The document discusses structural mechanics and principles of moments. It defines a moment as the product of a force and its perpendicular distance from a fulcrum. Moments can be used to calculate unknown support reactions in beams and other structural systems. Couples, consisting of two equal and opposite parallel forces, produce pure rotation and cannot be balanced by a single force. The degree of static indeterminacy refers to the number of redundant unknown internal forces and reactions that must be specified to fully describe the static behavior of a structure.

1. STRUCTURAL MECHANICS
LECTURE 3
STRUCTURAL DESIGN I

2. MOMENTS, PRINCIPLE OF THE LEVER
 The forces studied so far tend to either distort an
object or move bodies from one position to another.
 Forces can also have rotational effect on bodies if
these are hinged at a point.
 In the figure below the force F would rotate the body
around the point O.
 The point O is generally referred to as a fulcrum.
O
F F

3.  Try opening a door by pushing it near the hinge. It will not
open till force is applied at a distance from the hinge.
 Try opening the door by pulling it parallel to the door -
again same result.
 This suggests that forces have a best rotational affect
when these are at a distance from the hinge or fulcrum
and the perpendicular distance between the hinge and
the line of action of the force will determine the total
rotational affect.
 This phenomenon is determined by the principle of
“moments”.
 A moment is thus a product of the magnitude of force and
the perpendicular distance of the fulcrum from the line of
action of the force. This distance is also referred to as the
‘arm of the moment’.

4. Resultant Moments
 Moment = force x arm
 Moments can be clockwise or anti clockwise.
 These can be arithmetically added
 Find the resultant moment about the point O in the figure
above
4kN 6kN 2kN 5kN
O30 cm 30 cm 20 cm 40 cm

5. Calculating Beam Support Reactions
 This principle of moment can be used to calculate the
reactions at the support points in a structural system.
 Take the case of the following beam:
Assuming the system to be in equilibrium
Rl x 5m = (3kN x 1m) + (6kN x 3m)
Rr x 5m = (6kN x 2m) + (3kN x 4m)
6kN 3kN
2 m 2 m 1 m
Rl Rr

6.  Beam A of effective span 8m has a self load of 30N per meter and carries a wall load
of 710N per meter. It also carries the load of 3 beams B, C and D with point loads of
2000, 2000 and 4000N respectively. Beams B, C and D are equally spaced at 2
meters each. Find the reactions Rl and Rr.
 Hint: First draw the force system diagram.
WALL
B DC
A
OPENINGRl Rr

7. 2000 2000 4000
UDL
Rl
2m2m2m
Rr
2m

8. 2000 2000 4000
UDL
Rl
2m2m2m
Rr
2m
UDL = (30+710) x 8 N
Taking moments about Rr:
Rl x 8 = (4000x2) + (2000x4) +( (30+70)x8 x4) + (2000x6)
Taking moments about Rl:
Rr x 8 = (2000x2) + (2000x4) +( (30+70)x8 ) x4 + (4000x6)
Crosscheck: Rl + Rr = 2000 + 2000 + 4000 + UDL

9. 5kN 4kN10kN
Rl
8kN
9m
Rr
1m 1.5m2.5m5m1.5m
Find Rl and Rr in the loading system shown in the diagram above.

10. To find the reactions:
There are two unknown values. Rl and Rr
Taking the system to be in equilibrium ∑M = 0
And taking moments about one of the reaction points, the moment created by
the reaction force at that point becomes zero and there is hence only
one unknown value.
Now, taking moment about Rr we get:

11. Taking moments about Rr:
5kN force is creating a anticlockwise moment
Rl force is creating a clockwise moment
10kN force is creating a anticlockwise moment
8kN force is creating a anticlockwise moment
4kN force is creating a clockwise moment.
So
(Rl x 9) + (4 x 1.5) = (8 x 2.5) + (10 x 7.5) + (5 x 10)
9Rl + 6 = 20 + 75 + 50
Rl = 139 / 9 kN

