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BTECH MECHANICAL PRINCIPLES AND APPLICATIONS

Level 3 Unit 5

FORCES AS VECTORS

Vectors have a magnitude (amount) and a direction.

Forces are vectors

FORCES AS VECTORS (2 FORCES)

F1

F2

Forces F1 and F2

are in different

directions They

are NOT in

equilibrium


F1

F2

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

If the two

forces do not

meet, the

system is not in

equilibrium


F1

F2

F E

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)

This force is called the

EQUILIBRANT


F1

F2

FR

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

F3

F1

F2

F3 FE


F1

F2

F3

F1

F2

F3 FR

FORCES AS VECTORS ( 3 FORCES) 2ND EXAMPLE

F1

F2

F3

FE

F1

F2

F3

FR

Equilibrant

Resultant

F1

F2

F3

50o

50o 50o

FORCES AS VECTORS

F1

4N

12N

FE 10N

4N

12N

FR

Equilibrant Resultant

F1 = 10N

F2 = 4N

F3 = 12N

50o

50o 50o

40cm

60cm

Forces on a flat

rectangular plate

FINDING FE

IDENTIFY THE DIRECTION AND MAGNITUDE OF THE FORCES

THEN CONSTRUCT A VECTOR DIAGRAM

F1 = 10 N

F2 = 4N

F3 =12 N

50o

40cm

60cm

DRAW TO SCALE TO FIND THE MAGNITUDE AND DIRECTION OF FE (EQUILIBRANT)

10N

4N

12N

FE

50o

FE = 22N (Measured)

DRAW IT IN THE OPPOSITE DIRECTION TO FIND THE MAGNITUDE AND DIRECTION OF RESULTANT FORCE

10N

4N

12N

FR

50o

FE = 22N (Measured)

FINDING THE POSITION OF THE EQUILIBRANT (FE)

F1

4N

12N

22N 10N

4N

12N

50o

40cm

60cm

40cm

60cm

10N

4N

12N

50o

A x

Clockwise = Anticlockwise

22N

22N x X = 4N x 40cm + 10N x 0 + 12N x 0 = 160Ncm X = 160 ÷ 22 = 7.27 cm

The 10N and the 12N pass through the pivot A so the turning moment = 0

Put FE where you think it should be to balance the other forces

1.4kN

130o

35o A

2.6kN

1.4 kN

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)

B TECH Question example

4m

3m

1.4kN

130o

35o A

2.6kN

1.4 kN

Weight of plate = 200Kg

x 9,81 = 1.96kN

(acting from the centre of gravity of the uniform plate

1.96 kN

4m

3m

VECTOR DIAGRAM WITH RESULTANT

2.2kN

1.4kN

1.4kN

1.96kN

2.6kN

This shows a) the magnitude and direction of the resultant

VECTOR DIAGRAM WITH EQUILIBRANT

2.6kN

1.4kN

1.4kN

1.96kN

2.2kN

This shows b) the magnitude and direction of the equilibrant

1.4kN 35o A

2.6kN

1.4 kN

V2 = 2.6xsin 35

= 1.49kN V1 = 1.4xsin40 =

0.90kN H1 1.4 x cos40 =

1.07kN

1.96 kN

x

V2

H1

V1

H2 not needed , it passes through A

2.2kN

4m 40o

For explanation Click here

Resolve the diagonal forces 2.6kN and 1.4kN into vertical and horizontal components V1, H1 and V2 (H2 not needed)

130 - 90

3m

1.4kN 35o A

2.6kN

1.4 kN


1.96 x 2 + V2 x 4 + V1 x 4 + H1 x 3 2.2 x X 1.96 kN

x

V2

H1

V1


2.2kN

4m 40o

Resolve turning moments

0.9kN

1.49kN

1.07kN

3m

1.4kN 35o A

2.6kN

1.4 kN


1.96 x 2 + 1.49 x 4 + 0.9 x 4 + 1.07 x 3 2.2 x X 7.52 + 2.2X = 9.17 2.2X = 9.17 – 7.52 2.2X = 1.65X = 1.65 ÷ 2.2 = 0.75m

