1. BTECH Mechanical principles and applications
Level 3 Unit 5

2. Vectors have a magnitude (amount) and a direction.
Forces as vectors
Vectors have a magnitude (amount) and a direction.
Forces are vectors

3. Forces as vectors (2 forces)
Forces F1 and F2 are in different directions They are NOT in equilibrium F1 F2

4. Forces as vectors (2 forces)
The two forces can be drawn like this. (In the correct direction and the lengths should be drawn to scale to represent the magnitude of the forces) F1 F2

5. Forces as vectors (2 forces)
If the two forces do not meet, the system is not in equilibrium F1 F2

6. Forces as vectors (2 forces)
If a third force (FE) was added in the way shown the three would be in equilibrium (They are all joined up following each other, The force system is balanced)
F E
This force is called the EQUILIBRANT

7. Forces as vectors (2 forces)
If the line joining the two forces is in the opposite direction to the equilibrant it is the RESULTANT of the two forces
The forces area not in equilibrium and the resultant shows the direction and magnitude of the combination of the two forces F1 F2 FR

8. Forces as vectors (3 forces)FE F2 F1

9. Forces as vectors (3 forces)FR F2 F1

10. Forces as vectors ( 3 forces) 2nd exampleFE
Equilibrant
Resultant F1 F2 F3 FR
50o
50o

11. Forces on a flat rectangular plate
F1 = 10N
F2 = 4N
F3 = 12N
50o
Forces as vectors
40cm
60cm
Equilibrant
Resultant F1 4N
12N FE
10N 4N
12N FR
50o
50o

12. F1 = 10 N
F2 = 4N
F3 =12 N
50o
Finding FE Identify the direction and magnitude of the forces then construct a vector diagram
40cm
60cm

13. Draw to scale to find the magnitude and direction of FE (equilibrant)
FE = 22N (Measured)

14. 10N 4N
12N FR
Draw it in the opposite direction to find the magnitude and direction of resultant force
50o
FE = 22N (Measured)

15. Finding the position of the equilibrant (FE)
40cm
60cm

16. Put FE where you think it should be to balance the other forces
Clockwise = Anticlockwise
10N 4N
12N
50o
40cm
60cm
22N x X = 4N x 40cm
+ 10N x 0
+ 12N x 0
= 160Ncm
X = 160 ÷ 22 = cm x A
22N
The 10N and the 12N pass through the pivot A so the turning moment = 0

17. B TECH Question example
P1 The diagram shows a uniform rectangular plate supported in a vertical plane by forces acting at the three corners of the plate. The plate is 4m x 3m and has a mass of 200kg
a) Calculate the magnitude and direction of the resultant force
b) Show the magnitude and direction of the equilibrant force
c) Calculate the position of the resultant force with respect to the corner A (ie. Use A as a pivot)
1.4kN
130o
35o A
2.6kN
1.4 kN 4m 3m

18. Weight of plate = 200Kg x 9,81 = 1.96kN
(acting from the centre of gravity of the uniform plate
1.4kN
130o
35o A
2.6kN
1.4 kN 4m 3m
1.96 kN

19. Vector diagram with RESULTANT
2.2kN
1.4kN
1.96kN
This shows
a) the magnitude and direction of the resultant
2.6kN
1.4kN

20. Vector diagram with equilibrant
2.2kN
1.4kN
1.96kN
This shows
b) the magnitude and direction of the equilibrant
2.6kN
1.4kN

21. Resolve the diagonal forces 2. 6kN and 1
Resolve the diagonal forces 2.6kN and 1.4kN into vertical and horizontal components V1, H1 and V2 (H2 not needed) V2
1.4kN
35o A
2.6kN
1.4 kN
2.2kN
For explanation
Click here x
H2 not needed , it passes through A
V2 = 2.6xsin 35
= 1.49kN
V1 = 1.4xsin40 = 0.90kN
H1 1.4 x cos40 = 1.07kN 3m
1.96 kN H1
40o 4m V1

