Download - Chapter 2 basic structure concepts
![Page 1: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/1.jpg)
Chapter 2
Basic Structure Concepts
Jan
uar
y 20
15
1
![Page 2: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/2.jpg)
2.0 : BASIC STRUCTURE CONCEPT
2.1 Forces
2.2 Equilibrium and Reactions
2.3 Moments
2.4 Stress and Strain
2.5 Elastic and plastic range
2.6 Primary Loads & Secondary Loads
Jan
uar
y 20
15
2
![Page 3: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/3.jpg)
2.1 Forces
A force is that which tends to exert motion, tension or compression on an object.
The loads acting on a structure have mass which is usually measured in kilograms(kg).
The basic unit of force is the newton (N). The force exerted on a structure by a static load is dependent upon both its mass and the force of gravity.
The force exerted by a body as a result of gravity can be described as its weight.
F =W= m x g
Therefore, the force exerted by a mass of 1 kg is:
F = m x g
= 1 x 9.81
=9.81 N @ 0.00981 kN
1 kN = 1000 N
Jan
uar
y 20
15
3
![Page 4: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/4.jpg)
2.1 Forces
Scalars and Vectors
Scalars Vectors
Examples: mass, volume force, velocity
Characteristics: It has a magnitude It has a magnitude
(positive or negative) and direction
Addition rule: Simple arithmetic Parallelogram law
Special Notation: None Bold font, a line, an
arrow or a “carrot”
Jan
uar
y 20
15
4
![Page 5: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/5.jpg)
2.1 Forces
Application of Vector Addition
There are four
concurrent cable
forces acting on the
bracket.
How do you
determine the
resultant force acting
on the bracket ?
Jan
uar
y 20
15
5
![Page 6: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/6.jpg)
2.1 Forces
Vector Operations
Scalar Multiplication
and Division
Jan
uar
y 20
15
6
![Page 7: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/7.jpg)
2.1 Forces
Vector Addition Using Either the Parallelogram or Triangle
Parallelogram Law:
Triangle method (always
‘tip to tail’):
How do you subtract a vector? How can you add
more than two concurrent vectors graphically ?
Jan
uar
y 20
15
7
![Page 8: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/8.jpg)
2.1 Forces
Resolution of a vector
“Resolution” of a vector is breaking up a vector into components. It is kind of like using the parallelogram law in reverse.
Jan
uar
y 20
15
8
![Page 9: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/9.jpg)
2.1 Forces
Each component of the vector is
shown as a magnitude and a
direction.
Cartesian Vector Notation
We ‘ resolve’ vectors into components
using the x and y axes system
The directions are based on the x
and y axes. We use the “unit vectors”
i and j to designate the x and y axes.
Jan
uar
y 20
15
9
![Page 10: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/10.jpg)
2.1 Forces
For example,
F = Fx i + Fy j or F' = F'x i + F'y j
The x and y axes are always perpendicular to each
other. Together, they can be directed at any
inclination.
Jan
uar
y 20
15
10
![Page 11: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/11.jpg)
2.1 Forces
Addition of Several Vectors
Step 1 is to resolve each force into its
components
Step 2 is to add all the x components
together and add all the y components
together. These two totals become the
resultant vector.
Step 3 is to find the magnitude and
angle of the resultant vector.
Jan
uar
y 20
15
11
![Page 12: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/12.jpg)
2.1 Forces
Jan
uar
y 20
15
12
![Page 13: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/13.jpg)
2.1 Forces
Jan
uar
y 20
15
13
![Page 14: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/14.jpg)
2.1 Forces
Example 1
The screw eye in Figure below is subjected to forces F1 and F2. Determine the magnitude and direction (measured from x-positive axis) of the resultant force.
F2= 150 N
F1= 100 N 10o
15o
Jan
uar
y 20
15
14
![Page 15: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/15.jpg)
2.1 Forces Solution
10o
15o
15o + 90o + 10o
= 115o
R =√(1002 +1502 – (2x 100 x 150 x cos 115 )
= 212.55 N
Sin /150 = sin 115/212.55
Sin = (sin 115/212.55) x 150
= 39.76o
Direction = 15o + 39.76o
=54.76o
Jan
uar
y 20
15
15
![Page 16: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/16.jpg)
2.1 Forces
Example 2 Determine the magnitude of the resultant force and its direction, measured clockwise from the positive x-axis
Solution
a) Resolve the forces in their x-y components.
b) Add the respective components to get the resultant vector.
c) Find magnitude and angle from the resultant components.
