Downloaded 18 times
![Derivation of the location of the centroids of a plane triangle:
𝑌𝐴 = 𝑌𝑑𝐴 = 𝑦 𝑥2 − 𝑥1 𝑑𝑦
∆ CEF & ∆ CAB similar,
𝑥2 − 𝑥1
𝑏
=
ℎ−𝑦
ℎ
=> 𝑥2 − 𝑥1 =
𝑏
ℎ
(h-y)
𝑌𝐴 = 0
ℎ 𝑏
ℎ
(h−y) ydy =
𝑏
ℎ 0
ℎ
(hy−𝑦2
) dy
=
𝑏
ℎ
[ℎ
𝑦2
2
−
𝑦3
3
]0
ℎ
=
𝑏
ℎ
[ℎ
ℎ2
2
−
ℎ3
3
] =
𝑏
ℎ
x
ℎ3
3
=
𝑏ℎ2
6
𝑌 =
𝑌𝐴
𝐴
=
𝑏ℎ2
6
x
2
𝑏ℎ
=
ℎ
3
A
B
C
dy
𝑥1
𝑥2
h
X
Y
Z
b
y
D E
𝑋 =
𝑋𝐴
𝐴
=
𝑏
2
A =
1
2
bh](https://image.slidesharecdn.com/ce101-200414060557/75/Engineering-Mechanics-63-2048.jpg)
![Segment Area (𝑚𝑚2) 𝑦 (mm) A 𝑦 (𝑚𝑚3)
1 100 x 60 = 6000 30 180000
2 280 x 80 = 22400 60 + 140 = 200 4480000
3 300 x 60 = 18000 400 – 30 = 370 6660000
∑ = 46400 ∑ = 11320000
For Symmetry,
X = 0 mm.
Y =
AY
A
=
11320000
46400
= 244 mm.
y =
60 𝑥 100 𝑥 30+80 𝑥 400−60−60 𝑥 60+
400−60−60
2
+60 𝑥 300 𝑥 [400−30]
60 𝑥 300+80 𝑥 400−60−60 +60 𝑥 100
= 244 mm
60 mm
60 mm
80 mm
300 mm
280mm
100 mm X
2
1
3
400mm
YSolution:](https://image.slidesharecdn.com/ce101-200414060557/75/Engineering-Mechanics-83-2048.jpg)
![Derivation of the moment of inertia of a rectangle with respect to centre of gravity of
the rectangle.
dA = bdy
Ix =
−
ℎ
2
+
ℎ
2
𝑦2 𝑑𝐴 =
−
ℎ
2
ℎ
2
𝑦2 𝑏𝑑𝑦 = b
−
ℎ
2
ℎ
2
𝑦2 𝑑𝑦
= b x [
𝑦3
3
]
−
ℎ
2
ℎ
2
=
𝑏
3
x [
ℎ3
8
- (-
ℎ3
8
)]
=
𝑏
3
x
ℎ3
4
=
𝑏ℎ3
12
∴ Ix =
𝑏ℎ3
12
X
A
B C dy
h
Y
b
D
y](https://image.slidesharecdn.com/ce101-200414060557/75/Engineering-Mechanics-86-2048.jpg)
![Y
dA = hdx
Iy =
−
𝑏
2
+
𝑏
2
𝑥2 𝑑𝐴 =
−
𝑏
2
𝑏
2
𝑥2ℎ𝑑𝑥 = h
−
𝑏
2
𝑏
2
𝑥2 𝑑𝑥
= b x [
𝑥3
3
]
−
𝑏
2
𝑏
2
=
𝑏
3
x [
𝑏3
8
- (-
𝑏3
8
)]
=
ℎ
3
x
𝑏3
4
=
ℎ𝑏3
12
∴ Iy =
ℎ𝑏3
12
A
B C
h
X
b
D
dx
x
Derivation of the moment of inertia of a rectangle with respect to centre of gravity of
the rectangle.](https://image.slidesharecdn.com/ce101-200414060557/75/Engineering-Mechanics-87-2048.jpg)
Engineering Mechanics
The document summarizes problems involving calculating reaction forces at supports of structures using Lame's theorem and the principles of equilibrium. It provides 10 example problems showing the application of these principles to determine unknown reaction forces and loads. Diagrams accompany each problem showing the free body diagram and relevant forces and dimensions. Step-by-step solutions are provided for each problem applying the equations of equilibrium.
In this document
Powered by AI
Introduction to the lecture by Faruq Abdullah on Engineering Mechanics teaching at Dhaka International University.
Discussion on Free Body diagrams, moments, types of supports, and equations by Lame’s Theory.
Problem-solving for reaction forces at supports C, D, A. Involves calculations based on Lame’s theory.
Determines forces T1, T2, T3 in a truss system using Lame's theory with computed angles.
Problem-solving for determining the reaction forces at points A and B in a given structure.
Calculating reaction forces and tension in a structural scenario with multiple supports.
Analysis of reactions at multiple points in a truss and determining forces using Lame's theory.
Interaction of different reactions in a complex truss system and calculations of tension and forces.
Basic definition of truss, assumptions in frame analysis, and initial problem set for joint methods.
Application of the method of joints to find member forces in a given truss system.
Discussion on non-coplanar forces and the components of reactions at different points in 3D structures.
Finding tensions in cables and evaluating the static equilibrium at joints in a hanging structure.
Definition of centroid, derivation of centroid locations for common shapes, including triangles and rectangles.
Solving for centroids of various geometric shapes and understanding the method to calculate them.
Introduction to the concept of moments of inertia and its derivation for different geometries.
Deriving and calculating moments of inertia for multiple sections and composite areas with examples.
Analyzing cables' behavior under load, including determination of tensions, deflections and equations.
More Related Content
Engineering Mechanics
- 1. Engineering Mechanics Faruque Abdullah Lecturer Dept.of Civil Engineering Dhaka International University
- 2.
