3. Draw the moment diagram from left to right.
2 m 3 m 3 m 2 m
30 kN-m M
6 kN/m
6 kN/m
M
1 kN/m
1 kN/m
15 kN-m 12 kN-m
–12 kN-m –15 kN-m
Drawing moment diagram from left to right.
The computation under each FBD is self-explanatory. We start from the right FBD because it contains only two unknowns and we have exactly two equations to use. The third equation of equi- librium is the balance of forces in the horizontal direction, which produces no useful equation since there is no force in the horizon- tal direction.
Σ MC = 0, VB (3) – 3(1.5) + 6(1) = 0 VB = –0.5 kN
Σ Fy = 0, –0.5 – 3 – 6 + RC = 0 RC = 9.5 kN
Σ MA = 0, MA + 3(1.5) – 0.5(3) = 0 MA = –3 kN-m
Σ Fy = 0, 0.5 – 3 + RA = 0 RA = 2.5 kN
2. Draw the FBD of the whole beam and then shear and moment diagrams.
3 m 3 m 1 m
6 kN 1 kN/m
A B C D
2.5 kN
9.5 kN 3 kN-m
2.5 kN
–3.5 kN 6 kN
2.5 m V
0.125 kN-m 2 m
M
–6 kN-m x
–3 kN-m 3 m
Shear and moment diagrams drawn from the force data of the FBD.
Note that the point of zero moment is determined by solving the second order equation derived from the FBD shown next.
x 1 kN/m A
2.5 kN
3 kN-m Mx
Vx FBD to determine moment at a typical section x.
M(x) = –3 + 2.5 x – 0.5 x2 = 0, x = 2 m, 3 m
The local maximum positive moment is determined from the point of zero shear at x = 2.5 m, from which we obtain M(x = 2.5) = –3 + 2.5 (2.5) – 0.5 (2.5)2 = 0.125 kN-m.
Example 4.8
Analyze the loaded beam shown next and draw the shear and moment diagrams.
3 m 3 m
6 kN-m 3 kN/m
A beam loaded with a distributed force and a moment.
Solution
The problem is solved using the principle of superposition, which states that for a linear structure the solution of the structure under two loading systems is the sum of the solutions of the structure under each force system.
The solution process is illustrated in the following self-explanatory sequence of figures.
V(x) = 3 – 0.5 x2 = 0, x = 2.45 m
M(x = 2.45) = 3 x – 0.5 x2 (x/3) = 4.9 kN-m
3 m 3 m
3 kN/m
3 m 3 m
3 kN/m 1 m 4.5 kN 6 kN-m
1.5 kN 3.0 kN 1 kN 1 kN
x x
1.5 kN
x 4.9 kN-m 4.5 kN-m
–1 kN 3.0 kN-m
–3.0 kN-m M
V
3.0 kN
2.45 m –3.0 kN
Solving two separate problems.
The superposed shear and moment diagrams give the final answer.
V(x) = 4 – 0.5 x2 = 0, x = 2.83 m x = 2.83
–4.0 kN 0.5 kN
7.55 kN-m
1.5 kN-m 7.5 kN-m
M V
Combined shear and moment diagrams.
Example 4.9
Analyze the loaded frame shown next, and draw the thrust, shear, and moment diagrams.
P P
P P
N1
A statically determinate frame.
Solution
The solution process for a frame is no different from that for a beam.
1. Define FBDs and find reactions and internal nodal forces. Many different FBDs can be defined for this problem, but they may not lead to simple solutions. After trial-and-error, the follow- ing FBD offers a simple solution for the axial force in the two columns.
10 kN
c T T
4 m 4 m
3 m
FBD to solve for the axial force in columns.
Σ Mc = 0, T(8) = 10(3), T = 3.75 kN
Once the axial force in the two columns is known, we can proceed to define four FBDs to expose all the nodal forces at the internal hinges, as shown in the upcoming figure, and solve for any unknown nodal forces one by one using equilibrium equations of each FBD. The solution sequence is shown by the numbers attached with each FBD. Within each FBD, the bold-faced force values are those that are known from previous calculations and the other three unknowns are obtained from the equilibrium equa- tions of the FBD itself.