Jan
uar
y 20
15
16
![Page 17: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/17.jpg)
2.1 Forces
F1 = { 15 sin 40° i + 15 cos 40° j } kN
= { 9.642 i + 11.49 j } kN
F2 = { -(12/13)26 i + (5/13)26 j } kN
= { -24 i + 10 j } kN
F3 = { 36 cos 30° i – 36 sin 30° j } kN
= { 31.18 i – 18 j } kN
Jan
uar
y 20
15
17
![Page 18: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/18.jpg)
2.1 Forces
Summing up all the i and j components respectively, we get,
FR = { (9.642 – 24 + 31.18) i + (11.49 + 10 – 18) j } kN
= { 16.82 i + 3.49 j } kN
FR = ((16.82)2 + (3.49)2)1/2 = 17.2 kN
= tan-1(3.49/16.82) = 11.7°
Jan
uar
y 20
15
18
![Page 19: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/19.jpg)
2.1 Forces
Example 3 Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis
Solution
a) Resolve the forces in their x-y components.
b) Add the respective components to get the resultant vector.
c) Find magnitude and angle from the resultant components.
Jan
uar
y 20
15
19
![Page 20: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/20.jpg)
2.1 Forces
F1 = { (4/5) 850 i - (3/5) 850 j } N
= { 680 i - 510 j } N
F2 = { -625 sin(30°) i - 625 cos(30°) j } N
= { -312.5 i - 541.3 j } N
F3 = { -750 sin(45°) i + 750 cos(45°) j } N
{ -530.3 i + 530.3 j } N
Jan
uar
y 20
15
20
![Page 21: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/21.jpg)
2.1 Forces
Summing up all the i and j components respectively, we get,
FR = { (680 – 312.5 – 530.3) i + (-510 – 541.3 + 530.3) j }N
= { - 162.8 i - 521 j } N
FR = ((162.8)2 + (521)2) ½ = 546 N
= tan–1(521/162.8) = 72.64° or
From Positive x axis
= 180 + 72.64 = 253 °
Jan
uar
y 20
15
21
![Page 22: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/22.jpg)
TUTORIAL 1 (FORCES)
Question 1
Two forces are applied to an eye bolt fastened to a beam. Determine the magnitude and direction of their resultant.
6 kN
4.5 kN
25o
50o
Jan
uar
y 20
15
22
![Page 23: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/23.jpg)
Question 2
Two forces P and Q are applied as shown at point A of a hook support. Knowing that P = 60 kN and Q = 100 kN. Determine the magnitude and direction of their resultant.
(Ans: R = 150 kN, 76o towards x axis positive)
P Q
15o 30o
Jan
uar
y 20
15
24
![Page 24: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/24.jpg)
Question 3
The cables AB and AD help support pole AC. Knowing that the tension is 500 N in AB and 160 N in AD. Determine the magnitude and direction of the resultant of the forces exerted by cables at A.
(Ans: R = 575 N, 113o towards x axis positive)
A
B C D
2 m 1.5 m
2.5 m
Jan
uar
y 20
15
25
![Page 25: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/25.jpg)
Question 4
Determine the resultant and direction from x-axis positive of the five forces shown in Figure below by the graphical method.
(Ans: R = 32.5 kN, = 124o)
y
x
8 kN 9 kN
4 kN
60o
25 kN
3 kN
15o
30o
20o
Jan
uar
y 20
15
26
![Page 26: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/26.jpg)
Question 5
Determine the magnitude and direction measured counterclockwise from the positive x axis of the resultant force of the three acting on the ring A. Take F1 = 500 N and = 20o
(Ans: R = 1.03 kN, = 87.9o)
y
x
600 N F1
400 N
30o 3
4
5
Jan
uar
y 20
15
27
![Page 27: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/27.jpg)
Question 6
Three forces are applied at the end of the boom O. Determine the magnitude and orientation of the resultant force.
