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1. Deflection of Beams
Mechanics of Materials-I
1
Waqas Asghar
waqas.asghar@uettaxila.edu.pk
Lecturer, Mechanical Dept.
UET Taxila

2. Deflection Curve of a Beam
2
When a beam having straight longitudinal axis is
loaded by lateral forces, the axis of beam deforms
into a curve, called the deflection curve of beam.

3. Deflection
3
Upward deflections are taken as positive because
𝑦 − 𝑎𝑥𝑖𝑠 is positive upward
Downward deflections are taken as negative because
𝑦 − 𝑎𝑥𝑖𝑠 is negative downward
Deflection ( 𝒚) is the displacement of beam or any
point on axis of beam from its original position, due
to the applied forces and loads.
∆ 𝒐𝒓 𝒚

4. 4
Angle of Slope
Slope (𝜽 𝒐𝒓 𝒚′) is the angle between the x axis and the tangent to the deflection curve.
Counterclockwise angle of slope
is taken as positive
Clockwise angle of slope is taken
as negative
Other names are angle of inclination and angle of rotation
𝒚

5. 5
Atomic Force Microscopy
(commercially available since 1989)
Design of High Rise Buildings
+ Analysis of statically indeterminate structures!!
+Dynamic analysis (aircraft and m/c vibrations or response of buildings to earthquakes).
Design of Fishing Poles
Why We Study Beam Deflections???
Design of Diving Boards
Design of Jumping Poles

6. 6
Curvature of Beam
𝜥 =
𝟏
𝝆
=
𝑴
𝑬𝑰
=
𝒅𝟐𝒚
𝒅𝒙𝟐
(𝑮𝒆𝒓𝒆 𝑬𝒒. 𝟗 − 𝟓 & 𝟗 − 𝟔)
• Consider a cantilever beam, subjected to load P at its free end.
• Consider two points 𝑚1 & 𝑚2 , at any location on its deflection
curve.
• At 𝑚1 and 𝑚2 , draw normal lines ┴ to the tangents, drawn at
deflection curve.
• These lines intersect at point 𝑶′ .
Let,
𝑶′ = Center of curvature of the deflection curve
𝛒 = Radius of Curvature = Distance 𝒎𝟏𝑶′
𝚱 = Curvature (Greek letter kappa) = reciprocal of radius of curvature
Curvature is a measure of how sharply a beam is bent.
Relation B/w Curvature of Beam & its Deflection, when beam follows Hooke’s law

7. 7
If the load on a beam is small, → beam will be nearly straight,
→ point 𝑶′ will be located farther away from the beam → 𝛒 will be
very large → 𝚱 will be very small.
If the load on beam is increased, → amount of bending will
increase → point 𝑶′ will be located closer to the beam → 𝛒 will
become smaller → 𝚱 will become larger.
Rearranging Eq. (9-5) & Eq. (9-6) gives the basic differential equation of the deflection
curve of a beam:
⇨ 𝑴 = 𝑬𝑰
𝒅𝟐
𝒚
𝒅𝒙𝟐
(𝑮𝒆𝒓𝒆 𝑬𝒒. 𝟗 − 𝟕)
𝜥 =
𝟏
𝝆
=
𝑴
𝑬𝑰
=
𝒅𝟐𝒚
𝒅𝒙𝟐
(𝑮𝒆𝒓𝒆 𝑬𝒒. 𝟗 − 𝟓 & 𝟗 − 𝟔)
𝑜𝑟 𝑴 = 𝑬𝑰. 𝒚 ′′ (𝑮𝒆𝒓𝒆 𝑬𝒒. 𝟗 − 𝟕)
Sign Convention
(1) The x and y axes are positive to the right and upward, respectively;
(2) Deflection (∆ 𝒐𝒓 𝒚) is positive upward & vice versa.
(3) Slope (𝜽) is positive when taken counterclockwise w.r.t +x axis.
(4) Curvature k is positive when beam is bent concave upward.
(5) Bending moment M is positive when it produces compression in
upper part of beam.