22. V1 x 4 + H1 x 3 2.2 x X Resolve turning moments
1.49kN V2
1.4kN
35o A
2.6kN
1.4 kN
2.2kN x
Clockwise = Anticlockwise x V2 x 4 +
V1 x H1 x 3
2.2 x X
H2 not needed , it passes through A 3m
1.96 kN H1
1.07kN
40o 4m V1
0.9kN

23. Resolve turning moments
1.49kN V2
1.4kN
35o A
2.6kN
1.4 kN
2.2kN x
Clockwise = Anticlockwise x x 4 +
0.9 x x 3
2.2 x X
X =
2.2X = – 7.52
2.2X = X = 1.65 ÷ 2.2 = 0.75m
H2 not needed , it passes through A
1.96 kN H1
1.07kN
40o 4m V1
0.9kN

24. Resolving forces
2000 Newtons

25. Weight suspended by two ropes
Draw the perpendicular
Identify the angles between the forces A and B and the perpendicular
Draw the triangle using the angles A B
20o A
55o
55o
35o
70o
2000 Newtons
105o
The length of the sides of the triangle represent the magnitude of the forces NOT the length of rope
2000 N B
20o

26. Using the sine rule (if you know the angles)
a/sin A = b/Sin B = c/sin C
angle A = 20o angle B = 55o (opposites to sides a & b)
20o
55o
105o a
Angle C = 105o and side c represents 2000N
2000 N
(c) b

27. Using the sine rule (if you know the angles)
a/sin A = c/sin C therefore a/sin 20o = 2000/sin105o
a = 2000 x sin 20o/sin105o
708.17N
20o
55o
105o a b
2000 N
(c)
b/sin B = c/sin C therefore b/sin 55o = 2000/sin105o
a = 2000 x sin 55o/sin105o
1696.1N

28. Using the cosine rule ( if you know one angle and two sides)
F2 = 60N
70o
F1 = 30N F3

29. Using the cosine rule ( if you know one angle and two sides)
A2 = B2 + C2 -2BCcosA
70o
F2 = 60N (C)
F1 = 30N (B)
F3 (A)
A =110o
(F3)2 = – 2x60x30x cos110o
= 75.7N

30. Vertical and horizontal components of forces
Sketch the diagram
Fv can be drawn at the other
end of the sketch F Fv FH θ

31. Vertical and horizontal components of forces
Sketch the diagram
sin θ = Fv/F
F.sin θ = Fv F Fv FH θ Fv
cos θ = FH/F
F.cos θ = FH
back

32. Restoring force of two forces
F3 is the restoring force of F1 and F2
25o
70o
F1(55N)
F2 (25N) F3
25o
70o
Can be drawn to scale
74.8N

33. Restoring force of two forces
F1(55N)
F2 (25N) F3
F3 is the restoring force of F1 and F2
Can be solved by resolving the horizontal and vertical components of F1 and F2

34. Restoring force of two forces
F1(55N)
F2 (25N) F3
F1v = F1.sin70o
55sin70o
= 51.68N
F1h = F1.cos70o
55cos70o
= 18.81N

35. Restoring force of two forces
F1(55N)
F2 (25N) F3
F2v = F2.sin25o
25sin25o
= 10.57N
F2h = F2.cos25o
25cos25o
= 22.66N

36. Restoring force of two forces
F1(55N)
F2 (25N) F3
F3v = F1v + F2v
= 62.25N
F3h = F1h +F2h
= 41.47N

37. Restoring force of two forces
F3 = 74.80N
41.47N

38. Resultant of two forces
Tan θ = opposite/adjacent
Tan θ = 62.25/41.47
Tan θ = 1.5
θ = 56.33o F3
62.25N θ
41.47N
Direction of F3 = = o