1.96 kN

x

V2

H1

V1


2.2kN

4m 40o

Resolve turning moments

0.9kN

1.49kN

1.07kN

RESOLVING FORCES

2000 Newtons

70o 35o

20o 55o

20o

55o

105o

Draw the perpendicular

Identify the angles between the forces A and B and the perpendicular

2000 Newtons

Weight suspended by two ropes

Draw the triangle using the angles

2000 N

A B

A

B

The length of the sides of the triangle represent the magnitude of the forces NOT the length of rope

USING THE SINE RULE (IF YOU KNOW THE ANGLES)

20o

55o

105o

a

b

2000 N (c)

a/sin A = b/Sin B = c/sin C

angle A = 20o angle B = 55o (opposites to sides a & b)

Angle C = 105o and side c represents 2000N

USING THE SINE RULE (IF YOU KNOW THE ANGLES)

20o

55o

105o

a

b

2000 N (c)

a/sin A = c/sin C therefore a/sin 20o = 2000/sin105o

a = 2000 x sin 20o/sin105o

708.17N

b/sin B = c/sin C therefore b/sin 55o = 2000/sin105o

a = 2000 x sin 55o/sin105o

1696.1N

USING THE COSINE RULE ( IF YOU KNOW ONE ANGLE AND TWO SIDES)

F2 = 60N

70o

F1 = 30N

F3

USING THE COSINE RULE ( IF YOU KNOW ONE ANGLE AND TWO SIDES)

70o

F2 = 60N (C)

F1 = 30N (B)

F3 (A)

A =110o

A2 = B2 + C2 -2BCcosA

(F3)2 = 302 + 602 – 2x60x30x cos110o

= 75.7N

VERTICAL AND HORIZONTAL COMPONENTS OF FORCES

F Fv

FH

θ

Sketch the diagram

Fv can be drawn at the other

end of the sketch

VERTICAL AND HORIZONTAL COMPONENTS OF FORCES

F Fv

FH

θ

Sketch the diagram

Fv

sin θ = Fv/F

F.sin θ = Fv

cos θ = FH/F

F.cos θ = FH back

RESTORING FORCE OF TWO FORCES

25o

70o

F1(55N)

F2 (25N)

F3

F3 is the restoring force of F1 and F2

25o

70o

Can be drawn to scale

74.8N


25o

70o

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


25o

70o

F1(55N)

F2 (25N)

F3

F1v = F1.sin70o

55sin70o

= 51.68N

F1h = F1.cos70o

55cos70o

= 18.81N


25o

70o

F1(55N)

F2 (25N)

F3 F2v = F2.sin25o

25sin25o

= 10.57N

F2h = F2.cos25o

25cos25o

= 22.66N


25o

70o

F1(55N)

F2 (25N)

F3 F3v = F1v + F2v

51.68 +10.57

= 62.25N

F3h = F1h +F2h

18.81 + 22.66

= 41.47N


F3

62.25N

41.47N

(F3)2 = 62.252 + 41.472

(F3)2 = 5594.82

F3 = 74.80N

RESULTANT OF TWO FORCES

F3

62.25N

41.47N

θ

Tan θ = opposite/adjacent

Tan θ = 62.25/41.47

Tan θ = 1.5

θ = 56.33o

Direction of F3 = 180 + 56.33 = 236.33o

MOMENTS OF FORCE 2m

4N

4m

2N

Total Anticlockwise moments = Total Clockwise moments

8Nm 8Nm

MOMENTS OF FORCE 3m 4m

2m

4N

2N 2N

Total Anticlockwise moments = Total Clockwise moments

8Nm + 4Nm = 12Nm = 12 Nm

FORCE ON A AND B

4m

10m

3m

4N

2N 2N

Take A as the pivot

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

2m 1m A B

FORCE ON A AND B

4m

10m

2.5m

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

2.5m 1m A B

4 N/m uniformly distributed load

UDL 4N/m x length 5m acting from centre of UDL

B A

4kN/m

(uniform

distributed

load)