- 3. Moment: The bending momentat a section of a structure is the algebraic sum of the moments produced by all the external forces on one side of the section. M1 M2 M3M4 P1 P2 P3 P4 P5 P5y P5x P6 P6y P6x P7 P8 ∑ MA = 0 => P1 x X - RB x L = 0 A B X RB L X ∑ MA = 0 => -P2 x X - RB x L = 0 ∑ MA = 0 => P3 x X - RB x L = 0 X ∑ MA = 0 => P4 x X - RB x L = 0
- 4. Types of Supportsand Reaction Forces (2D)
- 5. Puzzle Clue ∑ Fx= 0 ∑ Fy = 0 ∑ M = 0 Lame’s Theory
- 6. Problem-1: Find thereaction forces at support C and D. 600 300 Q = 12 lb D B C
- 7. Solution: According toLame’s theory, 𝑅 𝐶 𝑆𝑖𝑛(900+300) = 𝑅 𝐷 𝑆𝑖𝑛(900+600) = 𝑄 𝑆𝑖𝑛(900) = 12 1 =12 ∴ 𝑅 𝐶 = 12 * 𝑆𝑖𝑛(900 + 300) = 6 3 lb ∴ 𝑅 𝐷 = 12 * 𝑆𝑖𝑛(900 + 600) = 6 lb 600300 Q = 12 lb 𝑅 𝐷 𝑅 𝐶 600300 𝑅 𝐶𝑅 𝐷 Q = 12 lb
- 8. Problem-2: Find thereaction forces at support A if the weight of joist is 20 lb T 300 450 20 lb A RA RAx RA𝑦 T Sin300 B
- 9. Solution: ∑ MA= 0 => 20 x 6 Cos 450 - T Sin300 x 12 = 0 => T = 20 x 6 Cos 450 12 x Sin300 => T = 10 2 6 Cos 450 T 300 450 20 lb A RA RAx RA𝑦 T Sin300 B
- 10. ∑ Fx =0 => T Cos150 - RAx = 0 => RAx = 10 2 * Cos150 = 13.66 lb ∑ Fy = 0 => T Sin150 + 20 - RA𝑦 = 0 => RA𝑦 = 10 2 * Sin150 +20 => RA𝑦 = 23.66 lb 𝑅 𝐴 = RAx 2 + RA𝑦 2 = 13.662 + 23.662 = 27.32 lb T 300 450 20 lb A RA RAx RA𝑦 T Cos150 T Sin150 B 150
- 11. Problem-3: Find outthe value of the forces T1, T2, T3 & the angle θ. 40 N 50 N A B T1 T2 T3 350 θ
- 12. Solution: According toLame’s theory, 𝑇1 𝑆𝑖𝑛(900) = 𝑇2 𝑆𝑖𝑛(900+550) = 40 𝑆𝑖𝑛(900+350) = 48.8 ∴𝑇1 = 48.83 * 𝑆𝑖𝑛(900 ) = 48.83 N ∴𝑇2 = 48.83 * 𝑆𝑖𝑛(900 + 550) =28 N 40 N 50 N A B T1 T2 T3 350 θ350𝑇1 40 N 550 𝑇2
- 13. Solution: According toLame’s theory, 𝑇2 𝑆𝑖𝑛(900+900 − 𝜃) = 𝑇3 𝑆𝑖𝑛(900) = 50 𝑆𝑖𝑛(900+ 𝜃) => 28 𝑆𝑖𝑛𝜃 = 𝑇3 1 = 50 𝐶𝑜𝑠𝜃 Now, 28 𝑆𝑖𝑛𝜃 = 50 𝐶𝑜𝑠𝜃 or, 𝑆𝑖𝑛𝜃 𝐶𝑜𝑠𝜃 = 28 50 or, tanθ = 28 50 or, θ = 𝑡𝑎𝑛−1( 28 50 ) = 29.240 ∴ 𝑇3 = 50 𝐶𝑜𝑠(29.240) * 1 = 57.3 N 40 N 50 N A B T1 T2 T3 350 θ 𝑇3 50 N 900 − 𝜃 𝑇2 θ
- 14. Problem-4: Find outthe reaction at A and B
- 15. ∑ Fy =0 => 𝑅 𝐴 + 𝑅 𝐵 - 40 = 0 => 𝑅 𝐴 + 30 - 40 = 0 => 𝑅 𝐴 = 10 N Solution: ∑ MA = 0 => 40 x 3 4 L - 𝑅 𝐵 x L = 0 => 𝑅 𝐵 = 40 x 3 4 L 𝐿 => 𝑅 𝐵 = 40 x 3 4 = 30 N
- 16. T 2 1 2′ 2′ 2′ 200lb 100 lb A B Problem-5: Find out the reaction at A and also find the value of the force T.
- 17. ∑ Fy =0 => 𝑅 𝐴𝑦 + T x 2 5 - 200 -100 = 0 => 𝑅 𝐴𝑦 + 279.51 x 2 5 - 200 -100 = 0 => 𝑅 𝐴𝑦 = 50 N ∑ Fx = 0 => 𝑅 𝐴𝑥 - T x 1 5 = 0 => 𝑅 𝐴𝑥 + 279.51 x 1 5 = 0 => 𝑅 𝐴𝑥 = 125 N Solution: ∑ MA = 0 => 200 x 2 + 100 x 6 - T x 2 5 x 4 = 0 => T = 400+600 2 5 x 4 =>T = 279.51 N T x 2 5 T x 1 5 T 2 1 2′ 2′ 2′ 200 lb 100 lb A B 5 RAx RA𝑦
- 18. Problem-6: Find outthe reaction at A and also find the value of the force T. 30 N A B C 300
- 19. Solution: According toLame’s theory, 𝑇 𝑆𝑖𝑛(900) = 𝑅 𝐵 𝑆𝑖𝑛(900+ 600) = 30 𝑆𝑖𝑛(900+300) = 20 3 ∴𝑇1 = 20 3 * 𝑆𝑖𝑛(900) = 20 3 N ∴𝑇2 = 20 3 * 𝑆𝑖𝑛(900 + 600) = 10 3 N 30 N A B C 300 30 N T 𝑅 𝐵 C 600 300 T 30 N 600 𝑅 𝐵
- 20. Problem-7: Find outthe reaction at A, B, C & F. 600D E 100 N 100 N A B CF
- 21. Solution: According toLame’s theory, 𝑅 𝐶 𝑆𝑖𝑛(900+ 600) = 𝑅 𝐹 𝑆𝑖𝑛(900+ 300) = 100 𝑆𝑖𝑛(900) = 100 ∴𝑅 𝐶 = 100 * 𝑆𝑖𝑛(900 + 600 ) = 50 N ∴𝑅 𝐹 = 100 * 𝑆𝑖𝑛(900 + 300) = 50 3 N 600D E 100 N 100 N A B CF 𝑅 𝐶 100 N 300 𝑅 𝐹 600
- 22. 600D E 100 N 100 N A B CF ∑Fy = 0 => 𝑅 𝐵 x Sin300 - 𝑅 𝐹 x Sin600 -100 = 0 => 𝑅 𝐵 x Sin300 -50 3 x Sin600 -100 = 0 => 𝑅 𝐵 = 350 N ∑ Fx = 0 => 𝑅 𝐵 x Cos300 + 𝑅 𝐹 x Cos 600 - 𝑅 𝐴 = 0 => 𝑅 𝐴 = 350 x Cos300 +50 3 x Cos 600 => 𝑅 𝐴𝑥 = 346.4 N 𝑅 𝐵 100 N 300 𝑅 𝐴 𝑅 𝐹 600
- 23. Problem-8: A rollerweighting 2000 N rests on a inclined bar weighting 800 N as shown in Figure. Assuming the weight of bar negligible, determine the reactions at D and C and reaction in bar AB. A B 300 C D E
- 24. A B 300 C D E Solution: Accordingto Lame’s theory, 𝑅 𝐴 𝑆𝑖𝑛(900+ 600) = 𝑅 𝐸 𝑆𝑖𝑛(900) = 2000 𝑆𝑖𝑛(900+300) = 2309.4 ∴𝑅 𝐴 = 2309.4 * 𝑆𝑖𝑛(900 + 600 ) = 1154.7 N ∴𝑅 𝐸 = 2309.4 * 𝑆𝑖𝑛(900) = 2309.4 N 𝑅 𝐸 2000 N 600 𝑅 𝐴
