Moment of Force about a Specific Axis | Engineering Mechanics - Civil Engineering (CE) PDF Download

Moment of a force about an axis

The moment of a force about an axis p is defined as the projection OB on p of the moment  (see Fig. 3.5).

Figure 3.5: Moment of the force F about the axis p.

Denoting λ the unit vector of p, the moment Mp of force F about an axis p can be expressed as scalar product (dot product)

Mp = λ . M_o      (3.22)

or as the mixed triple product of the unit vector λ , the position vector r, and the force F:

Mp = λ . ( r *F)     (3.23)

Using the determinant form for the mixed triple product, we have the magnitude of the moment

Moment of a force about an axis

  (3.24)

λ_x,

where λ_y, are direction cosines of axis p

λ_z

x, y, z are components of r

F_x,

F_y, F_z are components of F

Exercise 3.2.1

Sample problem A force F = 40 N is applied at a point M(x_M;y_M;z_M) M(3;2;4) [m] (see Fig. 3.6). The line o_F of action of the force F is described by direction angles (α_F ;β_F;C_r ) (80^o;60^o; acute angle). An axis a which passes

through the origin O has direction angles (αa;βa;γa) ≡ (60^o;100^o; acute angle).

Determine:

• The moment M_o of the force F about O.
• The moments M_x, M_y, and M_z of the force F about axes x, y, and z, respectively.
• The moment M_a of the force F about a axis.

 Moment of a force about an axis

Figure 3.6: Exercise 3.2.1. Moment of the force F about the axis a.

Notice: An acute angle is such an angle for which the condition 0 < α < 90^o is valid. 

Solution

We determine the angle C_r first. Since the expression

is valid for any set of direction cosines, we have

According to the definition of the moment M_o of the force F, we may write

Notice: Matlab works with radians and not with degrees. 

The magnitude of M_o is

88.06 Nm

The direction cosines of M_o are

As the components M_Ox, M_Oy, and MOz of M_o are equal to the moments M_x, M_y, and M_z of F about x, y, and z axes respectively, the following is valid:

M_x = M_Ox,    M_y = M_Oy,    M_z = M_Oz

According to the definition of the moment M_a of the force F about an axis a, we have

M_a = λ^T M_o

where the unit vector λ of a is

λ = [Cos α_a, cosβ_a, cosγ_a]^T

As

we conclude that

M_a = [0.5, 0.1736, 0.8484]    = 45.92 Nm

Using MATLAB the solution of the problem is much more convenient. The program for the purpose is called smoment.m. It can be found in program package.

Exercise 3.2.2 Moment of a force A force F = 50 N is applied at a point M(8;4;4) [m] and its line of action is described by direction angles (60^o; 60^o; acute angle). Determine the moments M_x, M_y, and M_z of the force F about the axes x, y, and z respectively and the moment M_a about the axis a passing through the point A(0;2;2) and having direction angles (30^o;0^o; acute angle).

Figure 3.7: Exercise 3.2.2. Moment of the force F about the axis a.

Solution

M_x = 41.4 Nm, M_y = -182.8 Nm, M_z = 100 Nm, M_a = -98.47 Nm

Exercise 3.2.3 Moment of a force A force F = 50 N is applied at a point M(2;3;1) [m] and its line of action is described by direction angles (60^o; 60^o; acute angle). The axis a passes through the origin O and lies in xy-plane having the angle  δ= 30^o with y axis. Determine

• The moment M_o  of the force F about the origin O.
• The moments M_x, M_y, and M_z of the force F about the axes x, y, and z respectively.
• The moment M_a of the force F about the axsis a.

Figure 3.8: Exercise 3.2.3. Moment of the force F about the axis a. 

Solution 

M_x = 81.05 Nm, M_y = -45.7 Nm, M_z = -25 N m, M_O = 96.34 Nm, M_a = -52.07 Nm 

Exercise 3.2.4 Moment of a force A force F = 40 N is applied at a point M(4;3;-2) [m] and its line of action is described by direction angles (30^o; 90^o; acute angle).
Determine the moment M_A of the force F about the point A(-1;3;6) [m] and the moment M_a of the force F about the axis a passing through the point A and having direction angles (60^o; 80^o; acute angle).

Figure 3.9: Exercise 3.2.4. Moment of the force F about the axis a.

Solution

M_Ax = 0 Nm, M_Ay = -377.13 Nm, M_Az = 0 Nm, M_a = -65.49 Nm

Exercise 3.2.5 Moment of a force A tube is welded to the vertical plate yz at the point A. The tube is loaded by a force F at the point D. The line of action of the force  has an angle β with y-axis and an angle γ with z-axis. Determine the moment of the force

F about the point A. Further determine the moment M_x about x-axis and the moment M_BC about BC axis of the same force F . How many solutions does the task have? It is known that  F = 200 N; β = 60^o; γ= 80^o;a = 0.2 m;b = 0.4 m;c = 0.8 m.

Figure 3.10: Exercise 3.2.5. Moment of the force F.

Solution

M_A = 152.2 Nm, M_x = - 66.1 Nm, M_BC = 135.75 Nm 

Exercise 3.2.6 Moment of a force Determine the moment of the force F = 500 N about the axis AC. The angle  α = 30^o, A(5;0;0), C(0;12;0), E(0;0;20), AB = 6.5, BD = 10 (all dimensions in m).

Figure 3.11: Exercise 3.2.6. Moment of a force F.

Solution

M_AC = 4510 Nm

Exercise 3.2.7 Tangent force A force T = 60 N acts at a tangent of a helix (see Fig.3.12). Find a generic expression for moments M_x, M_y, and Mz of the force T about x, y, and z axes respectively as a function of angle . Numerically compute the φ Numerically compute the magnitudes of these moments for φ =750^o knowing that the radius of the helix is r =10 m and the pitch angle is  α =30^o. Write a MATLAB program which can be used to calculate the above moments and use it for creating plots of moments values versus angle 0 ≤ φ ≤ 6π.

Figure 3.12: Exercise 3.2.7. Tangent force. 

Solution

M_x = 2222 Nm, M_y = -519.6 Nm, M_z = 3248 Nm