(Ans: FR= 485 N, = 37.7o)
y
x
F3 = 200N F2= 250 N
F1 = 400 N
45o
3
4
5
Jan
uar
y 20
15
29
![Page 28: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/28.jpg)
• Question 7
• Two cables are attached to the frame shown in Figure below. Using trigonometry, determine :
(a) the required magnitude of the force P if the resultant R of the two forces applied at A is to be vertical.
(b) the corresponding magnitude of R.
(Ans: 489 N, 738 N)
A
A
25o 35o
360 N
P
Jan
uar
y 20
15
30
![Page 29: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/29.jpg)
2.2 Equilibrium and Reactions
Any structure subjected to loads must be provided with supports to prevent it from moving. The forces generated on the structure by these supports are called reactions. If the structure is in equilibrium (i.e. not moving) then the net forces from the loads and reactions must be zero in all directions.
Fx = 0
Fy = 0
M = 0
Jan
uar
y 20
15
31
![Page 31: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/31.jpg)
Example 1
Determine the magnitudes of F1 and F2 so that particle P is in equilibrium.
(Ans: F1 = 435 N, F2 = 171 N)
P
F1 F2
400 N
3
4
5
30o
60o
Jan
uar
y 20
15
33
![Page 32: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/32.jpg)
Solution
The two unknown magnitudes F1 and F2 can be obtained from the two
scalar equations of equilibrium, ∑Fx=0 and ∑Fy = 0. To apply these
equations, the x, y axes are established on the free body diagram and
forces must be resolved into its x and y components.
P
F1 F2
400 N
3
4
5
30o
60o
400 cos 30o
400 sin 30o
F2 cos 60o
F2 sin 60o
F1 (4/5)
F1 (3/5)
Jan
uar
y 20
15
34
![Page 33: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/33.jpg)
Solution
∑ Fx =0 ; -400 sin 30o + F1(4/5) –F2sin 60o=0
∑ Fy =0; 400 cos 30o -F1(3/5) –F2cos 60o=0
(1)
(2)
Simplify : 0.8 F1 – 0.866 F2 = 200 (3)
-0.6 F1 – 0.5 F2 = -346.61 (4)
Solving Eqs. (3) & (4) by simultaneous equations to determine F1 & F2.
Ans : F1 = 435.08 N
F2 = 170.97 N
Jan
uar
y 20
15
35
![Page 34: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/34.jpg)
Application of equilibrium concepts in truss
A truss is a structure composed of slender
members joined together at their end
points.
The members commonly used in
construction consist of wooden struts
or metal bars
The joint connections are usually formed
by bolting or welding the ends of the
members to a common plate, called a
gusset plate, as shown in Fig below.
•
Jan
uar
y 20
15
36
![Page 35: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/35.jpg)
Jan
uar
y 20
15
37
![Page 36: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/36.jpg)
38
Timber Roof Trusses
Steel Bridge Trusses
Jan
uar
y 20
15
![Page 37: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/37.jpg)
Example • Determine the force in each member of the Pratt bridge truss
shown. State whether each member is in tension or compression.
6 kN 6 kN 6 kN
4 m
3 m 3 m 3 m 3 m
A
B
C
D
E
F
G H
Jan
uar
y 20
15
39
![Page 38: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/38.jpg)
2.3 Moments
APPLICATION
What is the net effect of the two
forces on the wheel?
Jan
uar
y 20
15
40
![Page 39: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/39.jpg)
APPLICATION
What is the effect of the 30 N force on
the lug nut?
Jan
uar
y 20
15
41
![Page 40: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/40.jpg)
The moment of a force about a point or axis provides a measure of the tendency of the force to cause a body to rotate about the point or axis.
This tendency for rotation caused by force is sometimes called a torque, but most often it is called the moment of a force or simply the moment
The magnitude of a moment about a point is the value of the force multiplied by the perpendicular distance from the line of action of the force to the point
M= T = F x d
The unit is Nm or kNm
The typical sign convention for
moment is that counter-clockwise
is considered positive.
Jan
uar
y 20
15
42
![Page 41: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/41.jpg)
Example 1
From Figure below, determine the moment of the force about point O. If the moment is increased to 5 kNm, what is the maximum mass can be supported by the diving board?