- 25. Solution: ∑ Fx =0 => 𝑅 𝐵Cos600 - 𝑅 𝐶𝑥 = 0 =>𝑅 𝐶𝑥 = 2309.4 x Cos600 = 1154.7 N ∑ MC = 0 => 𝑅 𝐷 x 5 Cos300 - 800 x 2.5 Cos300 - 𝑅 𝐸 x 2 = 0 =>𝑅 𝐷 = 1466.67 N ∑ Fy = 0 => 𝑅 𝐷 - 800 - 𝑅 𝐸 Sin600 + 𝑅 𝐶𝑦 = 0 =>𝑅 𝐶𝑦 = 1333.33 N 300 C D E 800 N 𝑅 𝐸 𝑅 𝐷 𝑅 𝐶𝑦 𝑅 𝐶𝑥600 𝑅 𝐸Cos600 𝑅 𝐸 2000 N 𝑅 𝐴 600
- 26. Problem-9: Determine thehorizontal and vertical force components acting at the pin connections B & C of the loaded frame as shown in the figure. 500 N 8′ 2′ 4′ 8′ 2′ A B C D E F
- 27. Solution: ∑ MA =0 => 500 x 10 - 𝑅 𝐷 x 14 = 0 => 𝑅 𝐷 = 357.14 N ∑ Fx = 0 => 𝑅 𝐴𝑥 - 𝑅 𝐷 = 0 => 𝑅 𝐴𝑥 = 𝑅 𝐷 = 357.14 N ∑ Fy = 0 => 𝑅 𝐴𝑦 - 500 = 0 => 𝑅 𝐴𝑦 = 500 N 500 N 8′ 2′ 4′ 8′ 2′ A B C D E F 𝑅 𝐴𝑦 𝑅 𝐴𝑥 𝑅 𝐷
- 28. ∑ MC =0 => 500 x 10 - 𝑅 𝐸𝑦 x 8 = 0 => 𝑅 𝐸𝑦 = 625 N ∑ Fx = 0 => 𝑅 𝐶𝑥 - 𝑅 𝐸𝑥 = 0 => 𝑅 𝐶𝑥 = 𝑅 𝐸𝑥 ∑ Fy = 0 => 𝑅 𝐶𝑦 + 𝑅 𝐸𝑦 - 500 = 0 => 𝑅 𝐶𝑦 + 625 - 500 = 0 => 𝑅 𝐶𝑦 = -125 N 500 N C E F 𝑅 𝐶𝑥 𝑅 𝐶𝑦 𝑅 𝐸𝑥 𝑅 𝐸𝑦 8′ 2′
- 29. ∑ MB =0 => - 𝑅 𝐴𝑥 x 4 - 𝑅 𝐶𝑥 x 8 - 𝑅 𝐷 x 10 = 0 => - 357.14 x 4 - 𝑅 𝐶𝑥 x 8 - 357.14x 10 = 0 => 𝑅 𝐶𝑥 = - 625 N ∑ Fx = 0 => 𝑅 𝐴𝑥 - 𝑅 𝐵𝑥 - 𝑅 𝐶𝑥 - 𝑅 𝐷 = 0 => 𝑅 𝐵𝑥 = 357.14 – (-625) -357.14 => 𝑅 𝐵𝑥 = 625 N ∑ Fy = 0 => 𝑅 𝐴𝑦 - 𝑅 𝐵𝑦 - 𝑅 𝐶𝑦 = 0 => 𝑅 𝐵𝑦 = 500 – (-125) = 625 N 4′ 8′ 2′ A B C D 𝑅 𝐴𝑦 𝑅 𝐴𝑥 𝑅 𝐷 𝑅 𝐵𝑥 𝑅 𝐵𝑦 𝑅 𝐶𝑥 𝑅 𝐶𝑦
- 30. Problem-10: Determine thehorizontal and vertical force components acting at the pin connections A & C of the loaded frame as shown in the figure. 2′ 8′ 10 Kip-ft A B C
- 31. Solution: ∑ MB =0 => - 10 - 𝑅 𝐶 x 4 = 0 => 𝑅 𝐶 = - 1.67 N ∑ Fy = 0 => 𝑅 𝐴𝑦 + 𝑅 𝐶 = 0 => 𝑅 𝐴𝑦 = - 𝑅 𝐶 = - (-1.67) = 1.67 N ∑ MB = 0 => 𝑅 𝐴𝑦 x 2 - 𝑅 𝐴𝑥 x 2 - 𝑅 𝐶 x 4 = 0 => 1.67 x 2 - 𝑅 𝐴𝑥 x 2 - (-1.67) x 4 = 0 => 𝑅 𝐴𝑥 = - 1.67 N 2′ 4′ 10 Kip-ft A B C 𝑅 𝐶 𝑅 𝐴𝑦 𝑅 𝐴𝑥 4′ 10 Kip-ft B 2′ 𝑅 𝐵𝑥 𝑅 𝐵𝑦 𝑅 𝐶 C
- 32.
- 33. Assumption Made InTruss Frame Analysis: Members are connected at the joints through pin connections. All the members have only axial force i.e. either tensile or compressive. All the external forces are applied only at the joints.
- 34. Problem-11: Determine allthe member forces by Method of Joint Method. 6′ 4 @ 8′ = 32′ 10 K 𝐿0 10 K15 K 𝐿1 𝐿2 𝐿3 𝐿4 𝑈1 𝑈2 𝑈3
- 35. 6′ 4 @ 8′= 32′ 10 K 𝐿0 10 K15 K 𝐿1 𝐿2 𝐿3 𝐿4 𝑈1 𝑈2 𝑈3 𝑅 𝐿4 𝑅 𝐿𝑜 Solution: ∑ ML0 = 0 => 10 x 8 + 15 x 16 + 10 x 24 - 𝑅 𝐶 x 32 = 0 => 𝑅 𝐿4 = 17.5 K = 𝑅 𝐿0 Free Joint of 𝐿0: ∑ Fy = 0 => 𝐿0 𝑈1 x 6 10 + 𝑅 𝐿0 = 0 =>𝐿0 𝑈1 = - 29.17 K (C) ∑ FX = 0 => 𝐿0 𝑈1 x 8 10 + 𝐿0 𝐿1 = 0 => 𝐿0 𝐿1 = - (- 29.17 x 8 10 ) = 23.33 K (T) 6′ 8′ 𝐿0 𝐿1 𝑅 𝐿0 𝑈1
- 36. 6′ 4 @ 8′= 32′ 10 K 𝐿0 10 K15 K 𝐿1 𝐿2 𝐿3 𝐿4 𝑈1 𝑈2 𝑈3 𝑅 𝐿4 𝑅 𝐿𝑜 Free Joint of 𝐿1: ∑ Fy = 0 => 𝐿1 𝑈1 = 0 ∑ FX = 0 => 𝐿0 𝐿1 - 𝐿1 𝐿2 = 0 =>𝐿1 𝐿2= 𝐿0 𝐿1 = 23.33 K (T) 𝑈1 6′ 8′ 𝐿0 𝐿2 𝐿1 8′
- 37. 6′ 4 @ 8′= 32′ 10 K 𝐿0 10 K15 K 𝐿1 𝐿2 𝐿3 𝐿4 𝑈1 𝑈2 𝑈3 𝑅 𝐿4 𝑅 𝐿𝑜 Free Joint of 𝑈1: ∑ Fy = 0 => - 𝐿0 𝑈1 x 6 10 - 10 - 𝐿2 𝑈1 x 6 10 = 0 => 𝐿2 𝑈1 x 6 10 = - (-29.17) x 6 10 -10 => 𝐿2 𝑈1 = 12.5 K (T) ∑ FX = 0 => 𝑈1 𝑈2 + 𝐿2 𝑈1 x 8 10 - 𝐿0 𝑈1 x 8 10 = 0 => 𝑈1 𝑈2 = -29.17 x 8 10 - 12.5 x 8 10 => 𝑈1 𝑈2 = -33.34 K (C) 𝑈2 6′ 8′ 𝐿1 𝐿0 𝐿2 8′ 𝑈1 10 K
- 38. 6′ 4 @ 8′= 32′ 10 K 𝐿0 10 K15 K 𝐿1 𝐿2 𝐿3 𝐿4 𝑈1 𝑈2 𝑈3 𝑅 𝐿4 𝑅 𝐿𝑜 Free Joint of 𝑈2: ∑ Fy = 0 => 𝐿2 𝑈2 + 15 = 0 => 𝐿2 𝑈2 = -15 K (C) ∑ FX = 0 => 𝑈1 𝑈2 - 𝑈2 𝑈3 = 0 => 𝑈2 𝑈3 = 𝑈1 𝑈2 = -33.34 K (C) 𝑈1 6′ 8′ 𝐿2 8′ 15 K 𝑈2 𝑈3
- 39. 6′ 4 @ 8′= 32′ 10 K 𝐿0 10 K15 K 𝐿1 𝐿2 𝐿3 𝐿4 𝑈1 𝑈2 𝑈3 𝑅 𝐿4 𝑅 𝐿𝑜 Free Joint of 𝐿2: ∑ Fy = 0 => 𝐿2 𝑈1 x 6 10 + 𝐿2 𝑈2 + 𝐿2 𝑈3 x 6 10 = 0 => 𝐿2 𝑈3 = (- 12.5 x 6 10 - 15) x 10 6 => 𝐿2 𝑈3 = -37.5 K (C) ∑ FX = 0 => -𝐿2 𝑈1 x 8 10 - 𝐿1 𝐿2 + 𝐿2 𝑈3 x 8 10 + 𝐿2 𝐿3= 0 => 𝐿2 𝐿3 = 12.5 x 8 10 + 23.33 – (-37.5) x 8 10 => 𝐿2 𝐿3 = 63.33 K (T) Proceed the same procedure to find out the other member forces. 𝑈2 6′ 8′ 𝐿1 𝐿3 𝐿2 8′ 𝑈3𝑈1
- 40. Problem-12: Determine theforce in members DF, DE and CE of the stadium truss shown. 2 K 1 K 2 K 2 K A B C D E F G I K M H J L N 8′8′ 6′ 8′ 6′ 6′ 15′
- 41. 2 K 1 K 2K 2 K A B C D E F G I K M H J L N 8′ 8′ 6′8′ 6′ 6′ 15′ 1 1 2 K 2 K 2 K A B C D E 8′ 8′ 4′ 1 F 1
- 42. Solution: ∑ MD =0 => -2 x 16 – 2 x 8 – c x 4 = 0 => c = - 12 K (C). ∑ MA = 0 => 2 x 8 + 2 x 16 + b x 16 = 0 => b = - 3 K (C). ∑ ME = 0 => a x 8 82+ 22 x 4 - 2 x 16 – 2 x 8 = 0 => a = - 12.37 K (T). 8′ 2′ 2 K 2 K 2 K A B C D E 8′8′ 4′ Fa b c 1 1
- 43.