39. Moments of force Total Anticlockwise moments = Total Clockwise moments
8Nm
8Nm

40. Moments of force 8Nm + 4Nm = 12Nm = 12 Nm
Total Anticlockwise moments = Total Clockwise moments
8Nm + 4Nm = 12Nm = Nm

41. Moments of force Total Anticlockwise moments = Total Clockwise moments
8Nm + 4Nm = 12Nm = Nm

42. Moments of force Total Anticlockwise moments = Total Clockwise moments
8Nm + 4Nm = 12Nm = Nm

43. Force on A and B 10m 3m 1m 4m 2m A B 2N 2N Take A as the pivot 4N
Anticlockwise ( B x 10) = (2 x1) + (2 x 5) + (4 x 8) = 44 Nm
Force on B = 44 ÷ 10 = 4.4N
Total downward force = 8 N Force on A = 3.6 N
Check this out using B as the pivot

44. Force on A and B Take A as the pivot
4 N/m uniformly distributed load
10m 1m 4m
2.5m
2.5m A B
20 N (UDL) 2N 2N
Take A as the pivot
Anticlockwise ( B x 10) = clockwise (2 x1) + (2 x 5) + (20 x 7.5) = 162 Nm
Force on B = 162 ÷ 10 = 16.2 N
Total downward force = 24 N Force on A = 7.8 N
Check this out using B as the pivot
UDL 4N/m x length 5m acting from centre of UDL

45. B TECH Question example
4kN/m (uniform distributed load)
4kN
6kN 2m 3m
5 m
P2 Calculate the support reactions A and B for the simply supported beam in the diagram

46. B TECH Question example solution
4kN/m (uniform distributed load)
4kN
6kN 2m 3m
5 m
40kN
Uniform load is 4kN/m. total UDL = 4 x 10 = 40kN ( acting from centre of beam)

47. B TECH Question example solution
4kN/m (uniform distributed load)
4kN
6kN 2m 3m
5 m
40kN
Clockwise moments
6 x 2 = 12kNm
4 x 5 = 20 kNm
40 x 5 = 200 kNm
total = 232kNm
Anticlockwise
B x 10 kNm
B = 232 ÷10
23.2kNm

48. B TECH Question example solution
4kN/m (uniform distributed load)
4kN
6kN 2m
5 m
40kN
Total upward force A + B = 50kN
A = 50kN
A = 50 – 23.2 = 26.8kN
Total downward force
= 50kN

49. B TECH Question example solution
4kN/m (uniform distributed load)
4kN
6kN 2m 3m
5 m
40kN
CW = A x10
ACW = 4 x 5 = 20kNm
40 x 5 = 200kNm
6 x 8 = 48kNm
= 268kNm
A = 26.8kN
Check using B as the pivot

50. Tensile Stress and strain
Necking
Strain hardening
Ultimate tensile strength
Yield strength
Fracture
Y (Stress)
X (Strain)
Stress
Strain

51. Force direction is perpendicular to cross sectional area
Tensile stress (σ)
FORCE
CROSS SECTIONAL AREA ( πr2)
Stress = Force ÷ Cross sectional area
FORCE
Force direction is perpendicular to cross sectional area

52. Tensile strain Young’s Modulus Stress ÷ Strain Strain = ∆L ÷ Lo
Lo (original length)
Increase in length ∆L
Young’s Modulus Stress ÷ Strain
Strain = ∆L ÷ Lo

53. Shear stress (τ) Force is parallel to cross sectional area of shear
Shear stress = Force ÷ cross sectional area of shear

