Strength of Materials - Lecture Notes / Mehmet Zor
1
23 Agust 2024
9. Deflection of Beams
(ELASTIC CURVE)
9.5 Superposition Method
(tvids: 9..5)
Strength of Materials - Lecture Notes / Mehmet Zor
2
23 Agust 2024
Region AB
Region BC
y
x
1
2
,
,
S1
,
1
2
y
x
,
S2
1
B
y
x
,
S3
Deflection of Beams / Superposition Method and Hyperstatic Systems
In this chapter; We will produce solutions by comparing the problem we are examining to one or a combination of the following different reference systems. The general equations of slope (𝜃) and deflection (𝑣) of each reference system are given next to them. We assume that these equations were obtained by previous methods. We will explain the subject through examples.
REFERANS SYSTEMS
9.5 Superposition Method and Hyperstatic Systems in Slope and Deflection Calculations
Strength of Materials - Lecture Notes / Mehmet Zor
3
23 Agust 2024
a
b
q
a
b
S4
S6
S5
S7
Deflection of Beams / Superposition Method and Hyperstatic Systems
S8
REFERANS SYSTEMS - Continue
Strength of Materials - Lecture Notes / Mehmet Zor
4
23 Agust 2024
y
x
1
2
S1
Example 9.5.1
P
P
S4
Solution: According to the principle of superposition, P and q are applied sequentially and the deflections can be added.
We can liken this singular loading above to the S1 reference system. According to this : a=b=L/2, x=xB=L/2
We can liken the distributed loading above to the S4 reference system. According to this;
If we substitute the results obtained above:
Figure 9.4.12
Deflection of Beams / Superposition Method and Hyperstatic Systems
Strength of Materials - Lecture Notes / Mehmet Zor
5
23 Agust 2024
y
x
1
2
S1
Example 9.5.2
Solution: Reaction force RB is placed instead of support B. This is a hyperstatic system. RB and q are applied sequentially. It can be obtained by combining the reference systems S1 and S4, as in the previous example. Since the RB force is in the opposite direction of P in reference S1, a minus sign is placed at the beginning of the equation.
S4
Due to the support B, the deflection at B is zero.
Support forces A and C can be calculated from static equations.
If we substitute the results obtained above:
Figure 9.4.13
Deflection of Beams / Superposition Method and Hyperstatic Systems
Strength of Materials - Lecture Notes / Mehmet Zor
6
23 Agust 2024
Example 9.5.3
S7
q
a
b
a
b
,
S6
q
a
L
q
a
L-a
tangent-B
L
Reaction force RC is placed instead of support C. The distributed load and the RC reaction force are applied sequentially.
We can simulate the above loading to the S7 reference system. Accordingly, the equivalents of the variables in the reference system are:
We can liken the above loading to the S6 reference system. Accordingly, point C in the upper system corresponds to point B in the reference system. Equivalents of other variables:
(or it could also be found from the S8 reference System.)
The total deflection at C due to the support is zero :
If the values above are substituted:
Solution:
Figure 9.4.14
Figure 9.4.15
Figure 9.4.16
Deflection of Beams / Superposition Method and Hyperstatic Systems
Strength of Materials - Lecture Notes / Mehmet Zor
7
23 Agust 2024
q
L/2
L/2
P
Example 9.5.4
Calculate the force on the spring in the system shown in the figure..
k
q
a=L/2
b=L/2
a=L/2
b=L/2
L
P
Solution:
S7
S6
S8
S7
q
a
b
a
b
,
S6
S8
Figure 9.4.17
Deflection of Beams / Superposition Method and Hyperstatic Systems