- 44.
- 45. It has componentsalong y and z axis because the member does not require x- axis to describe it’s position. Components of reaction at point A is 𝑅 𝐴𝑦 & 𝑅 𝐴𝑧. Components of reaction at point B is 𝑅 𝐵𝑦 & 𝑅 𝐵𝑧. X Y Z X′ Y′ Z′ A (4,0) B (0,5) 𝑅 𝐴𝑧 𝑅 𝐴𝑦 𝑅 𝐵𝑧 𝑅 𝐵𝑦 5′
- 46. It has componentsalong x, y and z axis because the member requires x, y & z-axis to describe it’s position. Components of reaction at point A is 𝑅 𝐴𝑥, 𝑅 𝐴𝑦 & 𝑅 𝐴𝑧 . Components of reaction at point B is 𝑅 𝐵𝑥, 𝑅 𝐵𝑦 & 𝑅 𝐵𝑧. X Y Z X′ Y′ Z′ A (2,4,0) B (0,0,5) 𝑅 𝐴𝑧 𝑅 𝐴𝑦 𝑅 𝐵𝑦 𝑅 𝐵𝑧 5′ 𝑅 𝐴𝑥 𝑅 𝐵𝑥
- 47. Problem-13: AE &CE steel frame is hanging with a cable DE. From E a load of 1000 lb has been suspended. The dimension of AB = BC = BD = 6′ & BE = 8′ . Find the tension in the cable and the force in each timber. X Y Z X′ Y′ Z′ 1000 lb 6′ 8′ E A B C D
- 48. Solution: Components of DE, 𝑇𝑥= T x 8 82+ 62 = 8 10 T 𝑇𝑦 = 0 𝑇𝑧 = T x 6 82+ 62 = 6 10 T Taking moment about y-axis at A, 𝑇𝑧 x 8 – 1000 x 8 = 0 => 6 10 T = 1000 => T = 1666.67 lb. XY Z 1000 lb E (8,0,0) A (0,6,0) B (0,0,0) C (0,6,0) D (0,0,6) 𝑅 𝐴𝑧 𝑅 𝐴𝑦 𝑅 𝐴𝑥 𝑅 𝐶𝑧 𝑅 𝐶𝑦 𝑅 𝐶𝑥 𝑅 𝐷𝑧 𝑅 𝐷𝑦 𝑅 𝐷𝑥 𝑇𝑧 𝑇𝑥 6′ 8′
- 49. Components of AE, 𝑅𝐴𝐸𝑥 = 𝑅 𝐴𝐸 x 8 82+ 62 = 8 10 𝑅 𝐴𝐸 𝑅 𝐴𝐸𝑦 = 𝑅 𝐴𝐸 x 6 82+ 62 = 6 10 𝑅 𝐴𝐸 𝑅 𝐴𝐸𝑧 = 0 Components of CE, 𝑅 𝐶𝐸𝑥 = 𝑅 𝐶𝐸 x 8 82+ 62 = 8 10 𝑅 𝐶𝐸 𝑅 𝐶𝐸𝑦 = 𝑅 𝐶𝐸 x 6 82+ 62 = 6 10 𝑅 𝐶𝐸 𝑅 𝐶𝐸𝑧 = 0 XY Z 1000 lb E (8,0,0) A (0,6,0) B (0,0,0) C (0,6,0) D (0,0,6) 𝑅 𝐴𝑧 𝑅 𝐴𝑦 𝑅 𝐴𝑥 𝑅 𝐶𝑧 𝑅 𝐶𝑦 𝑅 𝐶𝑥 𝑅 𝐷𝑧 𝑅 𝐷𝑦 𝑅 𝐷𝑥 𝑇𝑧 𝑇𝑥 6′ 8′
- 50. ∑ Fy =0 => 𝑅 𝐶𝐸𝑦 - 𝑅 𝐴𝐸𝑦 = 0 => 6 10 𝑅 𝐶𝐸 = 6 10 𝑅 𝐴𝐸 => 𝑅 𝐶𝐸 = 𝑅 𝐴𝐸 ∑ Fx = 0 => 𝑅 𝐴𝐸𝑥 + 𝑅 𝐶𝐸𝑥 - 𝑇𝑥 = 0 => 8 10 𝑅 𝐴𝐸 + 8 10 𝑅 𝐶𝐸 = 8 10 T => 2 x 8 10 𝑅 𝐴𝐸 = 8 10 T => 𝑅 𝐴𝐸 = 𝑅 𝐶𝐸 = 𝑇 2 = 1666.67 2 = 833.33 lb. XY Z 1000 lb E (8,0,0) A (0,6,0) B (0,0,0) C (0,6,0) D (0,0,6) 𝑅 𝐴𝑧 𝑅 𝐴𝑦 𝑅 𝐴𝑥 𝑅 𝐶𝑧 𝑅 𝐶𝑦 𝑅 𝐶𝑥 𝑅 𝐷𝑧 𝑅 𝐷𝑦 𝑅 𝐷𝑥 𝑇𝑧 𝑇𝑥 6′ 8′
- 51. Problem-14: The 1000lb mass steel pole is supported by a ball and socket joint at A and is retained by two cables shown. Determine the tensions in the cables and the total force acting on the joint A. X Y Z 𝑇1 𝑇2 10′ 10′ 20′ 16′ 2000 lb A
- 52. Solution: Components of 𝑇1, 𝑇1𝑥= 𝑇1 x 12 122+ 162+ 302 = 12 10 13 𝑇1 𝑇1𝑦 = 𝑇1 x 16 122+ 162+ 302 = 16 10 13 𝑇1 𝑇1𝑧 = 𝑇1 x 30 122+ 162+ 302 = 30 10 13 𝑇1 Components of 𝑇2, 𝑇2𝑥 = 0 𝑇2𝑦 = 𝑇2 x 20 202+ 202 = 20 20 2 𝑇2 = 𝑇2 2 𝑇2𝑧 = 𝑇2 x 20 202+ 202 = 20 20 2 𝑇2 = 𝑇2 2 X Y Z 𝑇1 𝑇2 10′ 10′ 20′ 16′ 2000 lb A 𝑇1 𝑧 𝑇1𝑥 𝑇1𝑦 𝑇2𝑧 𝑇2𝑦 1000 lb 𝑅 𝐴𝑧 𝑅 𝐴𝑥 𝑅 𝐴𝑦