8. 8
1. Macaulay’s Method
2. Method of Superposition
3. Double Integration Method
4. Moment Area Method
5. Strain Energy Method (Castigliano's Theorem)
6. Conjugate-beam method.
Methods of Determining Beam Deflections

9. 9

10. 10
∆ = 𝒚 = 𝒅𝒆𝒇𝒆𝒍𝒄𝒕𝒊𝒐𝒏
𝜽 = 𝒚′ = 𝒔𝒍𝒐𝒑𝒆
𝑽 = 𝒔𝒉𝒆𝒂𝒓 𝒇𝒐𝒓𝒄𝒆
𝑴 = 𝒊𝒏𝒕𝒆𝒓𝒏𝒂𝒍 𝒎𝒐𝒎𝒆𝒏𝒕
Boundary Conditions To Solve Constants of Integration of
Slope & Deflection Equations
• Beam having roller or pin support (1, 2,3, 4), then ∆ = 0 at these points.
• If roller or pin supports are located at the ends of the beam (1, 2), then 𝑀 = 0 at these points.
• Beam having fixed support (5), then 𝜃 = ∆ = 0 at fixed end.
• Beam having free end (6), then V = 𝑀 = 0 at that location.
• If two beam segments are connected by an “internal” pin or hinge (7), then 𝑀 = 0 at that
connection.

11. Macaulay’s Method to Find Deflections
11
Section AC (𝟎 ≤ 𝒙 ≤ 𝒂):
Beam has three sections, section AC, CD, DB.
Sectioning the beam one by one
B

12. 12
Section CD (𝒂 ≤ 𝒙 ≤ 𝒃):
Section DB (𝒃 ≤ 𝒙 ≤ 𝑳):
B
B

13. 13
Combining the terms 1, 2 & 3

14. 14
𝐶1 , 𝐶2 will always be determined by using boundary conditions.
Integrating moment equation gives slope equation:
In Macaulay’s Method,
• Always put the constants of integration
just after integrating the first term.
• Perform integration by following
Macaulay’s technique, not by the rules of
mathematics
• Negative value obtained inside bracket
of slope or deflection eq. just discard it
• Final moment equation in Macaulay’s
method should contain x.
Now integrating slope equation gives deflection
equation:
𝑅𝑒𝑎𝑐𝑡𝑖𝑜𝑛𝑠 𝑎𝑡 𝑠𝑢𝑝𝑝𝑜𝑟𝑡𝑠 → 𝑀𝑜𝑚𝑒𝑛𝑡 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑦′′ → 𝑆𝑙𝑜𝑝𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑦′ → 𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑦
𝐑𝐨𝐮𝐭𝐞 𝐟𝐨𝐥𝐥𝐨𝐰𝐞𝐝

15. 15
Problem: Using Macaulay's method, find the deflection of beam under load at pt.
C and maximum deflection.
Solution:
Putting in above equation, we have
𝐅𝐢𝐧𝐝𝐢𝐧𝐠 𝐑𝐞𝐚𝐜𝐭𝐢𝐨𝐧𝐬
Let section the beam from left side
Section AC (𝟎 ≤ 𝒙 ≤ 𝒂):

16. 16
Section CB (𝒂 ≤ 𝒙 ≤ 𝒃):
Lets combine the above terms
Integrating moment equation gives slope equation:
Double integrating moment equation gives deflection equation:

17. 17
Finding of 𝐶1 & 𝐶2 , by using boundary conditions.
1st boundary condition, At x =0, y=0 (deflection y=0 at support A). ⇨ C2 = 0
2nd boundary condition, At x =L, y=0 (deflection y=0 at support B). So C1 = ?
Finding final slope & deflection equation by
putting values of 𝐶1 & 𝐶2
At x = 0, we are in first portion of beam, take only first term of deflection equation.
At x = L, whole beam is covered, consider all the terms of deflection equation.