54. Shear strain = Change in length ÷ original length
Force
Shear strain = Change in length ÷ original length
∆l ÷ l

55. Shear Stress ÷ shear Strain
Force
Shear Modulus
Shear Stress ÷ shear Strain

56. a) Calculate the maximum direct stress in the connecting rods
20kN C B A
P3 The diagram shows a shackle joint subjected to a tensile load. The connecting rods A and B are made from steel and the pin c is made from brass. Young’s modulus is 210 GPa for steel and 100GPa foe brass. The shear modulus for steel is 140GPa and the shear modulus for brass is 70GPa. The smallest diameter for the connecting rods A and B is 20mm and the diameter of the pin C is 15 mm.
a) Calculate the maximum direct stress in the connecting rods
b) Calculate the maximum direct strain in the connecting rods
c) Calculate the change in length of a 500mm length of connecting rod.
d) Calculate the shear stress in the pin
e) Calculate the shear strain in the pin

57. Young’s modulus for steel = stress ÷ strain = 210GPa = 2.1 x10 11Pa
20kN C B A
Maximum direct stress is on the smallest connecting rod diameter which is 20mm (.02m), radius is 10mm (.01m). Cross sectional area = πr2 = π x = 3.14 x 10-4 m2 . Maximum stress = 20x103 ÷ 3.14 x 10-4 = 6.4 x 107 N/m2 (Pa)
Young’s modulus for steel = stress ÷ strain = 210GPa = 2.1 x10 11Pa
strain = stress ÷ Young’s modulus
Strain = 6.4 x 107 ÷ 2.1 x10 11
=3 x 10-4 m/m

58. strain = elongation ÷ original length
20kN C B A
Maximum direct stress is on the smallest connecting rod diameter which is 20mm (.02m), radius is 10mm (.01m). Cross sectional area = πr2 = π x = 3.14 x 10-4 m2 .
strain = elongation ÷ original length
elongation =strain x original length
= 3 x 10-4 m/m x 0.5
= 1.5x10-4 m = 0.15mm

59. Shear modulus for brass = 7 x 1010 Pa. Strain = stress ÷ modulus
20kN C B A
Shear stress in pin Force ÷ area = 20kN ÷ cross sectional area of pin (π x )
= 20x103 ÷ 1.77 x10 -4m2 = x108 Pa
Shear modulus for brass = 7 x 1010 Pa. Strain = stress ÷ modulus
strain = 1.13 x 108 ÷ 7 x =

60. a) Determine the operational factor of safety in tension.
F = 8kN F
In the diagram the diameter in the of the bolt shown for the angle is 12mm. It is made from a material with a tensile strength of 500MPa and a shear strength of 300 MPa
a) Determine the operational factor of safety in tension.
b) Determine the operational factor of safety in shear.

61. F = 8kN F 70o Direct force F1 Shear force F2 8kN
Sin70o = F1 ÷ 8kN F1 = 8kN x Sin70o
F1 = 7.5 kN
70o F2 F1
Cos70o = F1 ÷ 8kN F2 = 8kN x Cos70o
F2 = 2.7 kN

62. Cross sectional area of the bolt = πr2
F = 8kN F
Operational factor of safety = Tensile strength ÷ working stress
Cross sectional area of the bolt = πr2
= π x (.006)2 = 1.13 x10-4 m2

63. Tensile stress = F ÷ area
70o
F = 8kN F
Operational factor of safety = Tensile strength ÷ working stress
Tensile stress = F ÷ area
7.5 x 103 ÷ 1.14x 10-4
6.6 x 107 Pa

64. Operational factor of safety = Tensile strength ÷ working stress
F = 8kN F
Operational factor of safety = Tensile strength ÷ working stress
Shear stress = F ÷ area
2.7 x 103 ÷ 1.14x 10-4
2.4 x 107 Pa

65. Operational factor of safety = Tensile strength ÷ working stress
F = 8kN F
Operational factor of safety = Tensile strength ÷ working stress
operational factor of safety in tension.
500 x 106 Pa ÷ 6.6 x 107 Pa
= 7.6

66. Operational factor of safety = Tensile strength ÷ working stress
F = 8kN F
Operational factor of safety = Tensile strength ÷ working stress
operational factor of safety in shear.
300 x 106 Pa ÷ 2.4 x 107 Pa
= 12.5