- 53. Taking moment abouty-axis at A, ∑ MA−y = 0 => 2000 x 40 - 𝑇1𝑥 x 30 = 0 => 2000 x 40 - 12 10 13 𝑇1x 30 = 0 => 𝑇1 = 8013 lb Taking moment about x-axis at A, ∑ MA−𝑥 = 0 => 𝑇1𝑦 x 30 - 𝑇2𝑧 x 20= 0 => 16 10 13 𝑇1 x 30 - 𝑇2 2 x 20 = 0 => 𝑇2 = 7543 lb X Y Z 𝑇1 𝑇2 10′ 10′ 20′ 16′ 2000 lb A 𝑇1 𝑧 𝑇1𝑥 𝑇1𝑦 𝑇2𝑧 𝑇2𝑦 1000 lb 𝑅 𝐴𝑧 𝑅 𝐴𝑥 𝑅 𝐴𝑦
- 54. ∑ F 𝑍= 0 => 1000 + 𝑇1𝑧 + 𝑇2𝑧 - 𝑅 𝐴𝑧= 0 => 1000 + 13 10 13 𝑇1 + 𝑇2 2 - 𝑅 𝐴𝑧= 0 => 𝑅 𝐴𝑧 = 1000 + 30 10 13 x 8013 + 7543 2 => 𝑅 𝐴𝑧 = 13001 lb. ∑ F 𝑋 = 0 => 𝑅 𝐴𝑥 - 𝑇1𝑥 + 2000= 0 => 𝑅 𝐴𝑥 = 12 10 13 𝑇1 - 2000 => 𝑅 𝐴𝑥 = 12 10 13 x 8013 - 2000 => 𝑅 𝐴𝑥 = 666.67 lb X Y Z 𝑇1 𝑇2 10′ 10′ 20′ 16′ 2000 lb A 𝑇1 𝑧 𝑇1𝑥 𝑇1𝑦 𝑇2𝑧 𝑇2𝑦 1000 lb 𝑅 𝐴𝑧 𝑅 𝐴𝑥 𝑅 𝐴𝑦
- 55. ∑ FY =0 => 𝑅 𝐴𝑦 - 𝑇2𝑦 + 𝑇1𝑦= 0 => 𝑅 𝐴𝑦 = 𝑇2 2 - 16 10 13 𝑇1 => 𝑅 𝐴𝑦 = 7543 2 - 16 10 13 x 8013 => 𝑅 𝐴𝑦 = 1778 lb. ∴ 𝑅 𝐴 = 𝑅 𝐴𝑥 2 + 𝑅 𝐴𝑦 2 + 𝑅 𝐴𝑧 2 = 666.672 + 17782 + 130012 = 13139 lb. X Y Z 𝑇1 𝑇2 10′ 10′ 20′ 16′ 2000 lb A 𝑇1 𝑧 𝑇1𝑥 𝑇1𝑦 𝑇2𝑧 𝑇2𝑦 1000 lb 𝑅 𝐴𝑧 𝑅 𝐴𝑥 𝑅 𝐴𝑦
- 56. Problem-15: A 10m pile is acted upon by a force of 8.4 kN. It is held by a ball and socket at A and by the two cables BD and BE. Neglecting the weight of the pole, determine the tension in each cable and the reaction at A. X Y Z 𝑇1 𝑇2 7 m 8.4 kN A 3 m B C D E
- 57. Solution: Components of 𝑇1, 𝑇1𝑥= 𝑇1 x 6 62+ 62+ 72 = 6 11 𝑇1 𝑇1𝑦 = 𝑇1 x 6 62+ 62+ 72 = 6 11 𝑇1 𝑇1𝑧 = 𝑇1 x 7 62+ 62+ 72 = 7 11 𝑇1 Components of 𝑇2, 𝑇2𝑥 = 𝑇1 x 6 62+ 62+ 72 = 6 11 𝑇1 𝑇2𝑦 = 𝑇2 x 6 62+ 62+ 72 = 6 11 𝑇2 𝑇2𝑧 = 𝑇2 x 7 62+ 62+ 72 = 7 11 𝑇2 X Y Z 𝑇1 𝑇2 7 m 8.4 kN A 3 m B C D E 𝑇1 𝑧 𝑇1𝑥 𝑇1𝑦 𝑇2𝑧 𝑇2𝑦 𝑇2𝑥 𝑅 𝐴𝑧 𝑅 𝐴𝑥 𝑅 𝐴𝑦
- 58. Taking moment abouty-axis at A, ∑ MA−y = 0 => 𝑇1𝑥 x 7 - 𝑇2𝑥 x 7 = 0 => 6 11 𝑇1- 6 11 𝑇2= 0 => 𝑇1 = 𝑇2 Taking moment about x-axis at A, ∑ MA−𝑥 = 0 => 𝑇1𝑦 x 7 + 𝑇2𝑦 x 7 - 8.4 x 10= 0 => 6 11 𝑇1 x 7 + 6 11 𝑇2x 7 -84 = 0 => 2 x 6 11 x 7 x 𝑇1 = 84 => 𝑇1 = 11 kN = 𝑇2. X Y Z 𝑇1 𝑇2 7 m 8.4 kN A 3 m B C D E 𝑇1 𝑧 𝑇1𝑥 𝑇1𝑦 𝑇2𝑧 𝑇2𝑦 𝑇2𝑥 𝑅 𝐴𝑧 𝑅 𝐴𝑥 𝑅 𝐴𝑦
- 59. ∑ F 𝑍= 0 => 𝑇1𝑧 + 𝑇2𝑧 - 𝑅 𝐴𝑧= 0 => 7 11 𝑇1+ 7 11 𝑇2 - 𝑅 𝐴𝑧= 0 => 𝑅 𝐴𝑧 = 2 x 7 11 x 11 => 𝑅 𝐴𝑧 = 14 lb. ∑ F 𝑋 = 0 => 𝑅 𝐴𝑥 - 𝑇1𝑥 + 𝑇2𝑥= 0 => 𝑅 𝐴𝑥 = 0 X Y Z 𝑇1 𝑇2 7 m 8.4 kN A 3 m B C D E 𝑇1 𝑧 𝑇1𝑥 𝑇1𝑦 𝑇2𝑧 𝑇2𝑦 𝑇2𝑥 𝑅 𝐴𝑧 𝑅 𝐴𝑥 𝑅 𝐴𝑦
- 60. ∑ FY =0 => 𝑅 𝐴𝑦 + 𝑇2𝑦 + 𝑇1𝑦 - 8.4 = 0 => 𝑅 𝐴𝑦 = 6 11 𝑇2 + 6 11 𝑇1 - 8.4 => 𝑅 𝐴𝑦 = 2 x 6 11 x 11 – 8.4 => 𝑅 𝐴𝑦 = 3.6 lb. ∴ 𝑅 𝐴 = 𝑅 𝐴𝑥 2 + 𝑅 𝐴𝑦 2 + 𝑅 𝐴𝑧 2 = 02 + 3.62 + 142 = 14.46 lb. X Y Z 𝑇1 𝑇2 7 m 8.4 kN A 3 m B C D E 𝑇1 𝑧 𝑇1𝑥 𝑇1𝑦 𝑇2𝑧 𝑇2𝑦 𝑇2𝑥 𝑅 𝐴𝑧 𝑅 𝐴𝑥 𝑅 𝐴𝑦
- 61.
- 62. Centroid: The centerof mass of a geometric object of uniform density.