1.5 m
Mass = 250 kg
Jan
uar
y 20
15
43
![Page 42: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/42.jpg)
Solution
F = mg
= 250 (9.81)= 2452.5 N = 2.45 kN
1.5 m
Mo = ( 2.45 x 1.5) = 3.68 kNm
If Mo = 5 kNm, then new load is
Mo = F x d
5 = F x 1.5
F = 3.33 kN = 3333.33 N
F = m x g
3333.33 = m x 9.81
m = 339.79 kg
F
Jan
uar
y 20
15
44
![Page 43: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/43.jpg)
Example 2 For each case illustrated in Figure below, determine the moment of the force about point O.
5O kN
2 m
0.75 m
Solution
Mo = - (50 x 0.75) = - 37.5 kNm
Jan
uar
y 20
15
45
![Page 44: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/44.jpg)
Example 3 For each case illustrated in Figure below, determine the moment of the force about point O.
6O kN
3 m
45o
Solution
1 sin 45o
Mo = 60 x (1 sin 45o) = 42.43 kNm
Jan
uar
y 20
15
46
![Page 45: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/45.jpg)
Example 4 A 400 N force is applied to the frame and = 20o. Find the moment of the force at A.
Solution
400 sin 20o
400 cos 20o
MA = (400 sin 20ox 3) + (400 cos 20o x 2)
= 1162.18 Nm Jan
uar
y 20
15
47
![Page 46: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/46.jpg)
Figure above shows a bridge deck, of weight 500 kN, supporting a heavy vehicle
weighing 300 kN. Find the value of the support reactions at A and B when the load
is in the position shown.
Ans: RA = 430 kN, RB = 370 kN
Example 5
Jan
uar
y 20
15
48
![Page 47: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/47.jpg)
Solution
500 kN
5 m
300 kN
A B 4 m
∑ MA =0 ;
∑ Fy =0 ;
RB (10) – ( 300 x 4) – (500 x 5) = 0
RB = 370 kN
RA + RB – 300 – 500 = 0
RA = 430 kN
Jan
uar
y 20
15
49
![Page 48: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/48.jpg)
Example 6
Figure shows an anchorage for the
cable of a ski- lift, The cable produce a
total pull of 220 kN at the top of the
anchorage. Determine the support
reactions.
Ans: RAH = 191 kN, RAV = -196 kN
RB = 306 kN
Jan
uar
y 20
15
50
![Page 49: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/49.jpg)
Solution
A B
220 cos 30o
220 sin 30o
RB RAV
RAH
∑ MA =0 ;
RB (2.7) – ( 220 sin 30o x 1.8) – (220 cos 30o x
3.3) = 0
RB = 306 kN
∑ Fx =0 ;
-RAH + 220 cos 30o = 0
RAH = 191kN
RAV + RB – 220 sin 30o = 0
RAV = - 196 kN
∑ Fy =0 ;
Jan
uar
y 20
15
51
![Page 50: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/50.jpg)
2.4 Stress and Strain
Stress Concept of stress
To obtain distribution of force acting over a sectioned area
Assumptions of material:
1. It is continuous (uniform distribution of matter)
2. It is cohesive (all portions are connected together)
Normal stress
Intensity of force, or force per unit area, acting normal to ΔA
Symbol used for normal stress, is σ (sigma)
σz = lim
ΔA →0
ΔFz
ΔA
Jan
uar
y 20
15
52
![Page 51: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/51.jpg)
Tensile stress: normal force “pulls” or “stretches” the area element ΔA
Compressive stress: normal force “pushes” or “compresses” area
element ΔA
Shear stress
Intensity of force, or force per unit area, acting tangent to ΔA
Symbol used for normal stress is τ (tau)
τzx = lim
ΔA →0
ΔFx
ΔA
τzy = lim ΔA →0
ΔFy
ΔA
Units (SI system)
Newtons per square meter (N/m2) or a pascal (1 Pa = 1 N/m2)
kPa = 103 N/m2 (kilo-pascal)
MPa = 106 N/m2 (mega-pascal)
GPa = 109 N/m2 (giga-pascal)
Jan
uar
y 20
15
53
![Page 52: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/52.jpg)
Shear in Nature
Jan
uar
y 20
15
54
![Page 53: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/53.jpg)
Example 1
The 80 kg lamp is supported by two rods AB and BC as shown in
Figure below. If AB has a diameter of 10 mm and BC has a diameter
of 8 mm, determine the average normal stress in each rod.