18. 18
(a) Deflection under the load: Put x = a in deflection eq.
as load is acting at distance a from support A.
(b) Maximum deflection: It is obtained, when slope = 0. Max.
deflection occur b/w A & C so take only 1st term before separation line.
Putting value of x in
deflection equation, and
then solve to obtain
maximum deflection.

19. 19
Problem: Using Macaulay's method, find the deflection of beam under pt. loads
and maximum deflection. Take E = 210 GPa and I = 64 ×10-4 m4
Solution:
𝐅𝐢𝐧𝐝𝐢𝐧𝐠 𝐑𝐞𝐚𝐜𝐭𝐢𝐨𝐧𝐬
Let section the beam from left side
Section AC (𝟎 ≤ 𝒙 ≤ 𝟑𝒎)
Section CD (𝟑 ≤ 𝒙 ≤ 𝟗. 𝟓𝒎)
Section DB (𝟗. 𝟓 ≤ 𝒙 ≤ 𝟏𝟒𝒎)
Consider only 1st term
Consider first 2 terms only
Consider all 3 terms

20. 20
Integrating moment equation gives slope equation:
Now, integrating slope equation gives deflection equation:
Finding of 𝐶1 & 𝐶2 , by using boundary conditions.
1st boundary condition, At x =0, y=0 (deflection y=0 at support A). ⇨ C2 = 0
At x = 0, we are in first portion of beam, consider only first term of deflection equation.
2nd boundary condition, At x =14, y=0 (deflection y=0 at support B). So C1 = ?
At x = 14, whole beam is covered, consider all the terms of deflection equation.

21. 21
Putting 2nd boundary condition, At x =14, y=0. So C1 = ?
Finding final form of slope & deflection equation by putting values of 𝐶1 & 𝐶2
In deflection equation, if any term gives negative answer
after putting x & y values, just discard that term.

22. 22
(a-1) Deflection under Pt. loads: Deflection under 90 kN load i.e. at C where x = 3m
(a-2) Deflection under Pt. loads: Deflection under 60 kN load i.e. at D where x = 9.5m

23. 23
(b) Maximum deflection: It is obtained, when slope = 0.
Max. deflection occur b/w C & D so take only initial two
terms.
Equating slope equation = 0, we have
Putting x = 6.866 m in deflection equation, we
have

24. 24
Problem: Using Macaulay's method, find the deflection of beam at pt. D,
maximum deflection, and slope at pt. A. Take E = 200 GPa and I = 20 ×10-6 m4
Solution:
𝐅𝐢𝐧𝐝𝐢𝐧𝐠 𝐑𝐞𝐚𝐜𝐭𝐢𝐨𝐧𝐬
Let section the beam from left side
Section AC (𝟎 ≤ 𝒙 ≤ 𝟏𝒎)
Consider only 1st term
Section CD (𝟏 ≤ 𝒙 ≤ 𝟐𝒎)
Consider first 2 terms only
Section DB (𝟐 ≤ 𝒙 ≤ 𝟒𝒎)
Consider all 3 terms

25. 25
Integrating moment equation gives slope equation:
Now integrating slope equation gives deflection equation:
Finding of 𝐶1 & 𝐶2 , by using boundary conditions.
1st boundary condition, At x =0, y=0 (deflection y=0 at support A). ⇨ C2 = 0
At x = 0, we are in first portion of beam, consider only first term of deflection equation.
2nd boundary condition, At x =4, y=0 (deflection y=0 at support B). So C1 = ?
At x = 4, whole beam is covered, consider all the terms of deflection equation.