- 63. Derivation of thelocation of the centroids of a plane triangle: 𝑌𝐴 = 𝑌𝑑𝐴 = 𝑦 𝑥2 − 𝑥1 𝑑𝑦 ∆ CEF & ∆ CAB similar, 𝑥2 − 𝑥1 𝑏 = ℎ−𝑦 ℎ => 𝑥2 − 𝑥1 = 𝑏 ℎ (h-y) 𝑌𝐴 = 0 ℎ 𝑏 ℎ (h−y) ydy = 𝑏 ℎ 0 ℎ (hy−𝑦2 ) dy = 𝑏 ℎ [ℎ 𝑦2 2 − 𝑦3 3 ]0 ℎ = 𝑏 ℎ [ℎ ℎ2 2 − ℎ3 3 ] = 𝑏 ℎ x ℎ3 3 = 𝑏ℎ2 6 𝑌 = 𝑌𝐴 𝐴 = 𝑏ℎ2 6 x 2 𝑏ℎ = ℎ 3 A B C dy 𝑥1 𝑥2 h X Y Z b y D E 𝑋 = 𝑋𝐴 𝐴 = 𝑏 2 A = 1 2 bh
- 64. b h 𝑋 = 𝑏 2 𝑌 = ℎ 2 X Y X Y 𝑋= r 𝑌 = 4𝑟 3𝜋 d = 2r
- 65. X Y 𝑋 = 4𝑟 3𝜋 𝑌 = 4𝑟 3𝜋 r X Y α 𝑋= 2 3 𝑟 𝑆𝑖𝑛∝ ∝ r
- 66. Problem-16: Find outthe centroid of the following object. 2.5 m 5 m 2.5 m 2.5 m 5 m 5 m X Y
- 67. Segment Area (𝑚2 )𝑥 (m) 𝑦 (m) A 𝑥 (𝑚3 ) A 𝑦 (𝑚3 ) 1 𝜋𝑟2 2 = 𝜋 𝑥 2.52 2 = 9.82 -2.5 - 5 - 4𝑟 3𝜋 = -8.56 0 -84.06 0 2 10 x 5 = 50 -2.5 0 -125 0 3 0.5 x 5 x 5 = 12.5 0 2.5 + ℎ 3 = 4.17 0 52.125 ∑ = 72.32 ∑ = - 209.06 ∑ = 52.125 2.5 m 5 m 2.5 m 2.5 m 5 m 5 m X Y 𝑋 = 𝐴 𝑥 𝐴 = −209.06 72.32 = - 2.89 m 𝑌 = 𝐴 𝑦 𝐴 = 52.125 72.32 = 0.72 m 3 21 Solution:
- 68. Problem-17: Find outthe centroid of the following object. Y 12 m 24 m 12 m 600600 X 24 x Sin600 = 12 3 m
- 69. Segment Area (𝑚2 )𝑥 (m) 𝑦 (m) A 𝑥 (𝑚3 ) A 𝑦 (𝑚3 ) Half Circle 𝜋𝑟2 2 = 𝜋 𝑥 242 2 = 904.78 - 4𝑟 3𝜋 = - 10.19 0 - 9219.71 0 Triangle 0.5 x 24 x 12 3 = - 249.42 - 12 3 3 = - 4 3 0 + 1728.03 0 ∑ = 655.36 ∑ = - 7491.68 ∑ = 0 𝑋 = 𝐴 𝑥 𝐴 = − 7491.68 655.36 = - 11.43 m 𝑌 = 𝐴 𝑦 𝐴 = 0 m Y 12 m 24 m 12 m 600600 X 24 x Sin600 = 12 3 m Solution:
- 70. Problem-18: Find outthe centroid of the following object. 2 m 8 m 6 m 2 m 9 m 1 2 3 Y X
- 71. Segment Area (𝑚2 )𝑥 (m) 𝑦 (m) A 𝑥 (𝑚3 ) A 𝑦 (𝑚3 ) 1 - 𝜋𝑟2 2 = - 𝜋 𝑥 42 2 = - 25.13 - 4𝑟 3𝜋 = - 1.7 6 + 2 + 4 = 12 42.72 - 301.56 2 9 x 12 = 108 - 9 2 = - 4.5 6 + 12 2 = 12 -486 1296 3 0.5 x 9 x 6 = 27 - 2 3 x 9 = - 6 2 3 x 6 = 4 -162 108 ∑ = 109.87 ∑ = - 600.28 ∑ = 1102.44 𝑋 = 𝐴 𝑥 𝐴 = − 600.28 109.87 = - 5.46 m 𝑌 = 𝐴 𝑦 𝐴 = 1102.44 109.87 = 10.03 m 2 m 8 m 6 m 2 m 9 m 1 2 3 Y X Solution:
- 72. Problem-19: Find outthe centroid of the following object. 600 Y X 600
- 73. Segment Area (𝑚2)𝑥 (m) 𝑦 (m) A 𝑥 (𝑚3) A 𝑦 (𝑚3) Arc OAB 𝜋𝑟2 6 = 𝜋 𝑥 62 6 = 18.85 2 3 x 4 𝑥 𝑆𝑖𝑛300 30 𝑥 𝜋 180 = 3.82 0 72 0 Arc OCD - 𝜋𝑟2 6 = - 𝜋 𝑥 42 6 = - 8.38 2 3 x 4 𝑥 𝑆𝑖𝑛300 30 𝑥 𝜋 180 = 2.54 0 - 21.29 0 ∑ = 10.47 ∑ = 50.71 ∑ =0 𝑋 = 𝐴 𝑥 𝐴 = 50.71 10.47 = 4.84m 𝑌 = 𝐴 𝑦 𝐴 = 0 m Solution: 300 Y X 300 O A D C B
- 74. Problem-20: Find outthe centroid of the following object. 16" 20" 4" 4" 12" 6" 8"10" Y X A F H B E D C G J I
- 75. Segment Area (𝑖𝑛2)𝑥 (in) A 𝑥 (𝑖𝑛3) AKMB 24 x 16 = 384 -12 - 24 2 = - 24 - 9216 AEB - 0.5 x 16 x 10 = - 80 - 26 - 2 3 x 10 = - 32.67 2613.6 CFK & DLM 2 x 0.5 x 6 x 4 = 24 - 12 - 6 3 = - 14 - 336 FGJM 12 x 28 = 336 - 12 2 = - 6 - 2016 HIN - 𝜋𝑟2 2 = - 𝜋 𝑥 102 2 = - 157.08 - 4𝑟 3𝜋 = - 4.24 666.02 ∑ = 506.92 ∑ = 8288.38 16" 20" 4" 4" 12" 6" 8"10" Y X A F H B E D C G J I L K M N X = Ax A = 8288.38 506.92 = 16.35 ft. For Symmetry, Y = 0 ft. Solution:
- 76. Problem-21: Find outthe centroid of the following object. X A B D C G Y 4" 4" 4" 6" E F
- 77. X A B D C G Y 4" 4" 4"6" E F Segment Area (𝑖𝑛2 ) 𝑥 (in) A 𝑥 (𝑖𝑛3 ) ABCD 4 X 8 = 32 -2 - 64 ABE - 𝜋𝑟2 2 = - 𝜋 𝑥 22 2 = - 6.28 -2 12.56 CDF - 𝜋𝑟2 2 = - 𝜋 𝑥 22 2 = - 6.28 -2 12.56 BCG 0.5 x 6 x 8 = 24 6 3 = 2 48 ∑ = 43.44 ∑ = 9.12 X = Ax A = 9.12 43.44 = 0.21 in. For Symmetry, Y = 0 ft. Solution:
- 78. Problem-22: Find outthe centroid of the following object. A B D C E X Y 4" 3" 2" 3" 2.5" 2.5" F
- 79. A B D C E X Y 4" 3" 2"3" 2.5" 2.5" F Segment Area (𝑖𝑛2) 𝑥 (in) A 𝑥 (𝑖𝑛3) ABC 0.5 X 3 X 5 = 7.5 4 + 2 3 X 3 = 6 45 BDEC 5 X 5 = 25 4 + 3 + 2.5 = 9.5 237.5 DEF - 0.5 X 3 X 5 = - 7.5 4 + 5 + 2 3 X 3 = 11 - 82.5 ∑ = 25 ∑ = 200 X = Ax A = 200 25 = 8 in. For Symmetry, Y = 0 ft. Solution:
- 80. Problem-23: Find outthe centroid of the following object. X Y 40 mm 80 mm 60 mm 120 mm
- 81. Segment Area (𝑚𝑚2)𝑥 (mm) 𝑦 (mm) A 𝑥 (𝑚𝑚3) A 𝑦 (𝑚𝑚3) Semi-Circle 𝜋𝑟2 2 = 𝜋 𝑥 602 2 = 5754.87 60 80 + 4𝑟 3𝜋 = 105.46 345292 606909 Full-Circle -𝜋𝑟2 = -𝜋 𝑥 402 = -5026.55 60 80 - 301593 - 402124 Rectangle 120 x 80 = 9600 60 40 576000 384000 Triangle 0.5 x 120 x 60 = 3600 120 3 = 40 - 60 3 = - 20 144000 - 72000 ∑ = 13928 ∑ = 763699 ∑ = 516785 X Y 40 mm 80 mm 60 mm 120 mm 𝑋 = 𝐴 𝑥 𝐴 = 763699 13928 = 54.83 mm 𝑌 = 𝐴 𝑦 𝐴 = 516785 13928 = 37.10 mm Solution:
- 82. Problem-24: Find outthe centroid of the following object. 60 mm 60 mm 80 mm 300 mm 400 mm 100 mm X Y
- 83. Segment Area (𝑚𝑚2)𝑦 (mm) A 𝑦 (𝑚𝑚3) 1 100 x 60 = 6000 30 180000 2 280 x 80 = 22400 60 + 140 = 200 4480000 3 300 x 60 = 18000 400 – 30 = 370 6660000 ∑ = 46400 ∑ = 11320000 For Symmetry, X = 0 mm. Y = AY A = 11320000 46400 = 244 mm. y = 60 𝑥 100 𝑥 30+80 𝑥 400−60−60 𝑥 60+ 400−60−60 2 +60 𝑥 300 𝑥 [400−30] 60 𝑥 300+80 𝑥 400−60−60 +60 𝑥 100 = 244 mm 60 mm 60 mm 80 mm 300 mm 280mm 100 mm X 2 1 3 400mm YSolution:
- 84.