A
B
C
60o
3
4
Jan
uar
y 20
15
55
![Page 54: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/54.jpg)
A
B
C
60o
3
4
F = mg
= 80 x 9.81
= 784.8 N
= 0.78 kN
F1 F2
F1cos 60o
F1sin 60o
F2cos
F2sin
Solution
Jan
uar
y 20
15
56
![Page 55: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/55.jpg)
Solution
∑ Fx =0 ; – F1cos 60o + F2(4/5) =0
∑ Fy =0; F1sin 60o+ F2(3/5) – 0.78 =0
(1)
(2)
Simplify : - 0.5 F1 + 0.8 F2 = 0 (3)
0.866 F1 + 0.6 F2 = 0.78 (4)
Solving Eqs. (3) & (4) by simultaneous equations to determine F1 & F2.
0.6 x Eqs (3) - 0.3 F1 + 0.48 F2 = 0
0.8 x Eqs (4) 0.69 F1 + 0.48 F2 = 0.624
- 0.99 F1 = - 0.624
F1 = 0.63 kN
Subs F1 into Eqs (3)
-
- 0.5 F1 + 0.8 F2 = 0
- 0.5 x 0.63 + 0.8 F2 = 0
F2 = 0.315/0.8 = 0.39 kN
Jan
uar
y 20
15
57
![Page 56: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/56.jpg)
Solution
1 = F1 / A1
= 0.63 /( x 0.0052)
= 8021.41 kN/m2
2 = F2 / A2
= 0.39/( x 0.0042)
= 7758.8 kN/m2
Jan
uar
y 20
15
58
![Page 57: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/57.jpg)
Shear Stress
Shear stress is the stress component that act in the plane of the sectioned area.
Consider a force F acting to the bar
For rigid supports, and F is large enough, bar will deform and fail along the planes identified by AB and CD
Free-body diagram indicates that shear force, V = F/2 be applied at both sections to ensure equilibrium
Jan
uar
y 20
15
59
![Page 58: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/58.jpg)
Average shear stress over each section is:
P
A avg =
avg = average shear stress at section, assumed to
be same at each pt on the section
V = internal resultant shear force at section
determined from equations of equilibrium
A = area of section
Jan
uar
y 20
15
60
![Page 59: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/59.jpg)
Example 1
The bar shown in Figure below has a square cross section for which
the depth and thickness are 40 mm. If an axial force of 800 N is applied
along the centroidal axis of the bar’s cross-sectional area, determine
average normal stress and average shear stress acting on the material
along section planes a-a.
Jan
uar
y 20
15
61
![Page 60: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/60.jpg)
Internal loading
Based on free-body diagram, Resultant loading of axial force, P = 800 N
Average normal stress
= P/A = 800/(0.04)(0.04) = 500 kPa
Average shear stress
No shear stress exists on the section, since the shear force at the section is zero.
Jan
uar
y 20
15
62
![Page 61: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/61.jpg)
Example 2
Three plates are held together by two cyclindrical rivets. If a direct pull
of 5 kN is applied between one plate and the other two, estimate the
diameter of the rivets. The shear stress in the rivets is not to exceed 40
N/mm2.
5 kN
Jan
uar
y 20
15
63
![Page 62: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/62.jpg)
Solution
• 4 sliding areas (Double shear)
P
A =
40 = 5000 4(2/4)
= 6.3 mm
Jan
uar
y 20
15
64
![Page 63: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/63.jpg)
Question 1
A 50 kN axial load is applied to a short wooden post which is supported
by a square concrete footing resting on distributed soil. Determine
(a) The maximum bearing stress on the concrete footing
(b) The size of the footing for which the average bearing stress on the
soil is 150 kPa.