12. Taking moments about Rl
Force 4kn is creating a clockwise moment
Force Rr is creating a anticlockwise moment
Force 8kn is creating a clockwise moment
Force 10kn is creating a clockwise moment
Force Rl is creating a zero moment
Force 5kn is creating a anticlockwise moment
So
(Rr x 9) + (5 x 1 ) = ( 10 x 1.5) + (8 x 6.5) + (4 x 10.5)
9Rr + 5 = 15 + 52 + 42
9Rr = 104 or Rr = 104 / 9

13. To crosscheck;
When system is in equilibrium then ∑V= 0, ∑H= 0, ∑M= 0
There is no H (horizontal force)
To check that ∑V= 0
Upward forces are Rl + Rr = 139/9 + 104/9 = 243/9 = 27 kN
Downward forces are 5 + 10 + 8 + 4 = 27 kN

14. Parallel Force Systems
 These can be divided in two classes:
 Like parallel systems
 Unlike parallel systems

15. 10m
20kN
60kN
2m4m4m
50kN
40kN 10kN
Find the resultant of the unlike forces in the diagram above.
Magnitude of resultant = -50-60+20+40+10 = -40kN
Direction = Downward
Position= suppose x m distance from A
Taking moments about A
40 x x = (60 x 10) - (20 x 4) - ( 40 x 8) - (10 x 12)
x = 2m
R
A
2m

16. COUPLES
 In the previous examples the force system was of
parallel forces that were not in equilibrium because
the downward forces were more than the upward
forces.
 Supposing the sum of upward and downward forces
were equal, then two possibilities would be there:
 If the resultant of the upward and downward forces were in the
same line of action the system would be in equilibrium
 If these were not in a straight line the effect would of pure
rotation – which is known as a ‘couple’.
 Hence, a couple consists of two equal unlike parallel
forces acting out of line.

17. Anti clock wise couple Clock wise couple
Practical examples of couples can be seen within a beam that is loaded.
Within the beam the couple that counters the loading couple is known as the
“moment of resistance”

18. Properties of a couple:
A couple cannot be balanced by a single force. It requires another couple of
equal and opposite moment for a state of equilibrium.
A couple has the same moment about any point in a plane of forces which is
equal to the value of the couple itself.

19. 1.2m
20N
20N
20N
20N
20N
20N
0.2m
0.2m
0.2m
0.2m
Find the moment of the couple that is generated by the system of forces shown in the
figure by: i) finding the force and arm of the couple and ii) taking the moments of the
given forces about any convenient point.

20. 1.2m
A 20N
B 20N
D 20N
C 20N
E 20N
F 20N
0.2m
0.2m
0.4m
0.2m
0.4 + 0.2 + 0.2m
i) finding the force and arm of the couple:
Moment of resultant = Resultant moment
So, find the resultant of the two set of
Forces
Forces to the right:
Value of resultant = sum of all = 60N
Position: all the forces are equal and
equally spaced So, resultant will be in line
with the middle force
So, it will be in line with the middle force
So, the arm = 0.2 + 0.4 +0 .2 = 0.8 m
So the couple is 60 x 0.8 = 48 Nm
The couple will be acting CW
ii Finding the value of the couple by
taking moments about a convenient
point :
Take moment about the topmost force
Moment by B = ACW = 20 x 0.2 = 4 Nm
Moment by C = ACW = 20 x 0.4 = 8 Nm
Moment by D = CW = 20 x 0.8 = 16 Nm
Moment by E = CW = 20 x 1.0 = 20Nm
Moment by F = CW = 20 x 1.2m = 24Nm
So, the couple will be acting CW
Value of couple = (16 + 20 + 24) – (8 + 4)
= 60 – 12 = 48 Nm
60N
60N
0.2m

21. Degree of static indeterminacy
Degree of static indeterminacy is the number of
(independent) unknown static quantities (e.g. internal
member forces and support reactions) that must be
prescribed in addition to the use of equilibrium
equations to completely describe the static state of
the structure. The static state of the structure is known
if and only if all support reactions and the internal
member forces at any locations within the structure
can be determined. Those (independent) unknown
static quantities are termed as redundant unknowns.
The structure is said to be statically determinate if
and only if the degree of static indeterminacy is equal
to zero; in the contrary, the structure is said to be
statically indeterminate if and only if the degree of