a) Let the reactions at supports A and B caused by the load P be R_a and R_b, respectively.  Derive  the formulas 

R_a = Pb/L
R_b = Pa/L

using in each case a single moment equilibrium equation.  

SM_B = Pb - R_aL = 0,   R_a = Pb/L
SM_A = -Pa + R_bL = 0,   R_b = Pa/L

Can you verbalize the formulas so as to memorize their content without relying on symbols?

The reaction at a support is proportional to the distance of the load to the other support, relative to the distance between supports.

b) The formulas above are applicable  whether P is applied within or outside segment AB, provided a sign convention is adopted for R_a, R_b, and a or b.  With a and b related by a + b = L, find and state that sign convention.

R_a, R_b, > 0 upward
a > 0 if P is to the right of A, negative otherwise.

Consistently,

b = L-a > 0 if P is to the left of B, negative otherwise.

c) Show that the reactions to an applied moment M are R_a = M/L and R_b = -M/L. What is the sign convention in these formulas? Do the formulas depend on where the moment is applied?

SM_B = M - R_aL = 0
SM_A = M + R_bL = 0

M > 0 if counterclockwise.
R_a, R_b > 0 upward.

The formulas do not depend on where the moment is applied because the moment of a moment about any point is equal to that moment.

a) On a new sketch of the beam, replace the uniform load and the triangular load, respectively, by equivalent resultant forces, and draw a free-body diagram of the beam.  Apply the formulas of Sec. 1 to determine the reactions.

R_a = (pL)(1/2) - (qa/2)(a/3L) =  pL/2 - qa²/6L

R_b = (pL)(1/2) + (qa/2)(L + a/3)/L = pL/2 + qa/2 + qa²/6L

b) On a new sketch of the beam, replace the total applied loading by its resultants at B, and draw a free-body diagram of the beam. Obtain the reaction at A from that diagram, and check that it is the same as found in part a).

R_a = (pL²/2 - qa²/6)/L = pL/2 - qa²/6L

R_b = pL + qa/2  - (pL²/2 - qa²/6)/L = pL/2 + qa/2 + qa²/6L

c)  Determine the shear force to the immediate right of B by the force transmission approach.  Do the same to the immediate left of B using the reaction at A found earlier.  What is the difference between the two answers, and what is its physical significance? 

By the force transmission approach, shear to the right of B is

V_B+ = sum of forces to the right of the section, positive if downward = qa/2

Shear to the left of B using the reaction at A is

V_B- = sum of forces to the left of the section positive upward = R_a - pL  = - pL/2 - qa²/6L

The difference is

V_B+ - V_B- = qa/2  + pL/2 + qa²/6L

It is equal to R_b.  The shear force past a point of concentrated load jumps by the value of that load.

3. Resultants of a Spatial Force System

A road sign is formed of thin plate CDEF, parallel to the (x, z) plane, and attached to a cylindrical post of radius 1 ft centered on the z axis. The plate is subjected to wind pressure p, and its weight is W.  The post weighs w lb/ft.

a) Show on the figure the weight of the plate, and the resultant P of the wind load acting on it. Show a force perpendicular to the plane of the figure as a dot if it points out of the page (tip of the vector) or a cross if it points into the page (tail of the vector).
b) Determine the resultants of the forces acting on the plate, at point A on the z axis. (force resultant components F_x, F_y, F_z, and moment resultant components M_Ax, M_Ay, M_Az) .

a)  P = 8(15)p = 200p  as shown in figure.

b) Resultants of P and W at A:

F_x = 0

F_y = - P

F_z = -W

M_Ax = 14P - (1)W

M_Ay = 6.5W

M_Az = -6.5P

c) Determine by the force transmission approach the internal forces of the post at A, and show them on a sketch as acting on the bottom part.

The internal forces at A acting on the bottom part are equal to the resultants at A of the external forces on the top part.  Thus they are equal to the resultants computed in b) to which must be added the weight of the part of the post above A, which is 18w acting downward.

Forces and moments are shown in their actual sense of action.  It is assumed M_Ax = 14P - 1W > 0.

d) Use the results of part c) to determine the support reactions at B, and show them on a sketch.

 The reactions at B equilibrate the forces shown in the figure above:

R_Bx = 0

R_By = -F_y = 200p

R_Bz = -F_z = W + 18w

M_Bx = -M_Ax + 10F_y  = -14P + 1W - 2000p

M_By = -MAy = -6.5W

M_Bz = -M_Az = 6.5P

4. Internal Static Conditions
a) A bar such as AB, which is pinned at both ends, and is not subjected to any external load, is a two-force body subjected to a force at A and a force at B. If the bar is in equilibrium, and F is the force at A, the force at B is -F, and both forces act along line AB.  Why is that so?  What would happen if F and -F did not have a common line of action?

If the two opposite forces were not collinear, they would form a couple, and the bar would not be in equilibrium.  It would be spinning.

b) Determine the force in bar AB by a single equilibrium equation.

SM_D = Fasinq - p(a+b)²/2 = 0

F = p(a+b)²/2asinq

c) If bar AB is subjected to an external load, it ceases to be a two-force member.  Does the problem become statically indeterminate? Explain.

No.  Since the end moments of bar AB are zero,  its transverse end forces can be determined by two transverse equilibrium equations.  The axial equilibrium equation leaves only one axial force unknown, which can be determined, as in part b), by the moment equilibrium equation of the structure about point D.

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Force Transmission:

A system being cut into two parts, the internal forces acting on one part are equal, as vectors, to the resultants of the external forces acting on the other part.
Equivalently,
A system being cut into two parts, each part exerts on the other part the resultants of its own external forces.