(Ans: B = 4MPa, b= 577 mm)
Plan
125 mm
100 mm b
b
50 kN
Jan
uar
y 20
15
65
TUTORIAL 2 (STRESS
![Page 64: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/64.jpg)
Question 2
The column is subjected to an axial force of 8 kN at its top. If the cross
sectional area has the dimensions shown in the figure, determine the
average normal stress at section a-a. (Ans: = 1.74 MPa)
Plan
Front Elevation
160 mm
160 mm
a a
10 mm
10mm
10 mm
8 kN
160 m
m
Jan
uar
y 20
15
66
![Page 65: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/65.jpg)
Question 3
The 20 kg lamp is supported by two steel rods connected by a ring A.
Determine which rod is subjected to the greater average normal stress
and compute its value.
(Ans: = 2.33 N/mm2)
B
A
C
60o
45o
12 mm
10 mm
Jan
uar
y 20
15
67
![Page 66: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/66.jpg)
Question 4
A square hole having 12 mm sides is to be punched out of a metal
plate 1.6 mm thick. The shear stress required to cause fracture is 350
N/mm2. What force must be applied to punch die? What would be the
compressive stress in the punch?
(Ans: 26.88 kN, 0.19 kN/mm2) Jan
uar
y 20
15
68
![Page 67: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/67.jpg)
ALLOWABLE STRESS
When designing a structural member or mechanical element, the stress in it must be restricted to safe level
Choose an allowable load that is less than the load the member can fully support
One method used is the factor of safety (F.S.)
F.S. = Ffail
Fallow
If load applied is linearly related to stress developed within
member, then F.S. can also be expressed as:
F.S. = σfail
σallow F.S. =
fail
allow
In all the equations, F.S. is chosen to be greater than 1, to avoid
potential for failure
Specific values will depend on types of material used and its
intended purpose
Jan
uar
y 20
15
69
![Page 68: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/68.jpg)
Strain
Loads cause bodies to deform, thus points in the body will undergo displacements or changes in position
Normal strain () is a measure of elongation or contraction of small line segment in the body
Normal Strain () = Change in length = L
Original length L
Has no unit
Shear strain () is a measure of the change in angle that occurs between two small line segments that are originally perpendicular to each other.
Shear Strain () = x
L
L
x
Jan
uar
y 20
15
70
![Page 69: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/69.jpg)
Conventional stress-strain diagram Figure shows the characteristic stress-strain diagram for steel, a
commonly used material for structural members and mechanical elements
Jan
uar
y 20
15
71
![Page 70: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/70.jpg)
Elastic behavior. A straight line Stress is proportional to strain, i.e., linearly elastic Upper stress limit, or proportional limit; σpl
If load is removed upon reaching
elastic limit, specimen will return to its
original shape
Jan
uar
y 20
15
72
![Page 71: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/71.jpg)
Yielding. Material deforms permanently; yielding;
plastic deformation Yield stress, σY
Once yield point reached, specimen continues to elongate (strain) without any increase in load
Note figure not drawn to scale, otherwise induced strains is 10-40 times larger than in elastic limit
Material is referred to as being perfectly plastic
Jan
uar
y 20
15
73
![Page 72: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/72.jpg)
Strain hardening.
Ultimate stress, σu
While specimen is elongating, its x-
sectional area will decrease Decrease in area is fairly uniform over entire
gauge length
Jan
uar
y 20
15
74
![Page 73: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/73.jpg)
Necking.
At ultimate stress, x-sectional area
begins to decrease in a localized
region
As a result, a constriction or “neck” tends
to form in this region as specimen
elongates further
Specimen finally breaks at fracture stress, σf
Jan
uar
y 20
15
75
![Page 74: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/74.jpg)
Hooke’s Law
When a material is worked within its elastic limit, the extension is
proportional to the force.
Strain Stress
Stress = Constant (E)
Strain
This constant is known as the modulus of elasticity or Young’s
Modulus
Jan
uar
y 20
15
76
![Page 75: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/75.jpg)
Example 1
A tie-bar in a steel structure is of rectangular section 30 mm x 50
mm. The extension measured in a 250 mm length of the tie bar
when load is applied to the structure is 0.1 mm. Find :-
i) The tensile stress in the bar
ii) The tensile force
iii) The factor of safety used
Take E = 205 kN/mm2 and Ultimate stress = 460 N/mm2
Jan
uar
y 20
15
77
![Page 76: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/76.jpg)
Solution
i) The tensile stress in the bar
E = stress/strain
Strain (ε) = ∆L/L
= 0.1/250
= 4 x 10-4
E = stress/strain
205 = / 4 x 10-4
= 205 x 4 x 10-4 = 0.082 kN/mm2 = 82 N/mm2
ii) The tensile force
= F/A
82 = F/(30 x 50)
F = 82 x (30 x 50) = 123,000 = 123 kN
iiii) The factor of safety used
F.S. = Ultimate stress/ Tensile stress
= 460 /82 = 5.6
Jan
uar
y 20
15
78
![Page 77: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/77.jpg)
Example 2
The ultimate stress for a steel is 450 N/mm2. What is the maximum load which a rod 50 mm diameter can carry with a factor of safety of 5? If the rod is 1.5 m long, determine the extension under this loading.( E = 200 kN/mm2.)