26. 26
Putting 2nd boundary condition in deflection equation, At x =14, y=0. So C1 = ?
Finding final form of slope & deflection equation by putting values of 𝐶1 & 𝐶2
(a) Deflection at Pt. D: where x = 2m

27. 27
(b) Maximum deflection: It is obtained by putting slope eq. = 0. Max. deflection occur b/w C &
D so take only so take only initial two terms.
Equating slope equation = 0, we have
Putting x = 1.958 m in deflection equation, we have
(c) Slope at pt. A (x =0) : ⇨ 𝒚𝒂
′𝑜𝑟 𝜽𝒂= ? ? 𝒚′ is always in radians.

28. 28
Problem: Using Macaulay's method, find the deflection at midspan of beam,
maximum deflection, and slope at pt. A. Take E = 200 GPa and I = 120 ×10-6 m4
Solution:
𝐅𝐢𝐧𝐝𝐢𝐧𝐠 𝐑𝐞𝐚𝐜𝐭𝐢𝐨𝐧𝐬
Let section the beam from left side
• We want such moment equation, in which all
terms should contain x
• When 𝑈𝐷𝐿 → 𝑃𝑡. 𝑙𝑜𝑎𝑑 , x will not come
• So, as a solution, we will extend the UDL both on
top and bottom of beam

29. 29
Section AC (𝟎 ≤ 𝒙 ≤ 𝟐𝒎)
Consider only 1st term
Section CD (𝟐 ≤ 𝒙 ≤ 𝟒𝒎)
Consider first 2 terms only
Section DB (𝟒 ≤ 𝒙 ≤ 𝟖𝒎)
Consider all 3 terms
⇨
Integrating moment equation gives slope equation:
Double integrating moment eq. gives deflection equation:
Finding of 𝐶1 & 𝐶2 , by using boundary conditions.

30. 30
1st boundary condition, At x =0, y=0 (deflection y=0 at support A). ⇨ C2 = 0
At x = 0, we are in first portion of beam, consider only first term of deflection equation.
2nd boundary condition, At x =8, y=0 (deflection y=0 at support B). So C1 = ?
At x = 8, whole beam is covered, consider all the terms of deflection equation.
Finding final form of slope & deflection equation by putting
values of 𝐶1 & 𝐶2

31. 31
(a) Deflection at mid span: where x = 4m
(b) Maximum deflection: It is obtained by putting slope eq. = 0. Max. deflection occur b/w C &
D so take only so take only initial two terms.
⇨
Equating slope equation = 0, we have
Putting x = 3.756 m in deflection equation, we have
(c) Slope at pt. A (x =0) : ⇨ 𝒚𝒂
′𝑜𝑟 𝜽𝒂= ? ? 𝒚′ is always in radians.

32. 32
Problem: Using Macaulay's method, find the deflection at pt. C. Take EI = 8
×104 kN m4
Solution:
𝐅𝐢𝐧𝐝𝐢𝐧𝐠 𝐑𝐞𝐚𝐜𝐭𝐢𝐨𝐧𝐬
Let section the beam from left side
Section AC (𝟎 ≤ 𝒙 ≤ 𝟒𝒎)
Consider only 1st term
Section CB (𝟒 ≤ 𝒙 ≤ 𝟔𝒎)
Consider all 3 terms

33. 33
Integrating moment equation gives slope equation:
Double integrating moment equation gives deflection equation:
Finding of 𝐶1 & 𝐶2 , by using boundary conditions.
1st boundary condition, At x =0, y=0 (deflection y=0 at support A). ⇨ C2 = 0
At x = 0, we are in first portion of beam, consider only first term of deflection equation.
2nd boundary condition, At x =6, y=0 (deflection y=0 at support B). So C1 = ?
At x = 8, whole beam is covered, consider all the terms of deflection equation.
Finding final form of slope & deflection equation by putting values of 𝐶1 & 𝐶2

34. 34
(a) Deflection at pt. C: where x = 4m

35. 35
Practice Questions of Ch. # 12 (Hibbeler 8th ed.)
• Problems = 12- 1,2, 3, 4, 5, 6, 35, 37, 38, 39
• Examples = 12- 1,2, 3, 4, 5, 6

36. From Hibler

37. From Hibler