- 85. Moments of Inertia:It is defined as the sum of second moment of area of individual sections about an axis. X Y 𝑋1 𝑋3 𝑎1 𝑋2 𝑌3 𝑌1 𝑌2 𝑎2 𝑎3Total Area, A Let, 𝑎1, 𝑎2 & 𝑎3 = Small element areas in total area (A). Taking moments of all areas about x- axis once, 𝐼 𝑥 = 𝑎1 𝑦1 + 𝑎2 𝑦2 + 𝑎3 𝑦3 Similarly, Taking moments of area again about y-axis, 𝐼 𝑥 = 𝑎1 𝑦1 2 + 𝑎2 𝑦2 2 + 𝑎3 𝑦3 2 𝐼 𝑥 = ∑ a𝑦2 Similarly, 𝐼 𝑦 = ∑ a𝑥2
- 86. Derivation of themoment of inertia of a rectangle with respect to centre of gravity of the rectangle. dA = bdy Ix = − ℎ 2 + ℎ 2 𝑦2 𝑑𝐴 = − ℎ 2 ℎ 2 𝑦2 𝑏𝑑𝑦 = b − ℎ 2 ℎ 2 𝑦2 𝑑𝑦 = b x [ 𝑦3 3 ] − ℎ 2 ℎ 2 = 𝑏 3 x [ ℎ3 8 - (- ℎ3 8 )] = 𝑏 3 x ℎ3 4 = 𝑏ℎ3 12 ∴ Ix = 𝑏ℎ3 12 X A B C dy h Y b D y
- 87. Y dA = hdx Iy= − 𝑏 2 + 𝑏 2 𝑥2 𝑑𝐴 = − 𝑏 2 𝑏 2 𝑥2ℎ𝑑𝑥 = h − 𝑏 2 𝑏 2 𝑥2 𝑑𝑥 = b x [ 𝑥3 3 ] − 𝑏 2 𝑏 2 = 𝑏 3 x [ 𝑏3 8 - (- 𝑏3 8 )] = ℎ 3 x 𝑏3 4 = ℎ𝑏3 12 ∴ Iy = ℎ𝑏3 12 A B C h X b D dx x Derivation of the moment of inertia of a rectangle with respect to centre of gravity of the rectangle.
- 88. Y h X b Y h X b Origin passes throughthe c.g. of the rectangle: Ix = 𝑏ℎ3 12 & Iy = ℎ𝑏3 12 Origin passes through the corner of the rectangle: Ix = 𝑏ℎ3 3 & Iy = ℎ𝑏3 3
- 89. h b X Y h b X Y Origin passes throughthe c.g. of the triangle: Ix = 𝑏ℎ3 36 & Iy = ℎ𝑏3 36 Origin passes through the vertex of triangle: Ix = 𝑏ℎ3 12 & Iy = 𝑏ℎ3 12
- 90. Origin passes throughthe c.g. of the circle: Ix = Iy = 𝜋𝑟4 4X Y r
- 91. X Y 𝑋 = r𝑌 = 4𝑟 3𝜋 r Origin passes through the centre of the circle: Ix = Iy = 𝜋𝑟4 8
- 92. X Y 𝑋 = 4𝑟 3𝜋 𝑌 = 4𝑟 3𝜋 r Originpasses through the centre of the circle: Ix = Iy = 𝜋𝑟4 16
- 93. 𝑋 𝑌 r X Y Origin passes throughthe vertex of the quarter circular spandrel: Ix = (1 - 5𝜋 16 ) 𝑟4 & Iy =( 1 3 - 𝜋 16 ) 𝑟4
- 94. Parallel Axis Theorem: Originof axis passes through any ordinate: Ix = Ix𝑐 + A𝑑2 Iy = Iy𝑐 + A𝑑2 Where, Ix = Moment of inertia of the object w.r.to the given axis. Ix𝑐 = Moment of inertia w.r.to centroid of the object. A = Area of the object. d = center to center (c/c) distance between the axis and the cantorial axis.
- 95. Moment of inertialof the rectangle w.r.to the axis X′ & Y′: IX′ = 𝑏ℎ3 12 + A𝑑1 2 IY′ = ℎ𝑏3 12 + A𝑑2 2 Y h X b X′ Y′ 𝑑1 𝑑2
- 96. Problem-25: Find outthe moment of inertia of the following object w.r.to the x-axis & y-axis. X Y 120 mm 120 mm 120mm 120mm 30 mm 30 mm 30 mm 30 mm
- 97. For the squaresection, Ix = 𝑏ℎ3 12 = 300 𝑥 3003 12 = 675 x 106 𝑚𝑚4 Iy = ℎ𝑏3 12 = 300 𝑥 3003 12 = 675 x 106 𝑚𝑚4 Solution: X 120 mm 120 mm 120mm 120mm 30 mm 30 mm 30 mm 30 mm For the blank square section, Ix = 𝑏ℎ3 12 + A𝑑1 2 = 180 𝑥 1203 12 + (180 x 120) x 902= 200.88 x 106 𝑚𝑚4 Iy = ℎ𝑏3 12 + A𝑑2 2 = 120 𝑥 1803 12 + (180 x 120) x 902= 233.28 x 106 𝑚𝑚4 ∴ Ix = 675 x 106 - 2 x 200.88 x 106 = 273.24 x 106 𝑚𝑚4 ∴ Iy = 675 x 106 - 2 x 233.28 x 106 = 208.44 x 106 𝑚𝑚4 30 + 60 = 90 30 + 60 = 90 𝑑1 𝑑2
- 98. 3 m6 m3m 3 m Y X Problem-26: Find out the moment of inertia of the following object w.r.to the x-axis & y-axis. Also find out the radius of gyration.
- 99. Solution: For the semi-circularsection, Ix = Iy = 𝜋𝑟4 8 = 𝜋 𝑥 64 8 = 508.94 𝑚4 For the triangular section, Ix = Iy = 𝑏ℎ3 12 = 6 𝑥 33 12 = 13.5 𝑚4 ∴ Ix = Iy = 508.94 – 13.5 = 495.44 𝑚4 A = 𝜋𝑟2 2 - 0.5 x b x h = 𝜋 𝑥 62 2 - 0.5 x 6 x 3 = 47.55 𝑚2 Radius of gyration, r = 𝐼 𝑚𝑖𝑛 𝐴 = 495.44 47.55 = 3.22 m. 3 m6 m3 m 3 m Y X
- 100. Problem-27: Find outthe moment of inertia of the following object w.r.to the x-axis & y-axis. Also find out the radius of gyration. Y 5" X 10" 6"2" 6"8"
- 101. Solution: For thesemi-circular section, Ix = Iy = 𝜋𝑟4 8 = 𝜋 𝑥 54 8 = 245.44 𝑖𝑛4 For the rectangular section, Ix = 𝑏ℎ3 3 = 10 𝑥 63 3 = 720 𝑖𝑛4 Iy = ℎ𝑏3 12 = 6 𝑥 103 12 = 500 𝑖𝑛4 For the triangular section, Ix = 𝑏ℎ3 12 + A𝑑2 = 6 𝑥 63 12 + 0.5 x 6 x 6 x 22 = 396 𝑖𝑛4 Iy = ℎ𝑏3 36 = 6 𝑥 63 36 = 36 𝑖𝑛4 Y 5" X 10" 6"2" 6" 6"
- 102. Y 5" X 10" 6"2" 6" 6" ∴ Ix =245.44 + 720 - 396 = 569.44 𝑖𝑛4 ∴ Iy = 245.44 + 500 - 36 = 709.44 𝑖𝑛4 A = 𝜋𝑟2 2 + b x h - 0.5 x b x h = 𝜋 𝑥 52 2 + 10 x 6 - 0.5 x 6 x 6 = 81.27 𝑖𝑛2 ∴ Radius of gyration, r = 𝐼 𝑚𝑖𝑛 𝐴 = 569.44 81.27 = 2.64 in.
