5.Truss Deflection
Problem 1. Discuss the determinacy of the beams and frames shown
4. Statically Determinate Beams and Frames
Analysis of statically determinate beams and frames starts from defining the FBDs of members and then utilizes the equilibrium equations of each FBD to find the force unknowns. The process is best illustrated through example problems.
Example 7. Analyze the loaded beam shown and draw the shear and moment diagrams.
2m 3m 3m
6 kN
2m 6 kN
V
V
5 kN 5 kN
6 kN 1 kN
6 kN
2m 3m 3m 2m
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
A statically determinate beam problem.
Solution. The presence of an internal hinge calls for a cut at the hinge to produce two separate FBDs. This is the best way to expose the force at the hinge.
(1) Define FBDs and find reactions and internal nodal forces.
Two FBDs exposing all nodal forces and support reactions.
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 equilibrium is the balance of forces in the horizontal direction, which produces no useful equation since there is no force in the horizontal direction.
ΣMC =0, VB(3)−3(1.5)+6(1)=0 VB = −0.5 kN
ΣFy =0, −0.5−3−6+RC =0 RC = 9.5 kN ΣMA =0, MA +3(1.5) −0.5(3)=0
MA= −3 kN-m ΣFy =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
3 m 1 m
6 kN 1 kN/m
3 m 1 kN/m
A B C D
VB
VB
C
RC
RA
MA
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 below.
FBD to determine moment at a typical section x.
M(x) = –3 +2.5 x –0.5 x2 = 0, x =2m , 3m
The local maximum positive moment is determined from the point of zero shear at x=2.5m , from which we obtain M(x=2.5) = –3 +2.5 (2.5) –0.5 (2.5)2 = 0.125 kN-m.
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
6.0 kN
2.5 m
V
0.125 kN-m
2 m
M
-6 kN-m
x
3 m
x 1 kN/m
A 2.5 kN
3 kN-m Mx
Vx -3 kN-m
Example 8. Analyze the loaded beam shown and draw the shear and moment diagrams.
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
Solving two separate problems.
3 m 3 m
6 kN-m 3 kN/m
3 m 3 m
3 kN/m
3 m 3 m
6 kN-m
3 kN/m 6 kN-m
4.5 kN 1 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
The superposed shear and moment diagrams give the final answer.
V(x)= 4-0.5 x2 = 0, x=2.83 m
M(x)= 4 x – 0.5 x2(x/3), Mmax= M(x=2.83)= 7.55 kN-m Combined shear and moment diagrams.
Example 9. Analyze the loaded frame shown and draw the thrust, shear and moment diagrams.
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 following FBD offers a simple solution for the axial force in the two columns.
-4.0 kN 0.5 kN
7.55 kN-m
1.5 kN-m 7.5 kN-m
x=2.83
M V
3 m 3 m
4 m 4 m
10 kN
FBD to solve for the axial force in columns.
ΣMc =0, T(8)=10(3), T=3.75 kN
Once the axial force in the two columns are known, we can proceed to define four FBDs to expose all the nodal forces at the internal hinges as shown below 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 equations of the FBD itself.
Four FBDs exposing all internal forces at the hinges.
We note that we could have used the twelve 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.
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
10 kN
c
T T
2 1
5 kN
3.75 kN
3 2
4 m 4 m
3 m
4 m 4 m
3 m
3 m 3 m
3 m
3.75 kN
For the thrust diagram, we designate tension force as positive and compression force as nagative. For shear and moment diagrams, we use the same sign convention for both beams and columns. For the vertically orientated columns, it is customary to equate the
“in-side” of a column to the “down-side” of a beam and draw the positive and nagetive shear and moment diagrams accordingly.
Thrust, shear and moment diagrams of the example problem.
Example 10. Analyze the loaded frame shown and draw the thrust, shear and moment diagrams.
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
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.
FBD for finding reactions.
∑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 figure below reflects that step. Once the nodal forces are properly oriented, the drawing of the thrust, shear, and moment diagrams is achieved effortlessly.
3 m 5 m
4 m 15 kN/m
3 m 5 m
4 m 60 kN
R=15 kN
15 kN 60 kN
c
Thrust, shear, and moment diagrams of the example problem.
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/5m=9.6 kN/m
36 kN/5m=7.2 kN/m
0.94 m
5m (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
Problem 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)
Problem 2. Beam Problems.
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@2m=10m
10 kN
5@2m=10m 2 kN/m 4 m
(11) (12)
(13) (14)
(15) (16)
Problem 2. Frame problems.
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