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CONCURRENT FORCES & FORCE DIAGRAMS. A Concurrent F orce System, is a force system where all forces start from the same point. 100N. 2 00N. TERMS: Resultant, the single force which can replace a force system. Equilibrant, the single force which can bring a force system in to equilibrium.

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slide1

CONCURRENT FORCES & FORCE DIAGRAMS

  • A Concurrent Force System, is a force system where all forces start from the same point.

100N

200N

  • TERMS:
  • Resultant, the single force which can replace a force system.
  • Equilibrant, the single force which can bring a force system in to equilibrium.
  • To achieve equilibrium, all horizontal, vertical and moments must be balanced.
  • We can determine the value of the resultant and equilibrant through graphical
  • techniques.
slide2

CONCURRENT FORCES & FORCE DIAGRAMS

Lets consider the previous force system.

100N

Select a scale to draw the forces against.

In this case, lets say 5mm = 10N.

Draw the 200N force to scale, i.e. 100mm.

Draw the 100N force from the end of this

vertically upwards. i.e. 50mm.

4. Join the ends and measure this line.

200N

Line length = Approx. 110mm

So resultant = (110/5) x 10N

= Approx. 220N

Resultant

50mm

100mm

slide3

CONCURRENT FORCES & FORCE DIAGRAMS

The Equlibrant will always be in the opposite direction, but will always have the same value.

This technique works for force systems with more than 2 forces, as an example.

F3

F2

F1

F4

F4

F3

Force System

F2

R

F1

slide4

CONCURRENT FORCES & FORCE DIAGRAMS

Pupil problems:

100N @30

1)

50N @ 45

2)

65N

100N

30N

60N @ 45

50N @45

150N

200N

75N

3)

50N

slide5

STRESS & STRAIN

STRESS = Load/Cross Sectional Area (Nmm-2)

Strain = Change in Length/ Original Length (no units)

Example 1: A structural member is subjected to a compressive force of 20000N. If the diameter of the member is 250mm, calculate the stress within the member.

If the load reduces the member length by 0.5mm and the original length was 2m, calculate the strain.

solutions
Solutions

Stress

= 3.14 x 125 x 125

= 49062.5 mm2

Stress = Load/Cross Sectional Area

= 20000/49062.5

= 0.407Nmm-2

Strain

Strain = Change in Length/ Original Length

= 0.5/2000

= 0.00025