- 103. Problem-28: Find outthe moment of inertia of the following object w.r.to the x-axis. 3" 8" 6" X
- 104. Solution: For the semi-circularsection, Ix = 𝜋𝑟4 8 - A𝑑1 2 = 𝜋 𝑥 34 8 - 𝜋 𝑥 32 2 x ( 4 𝑥 3 3 𝑥 𝜋 )2 = 8.87 𝑖𝑛4 Ix = Ix + A𝑑2 2 = 8.87 + 𝜋 𝑥 32 2 x (8 − 4 𝑥 3 3 𝑥 𝜋 )2 = 648.57 𝑖𝑛4 For the rectangular section, Ix = 𝑏ℎ3 3 = 6 𝑥 83 3 = 1024 𝑖𝑛4 For the triangular section, Ix = 𝑏ℎ3 12 = 6 𝑥 63 12 = 108 𝑖𝑛4 ∴ Ix = 1024 + 108 – 648.57 = 483.43 𝑖𝑛4 3" 8" 6" X
- 105. Problem-29: Find outthe moment of inertia of the following object w.r.to the x-axis. 60 mm 60 mm 80 mm 300 mm 400 mm 100 mm X
- 106. Segment Area (𝑚𝑚2)𝑦 (mm) A 𝑦 (𝑚𝑚3) 1 100 x 60 = 6000 30 180000 2 280 x 80 = 22400 60 + 140 = 200 4480000 3 300 x 60 = 18000 400 – 30 = 370 6660000 ∑ = 46400 ∑ = 11320000 Y = AY A = 11320000 46400 = 244 mm. 60 mm 60 mm 80 mm 300 mm 280mm 100 mm X 2 1 3 400mm YSolution: I = 𝑏1ℎ1 3 12 + 𝐴1 𝑑1 2 + 𝑏2ℎ2 3 12 + 𝐴2 𝑑2 2 + 𝑏3ℎ3 3 12 + 𝐴3 𝑑3 2 = 100 𝑥 603 12 + (60 x 100) x (244−30) 2 + 80 𝑥 2803 12 + (80 x 280) x (200−244) 2 + 300 𝑥 603 12 + (60 x 300) x (400−244−30) 2 = 75.7 x 107 𝑚𝑚4
- 107.
- 108. Problem-30: A lightcable weighing 40 lb is attached to a support at A and passes over a small pulley at B and supports a load P. Find out the load P, The slope of the cable at B and total length of the cable from A to B. Neglect the weight of the portion of cable from B to P. P 80′ 1′ A B
- 109. P 80′ 1′ A B Solution: w = 40 80 =0.5 lb/ft L = 80 ft d = 1 ft 𝐹′ Sinθ = 𝑤𝐿 2 = 0.5 𝑥 80 2 = 20 lb 𝐹′Cosθ = Q = 𝑤𝐿2 8𝑑 = 0.5 𝑥 802 8 𝑥 1 = 400 lb 𝐹′ = P = 𝐹′Sinθ2 + 𝐹′Cosθ2 = 202 + 4002 = 400.5 lb 𝜃 = tan−1( F′Sinθ F′Cosθ ) = tan−1(20/400) = 2.860 Total length of the cable from A to B S = L + 8 𝑑2 3 𝐿 - 32 𝑑4 5 𝐿3 = 80 + 8 𝑥 12 3 𝑥 80 - 32 𝑥 14 5 𝑥 803 = 80.033 ft.
- 110. Problem-31: Cable ABsupports a load distributed uniformly along the horizontal. Determine the maximum and minimum values of the tension in the cable. 15′ 30′ A B 200′ 50 lb/ft.
- 111. Solution: ∑ MA =0 => 50 a x 𝑎 2 - Q x 15 = 0 => Q = 5 3 𝑎2 ∑ M 𝐵 = 0 => 50 a x 𝑎 2 - Q x 15 = 0 => Q = 5 3 𝑎2 15′ 30′ A B 200′ 50 lb/ft. Qa F′ 50 a A 15′
- 112. ∑ MB =0 => 50 (200 – a) x (200 − 𝑎) 2 - Q x 45 = 0 => 5 3 𝑎2 x 45 = 25 x (200 − 𝑎)2 => a = 73.21 ft. Minimum Tension, Q = 5 3 𝑎2 = 5 3 x 73.212 = 8933 lb F1 ′ Sin𝜃 F1 ′ Cos𝜃 𝐹1 ′ 𝜃 B Q 30′ (200 – a) 50 (200 – a) ∑ FX = 0 => F1 ′ Cos𝜃 - Q= 0 => F1 ′ Cos𝜃 = Q = 8933 lb. ∑ FY = 0 => F1 ′ Sin𝜃 – 50 (200-a) = 0 => F1 ′ Sin𝜃 = 50 (200 – 73.21) => F1 ′ Sin𝜃 = 6340 lb ∴ Maximum Tension, 𝐹1 ′ = F1 ′ Sin𝜃2 + F1 ′ Cos𝜃2 = 89332 + 63402 = 10953 lb.
- 113. Problem-32: Determine theequation of the curve assumed by the cable. T0 is the tension in the left joint. A B b h 𝑤0 - kx x
- 114. Solution: Given that, w= w0 - kx When, x = 0, w = w0 When, x = b, w = 0 Then, w0 = w0 - kx & 0 = w0 - kb or, k = w0 b From, equation 1 we get, w = w0 - w0 b x = w0 (1 - x b ) T0 b T Bh w b 3 2 3 b ∑ MB = 0 => T0 x h - 𝑤𝑏 2 x 2 3 b = 0 => T0 = w0 (1 − x L ) 𝑏2 3ℎ We know, y = 𝑤𝑥2 2T0 = w0 (1 − x b ) 𝑥2 ∗3ℎ 2w0 (1 − x b ) 𝑏2 = 3ℎ 2 ( 𝑥 𝑏 )2
- 115. Problem-33: If theslope of the cable at point A is zero, determine the deflection curve and the maximum tension developed in the cable. A B L h 𝑤0 x
- 116. Solution: ∑ MB =0 => T0 x h - w0 𝐿 2 x L 3 = 0 => T0 = w0 𝐿2 6 ℎ ∑ FX = 0 => 𝐹𝐵𝑥 - T0 = 0 => 𝐹𝐵𝑥 = w0 𝐿2 6 ℎ ∑ FY = 0 => 𝐹𝐵𝑦 - w0 𝐿 2 = 0 => 𝐹𝐵𝑦 = w0 𝐿 2 ∴ Maximum tension, 𝑇 𝑚𝑎𝑥 = 𝐹𝐵𝑥 2 + 𝐹𝐵𝑦 2 = ( w0 𝐿2 6 ℎ )2+ ( w0 𝐿 2 )2 = w0 𝐿 2 ( 𝐿 3ℎ )2+1 T0 L 𝐹𝐵 Bh w0 𝐿 2 L 3 2 3 L 𝐹𝐵𝑥 𝐹𝐵𝑦
- 117. Thank you For Taking theStress