of reaction at the bearing B.
Compute the x, y, z components of reaction at the bearing C.

Answer

General guidance

Concepts and reason
Principle of equilibrium: The principle of equilibrium states that, a stationary body will be in equilibrium when subjected to concurrent or parallel forces if the algebraic sum of all the external forces and moments of all the external forces is zero. Free body diagram: It is a graphical, symbolic illustration which is used to visualize the forces which are applied, moments and the reactions on a body for a particular condition.

Fundamentals

Action and Reaction: ACTION
AL SUPPORT
A
TRA
(a)
(c)
(b)
figure (1) On a horizontal surface, a ball is placed as shown in figure (1.a) which cannot move vertically downward but free to move along the horizontal surface. Therefore a force is exerted vertically downwards at the support by the ball as shown in figure (1.b) and this force is called as action. At the point of contact of the ball with the surface, the support will exert a force vertically upwards on the ball as shown in figure (1.c) and this force is called as reaction. Hence “any force on a support causes an equal and opposite force from the support so that action and reaction are two equal and opposite forces”. Resolution of forces: In resolving the forces as horizontal forces and vertical forces, when the angle of the force is made with the horizontal then the trigonometric angle of cosine is considered only in horizontal direction and the trigonometric angle of sine is considered only in vertical direction. The illustration is shown in Figure (2). Fsino
Fcoso
Figure (2) The vertical force is calculated as shown. F, = Fsino Here, force is F and angle of force with the normal is. The horizontal force is calculated as shown. F, = F cose Here, angle of force with the horizontal is. Principle of equilibrium: The equation for principal of equilibrium is written as, ΣF = 0 ΣΜ = 0 Here, algebraic sum of the forces is , algebraic sum of the moments is , force law of equilibrium is ΣF = 0, and moment law of equilibrium is ΣΜ = 0. In general, the forces are resolved into horizontal as well as vertical components. The forces in the vertical components is written as, ΣF = 0 The forces in the horizontal components is written as, ΣΕ, = 0 Here, algebraic sum of all horizontal components is , and algebraic sum of all vertical components is . General sign convention for moment: The moments which rotate the body in counter clockwise direction are taken as positive and the moments which rotate the body in clockwise direction are taken as negative. General sign convention for force: The forces in the rightward and upward direction are taken as positive while the forces in downward and leftward direction are taken as negative.

Step-by-step

Step 1 of 11

Draw the free body diagram of the system as shown in Figure (3). 1 ft| Ay
4 ft
Ax
2
ff
3 ft
/
450
F
Figure 3 Apply the force law of equilibrium along the x-direction. ΣF = 0 4+B+F, cos 45º sin 30º = 0 Here, reaction at A along x-direction is , and reaction at B along x-direction is . Substitute 450 lb for . A + B +450x cos45°sin 30° = 0
4+B4 +159.099 = 0 …… (1) Apply the force law of equilibrium along the y-direction. ΣΕ, = 0 A, +C, - F, cos 45° + F, cos 45ºcos 30º = 0 Here, reaction at A along y-direction is , and reaction at C along z-direction is . Substitute 800 lb for and 450 lb for . 4, +C, -800x cos 45º +450x cos 45ºcos 30º = 0
4,+C - 565.685+275.56=0
4, +C, - 290.11=0 …… (2) Apply the force law of equilibrium along the z-direction. Here, reaction at B along z-direction is , and reaction at C along z-direction is . Substitute 800 lb for and 450 lb for . B. = 883.875-C. …… (3)

The free body diagram of the system is drawn and the force law of equilibrium is applied along x-axis, y-axis, and z-axis.

Step 2 of 11

Apply the moment law of equilibrium along y-axis. …… (4) Apply the moment law of equilibrium along z-axis. ΣΜ. = 0
A (2+ 3) + Β. (3)-C, (5) = 0 …… (5) Apply the moment law of equilibrium along x-axis. Substitute 800 lb for . …… (6)

The equations are formed by applying the moment law of equilibrium along x-axis, y-axis, and z-axis which are used for calculating the various reactions in the system.

Step 3 of 11

Find the reaction at B along x-direction from equation (5) as follows, Substitute for from equation (1). Substitute for from equation (2). Substitute for from equation (6). Substitute for from equation (3). Substitute for from equation (4). Substitute for from equation (1).

The reaction at B along x-direction is .


The reaction at B along x-direction is calculated using the equations obtained by applying the principle law of equilibrium.

Step 4 of 11

Find the reaction at A along x-direction from equation (1) as follows, Substitute for .

The reaction at A along x-direction is .


The reaction at A along x-direction is calculated from the equation (1) by substituting the reaction at B along x-direction in the equation.

Step 5 of 11

Find the reaction at C along x-direction.

The reaction at C along x-direction is 0 lb.


The reaction at C along x-direction is calculated from the Figure (3).

Step 6 of 11

Find the reaction at C along z-direction from equation (4) as follows, Substitute 1851.184 lb for .

The reaction at C along z-direction is .


The reaction at C along z-direction is calculated from equation (4) by substituting the reaction at A along x-direction in the equation.

Step 7 of 11

Find the reaction at B along z-direction from equation (3) as follows, Substitute for .

The reaction at B along z-direction is .


The reaction at B along z-direction is calculated from equation (3) by substituting the value of the reaction at C along z-direction in the equation.

Step 8 of 11

Find the reaction at A along z-direction.

The reaction at A along z-direction is 0 lb.


The reaction at A along z-direction is obtained from Figure (3).

Step 9 of 11

Find the reaction at A along y-direction from equation (6) as follows, Substitute for .

The reaction at A along y-direction is .


The reaction at A along y-direction is calculated from equation (6) by substituting the reaction at B along z-direction.

Step 10 of 11

Find the reaction at C along y-direction from equation (5) as follows, Substitute for and for

The reaction at C along y-direction is .


The reaction at C along y-direction is calculated from equation (5) by substituting the reaction at A and B along x-direction.

Step 11 of 11

Find the reaction at B along y-direction.

The reaction at B along y-direction is 0 lb.


The reaction at B along y-direction is obtained from Figure (3).

Answer

The reaction at B along x-direction is .

The reaction at A along x-direction is .

The reaction at C along x-direction is 0 lb.

The reaction at C along z-direction is .

The reaction at B along z-direction is .

The reaction at A along z-direction is 0 lb.

The reaction at A along y-direction is .

The reaction at C along y-direction is .

The reaction at B along y-direction is 0 lb.

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