10 kN
3.75 kN 3.75 kN
5 kN 5 kN
3.75 kN 3.75 kN
5 kN
5 kN 5 kN
5 kN 5 kN
3.75 kN 15 kN-m 3.75 kN 15 kN-m
2 1
5 kN
3.75 kN
3 2
4 m 4 m
3 m
3 m 3 m
3 m
3.75 kN
Four FBDs exposing all internal forces at the hinges.
We note that we could have used the 12 equilibrium equations from the above four FBDs to solve for the twelve force unknowns without the aid of the previous FBD to find the axial force in the two columns first. But the solution strategy presented offers the simplest computing sequence without having to solve for any simultaneous equations.
2. Draw the thrust, shear, and moment diagrams. For the thrust dia- gram, we designate tension force as positive and compression force as negative. For shear and moment diagrams, we use the same sign convention for both beams and columns. For the verti- cally orientated columns, it is customary to equate the “inside” of a column to the “downside” of a beam and draw the positive and negative shear and moment diagrams accordingly.
3 m 3 m
4 m 4 m
3 m 3 m
4 m 4 m
3 m 3 m
4 m 4 m
T
V
M
3.75 kN –3.75 kN
–5 kN
–3.75 kN
5 kN 5 kN
15 kN-m
–15 kN-m
15 kN-m –15 kN-m
Thrust, shear, and moment diagrams of the example problem.
Example 4.10
Analyze the following loaded frame, and draw the thrust, shear, and moment diagrams. The intensity of the horizontal force is 15 kN per unit vertical length.
5 m 3 m
15 kN/m 4 m
Statically determinate frame example problem.
Solution
The inclined member requires a special treatment in finding its shear diagram.
1. Find reactions and draw FBD of the whole structure.
5 m 3 m
4 m 60 kN
R = 15 kN
15 kN 60 kN
c
FBD for finding reactions.
Σ Mc = 0, 60(2) –R(8) = 0, R = 15 kN
2. Draw the thrust, shear, and moment diagrams. Before drawing the thrust, shear, and moment diagrams, we need to find the nodal forces that are in the direction of the axial force and shear force.
This means we need to decompose all forces not perpendicular to or parallel to the member axes to those that are. The upper part of the following figure reflects that step. Once the nodal forces are properly oriented, the drawing of the thrust, shear, and moment diagrams is effortlessly achieved.
60 kN
48 kN 36 kN
15 kN 12 kN 9 kN
7.2 kN/m 9.6 kN/m
12 kN 9 kN
39 kN 48 kN
60 kN 15 kN 15 kN
60 kN
T
V
–12 kN
–48 kN
–60 kN 75 kN-m
9 kN
–39 kN
15 kN
M
–75 kN-m 75 kN-m 48 kN/5 m = 9.6 kN/m
36 kN/5 m = 7.2 kN/m
0.94 m 5 m (9)/(39+9) = 0.94 m
4.22 kN-m
M(x = 0.94) = (9)(0.94) – (9.6)(0.94)2/2 = 4.22 kN-m 0.94 m
5 m
Thrust, shear, and moment diagrams of the example problem.
PROBLEM 4.2
Analyze.the.beams.and.frames.shown,.and.draw.the.thrust.(for.frames.
only),.shear,.and.moment.diagrams.
(1) (2)
(3) (4)
(5) (6)
(7) (8)
(9) (10)
3 m 5 m
10 kN
3 m 5 m
10 kN
3 m 5 m
3 kN/m
3 m 5 m
3 kN/m
3 m 5 m
10 kN-m
3 m 5 m
10 kN-m
6 m 10 kN-m
6 m 4 m
2 kN/m
5@2 m = 10 m
10 kN
5@2 m = 10 m 2 kN/m 4 m
Problem.4.2.beam.problems.
(11) (12)
(13) (14)
(15) (16)
5 m
5 m 10 kN
5 m
5 m 10 kN-m
5 m
5 m 10 kN
5 m
5 m 10 kN-m
5 m
5 m 10 kN
5 m
5 m 10 kN-m
Problem.4.2.frame.problems.
117