Left part:     ?F_x = – F₁ + A_x = 0
 (OA)          ?F_y = P₁ + A_y – F₂ = 0
                  ?M₀ = P₁c + A_y(c + d) + M_A – F₂a = 0
Right side: (AB)        ?Fx = – Ax + P3 = 0
                  ?F_y = P₂ – A_y – F₃ = 0
                  ?M_A = -M_A + P₂e + M – F₃?  = 0
                 

Six unknowns (P₁, P₂, P₃, A_x, A_y, M_A) are in six equations.
Note: In the ?rst diagram (Figure 1.2.13) the couple M may be moved
anywhere from O to B. M is not shown in the second diagram (O to A)
because it is shown in the third diagram (in which it may be moved
anywhere from A to B).

Example 3

The arm of a factory robot is modeled as three bars (Figure 1.2.15) with
coordinates A: (0.6, 0.3, 0.4) m; B: (1, 0.2, 0) m; and C: (0.9, 0.1, 0.25) m.
The weight of the arm is represented by W_A = 60 Nj at A, and W_B = 40 Nj
at B. A moment M_C = (100i 20j + 50k) N ? m is applied to the arm at C.
Determine the force and moment reactions at O, assuming that all joints
are temporarily ?xed.

FIGURE 1.2.15 Model of a factory robot.

Solution. The free-body diagram is drawn in Figure 1.2.15b, showing the
unknown force and moment reactions at O. From Equation 1.2.13,

?F = 0
F₀ + W_A + W_B = 0
F₀ – 60 Nj – 40 Nj = 0
F_O = 100 Nj
?M₀ = 0
M₀ + M_C + (r_0A x W_A) + (r_0B x W_B) = 0

M₀ + (100i – 20j + 50k)N?m +(0.6i – 0.3j + 0.4k)m x(- 60 Nj) +
(i – 0.2j) m x(- 40 Nj) = 0

M₀ + 100 N?m i – 20 N?m j + 50 N?m k – 36 N?m k + 24 N?m i –
40 N?m k = 0

M₀ = (124i + 20j + 26k) N?m

Example 4

A load of 7 kN may be placed anywhere within A and B in the trailer of
negligible weight. Determine the reactions at the wheels at D, E, and F,
and the force on the hitch H that is mounted on the car, for the extreme
positions A and B of the load. The mass of the car is 1500 kg, and its
weight is acting at C (see Figure 1.2.16).

FIGURE 1.2.16 Analysis of a car with trailer.

Solution. The scalar method is best here.

Put the load at position A ?rst
For the trailer alone, with y as the vertical axis
?M_F = 7(1) H_y(3) = 0, H_y = 2.33 kN
On the car
H_y = 2.33 kN  Ans.
?F_y = 2.33 7 + F_y = 0, F_y = 4.67 kN  Ans.
For the car alone
?M_E = 2.33(1.2) D_y(4) + 14.72(1.8) = 0
D_y = 5.93 kN  Ans.
?F_y = 5.93 + E_y 14.72 2.33 = 0
E_y = 11.12 kN  Ans.
Put the load at position B next
For the trailer alone
?M_F = 0.8(7) H_y(3) = 0, H_y = 1.87 kN
On the car
H_y = 1.87 kN  Ans.
?F_y = 1.87 7 + Ey = 0
E_y = 8.87 kN  Ans.
For the car alone
?M_E = (1.87)(1.2) Dy(4) + 14.72(1.8) = 0
D_y = 7.19 kN  Ans.
?F_y = 7.19 + E_y 14.72 (1.87) = 0
E_y = 5.66 kN  Ans.

Forces and Moments in Beams

Beams are common structural members whose main function is to resist
bending. The geometric changes and safety aspects of beams are analyzed by
?rst assuming that they are rigid.  The preceding sections enable one to
determine (1) the external (supporting) reactions acting on a statically
determinate beam, and (2) the internal forces and moments at any cross
section in a beam.

Classi?cation of Supports

Common supports and  external reactions for two-dimensional loading of
beams are shown in Figure 1.2.17.

FIGURE 1.2.17 Common beam supports.

Internal Forces and Moments

The internal force and moment reactions in a beam caused by external
loading must be determined for evaluating the strength of the beam. If there
is no torsion of the beam, three kinds of internal reactions are possible:
a horizontal normal force H on a cross section,  vertical (transverse) shear
force V, and bending moment M. These reactions are calculated from the
equilibrium equations applied to the left or right part of the beam from the
cross section considered. The process involves free-body diagrams
of the beam and a consistently applied system of signs. The modeling is
illustrated for a cantilever beam in Figure 1.2.18.

FIGURE 1.2.18 Internal forces and moments in a cantilever beam.

Sign Conventions. Consistent sign conventions should be used in any
given problem. These could be arbitrarily set up, but the following is slightly
advantageous. It makes the signs of the answers to the equilibrium equations
correct for the directions of the shear force and bending moment.
A moment that makes a beam concave upward is taken as positive.
Thus, a clockwise moment is positive on the left side of a section, and a
counterclockwise moment is positive on the right side. Ashear force that
acts upward on the left side of a section, or downward on the right side, is
positive (Figure 1.2.19).

FIGURE 1.2.19 Preferred sign conventions.

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