4kN 6kN

P2 Calculate the support reactions A and B for the simply supported beam in the diagram

5 m 3m 2m

B TECH Question example

B A

4kN/m

(uniform

distributed

load)

4kN 6kN

Uniform load is 4kN/m. total UDL = 4 x 10 = 40kN ( acting from centre of beam)

5 m 3m 2m

B TECH Question example solution

40kN

B A

4kN/m

(uniform

distributed

load)

4kN 6kN

Clockwise moments 6 x 2 = 12kNm 4 x 5 = 20 kNm 40 x 5 = 200 kNm total = 232kNm

5 m 3m 2m


40kN

Anticlockwise B x 10 kNm B = 232 ÷10 23.2kNm

B A

4kN/m

(uniform

distributed

load)

4kN 6kN

Total upward force A + B = 50kN A + 23.2 = 50kN A = 50 – 23.2 = 26.8kN

5 m 40kN

2m


Total downward force 6 + 4 + 40 = 50kN

B A

4kN/m

(uniform

distributed

load)

4kN 6kN

CW = A x10 ACW = 4 x 5 = 20kNm 40 x 5 = 200kNm 6 x 8 = 48kNm = 268kNm A = 26.8kN

5 m 3m 2m


Check using B as the pivot

40kN

TENSILE STRESS AND STRAIN

Necking Strain hardening

Ultimate tensile strength

Yield strength

Fracture

Y (Stress)

X (Strain)

Stress

Strain

TENSILE STRESS (σ)

FORCE FORCE

CROSS SECTIONAL AREA ( πr2)

Stress = Force ÷ Cross sectional area

Force direction is

perpendicular to cross

sectional area

TENSILE STRAIN Lo (original length)

Increase in length ∆L

Strain = ∆L ÷ Lo

Young’s Modulus Stress ÷ Strain

SHEAR STRESS (τ)

Force

Shear stress = Force ÷ cross sectional area of

shear

Force is parallel to cross sectional area

of shear

SHEAR STRAIN

Force

Shear strain = Change in length ÷ original length

∆l ÷ l

SHEAR STRESS

Force

Shear Modulus Shear Stress ÷ shear Strain

20kN

C

B

A 20kN

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

20kN

C

B

A 20kN

Maximum direct stress is on the smallest connecting rod diameter which is 20mm (.02m), radius is 10mm (.01m). Cross sectional area = πr2 = π x .012 = 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

20kN

C

B

A 20kN

Maximum direct stress is on the smallest connecting rod diameter which is 20mm (.02m), radius is 10mm (.01m). Cross sectional area = πr2 = π x .012 = 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

20kN

C

B

A 20kN

Shear stress in pin Force ÷ area = 20kN ÷ cross sectional area of pin (π x .00752) = 20x103 ÷ 1.77 x10 -4m2 = 1.13 x108 Pa Shear modulus for brass = 7 x 1010 Pa. Strain = stress ÷ modulus strain = 1.13 x 108 ÷ 7 x 1010 = 0.0016

70o

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.

70o

F = 8kN F

Direct force F1

Shear force F2

8kN

F1 F2

70o Sin70o = F1 ÷ 8kN F1 = 8kN x Sin70o

F1 = 7.5 kN

Cos70o = F1 ÷ 8kN F2 = 8kN x Cos70o

F2 = 2.7 kN

70o

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

70o

F = 8kN F


Tensile stress = F ÷ area

7.5 x 103 ÷ 1.14x 10-4

6.6 x 107 Pa

70o

F = 8kN F


Shear stress = F ÷ area

2.7 x 103 ÷ 1.14x 10-4

2.4 x 107 Pa

70o

F = 8kN F


operational factor of safety in tension.

500 x 106 Pa ÷ 6.6 x 107 Pa = 7.6

70o

F = 8kN F


operational factor of safety in shear.

300 x 106 Pa ÷ 2.4 x 107 Pa = 12.5

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