Solution
The factor of safety
F.S. = Ultimate stress/ Tensile stress
5 = 450 / tensile stress
Tensile stress = 450/5 = 90 N/mm2
= F/A
90 = F/(π r2)
F = 90 x (π x 252) = 176,714.59 N = 176.71 kN
Jan
uar
y 20
15
79
![Page 78: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/78.jpg)
Solution
The tensile stress in the bar
E = stress/strain
200 x 103 = 90/ ε
ε = 90/ 200 x 103
= 4.5 x 10-4
Strain (ε) = ∆L/L
4.5 x 10-4 = ∆L/1500
∆L = 0.675 mm
Jan
uar
y 20
15
80
![Page 79: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/79.jpg)
Exercise 1
A flat steel tie-bar 4.5 m long, is found to be 2.4mm short. It is
sprung into place by means of drafts driven into holes in the end
of the bar. Determine:
(a) the stress in the bar
(b) the factor of safety if the material of the tie-bar has an
ultimate stress of 450 N/mm2. Take E for the material as
205 kN/mm2.
( Ans: 109.3 N/mm2, 4.117)
Jan
uar
y 20
15
81
TUTORIAL 3 (STRESS, STRAIN & FACTOR OF SAFETY)
![Page 80: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/80.jpg)
Exercise 2
A metal tube of outside diameter 75 mm and length 1.65 m is to
carry a compressive load of 60 kN. If the allowable axial stress is
75 N/mm2 , calculate the inside diameter of the tube. If E of the
material is 90 kN/mm2., by how much will the tube shorten under
this load?
Ans: 67.87 mm, 1.375 mm)
Jan
uar
y 20
15
82
![Page 81: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/81.jpg)
2.5 Primary Loads & Secondary Loads
DEADLOADS Dead Loads are those loads which are considered to act permanently; they are "dead," stationary, and unable to be removed. The self-weight of the structural members normally provides the largest portion of the dead load of a building. This will clearly vary with the actual materials chosen. Permanent non-structural elements such as roofing, concrete, flooring, pipes, ducts, interior partition walls, Environmental Control Systems machinery, elevator machinery and all other construction systems within a building must also be included in the calculation of the total dead load. These loads are represented by the red arrow in the illustration.
Primary Loads are divided into three broad categories according to the way in which they act upon the structure or structural element. These are DEAD LOADS, LIVE LOADS(IMPOSED LOADS) and WIND LOADS
Jan
uar
y 20
15
83
![Page 82: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/82.jpg)
2.5 Primary Loads
Unit weights of various building materials (kN/m3)
Materials Unit Weight
Aluminium 24
Bricks 22*
Concrete 24
Concrete blocks (lightweight) 12*
Concrete blocks (dense) 22*
Glass fibre composite 18
Steel 70
Timber 6*
* Subject to considerable variation
Jan
uar
y 20
15
84
![Page 83: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/83.jpg)
2.5 Primary Loads
Unit weights of various sheet materials (kN/m2)
The dead load of a floor or a roof is generally evaluated for one square meter of floor or roof area
Sheet materials kN/m2
Acoustic ceiling tiles 0.1
Asphalt (19 mm) 0.45
Aluminium roof sheeting 0.04
Glass (single glazing) 0.1
Plaster (per face of wall) 0.3
Plasterboard 0.15
Rafters, battens and felt 0.14
Sand/cement screed (25 mm) 0.6
Slates 0.6
Steel roof sheeting 0.15
Timber floorboards 0.15
Vinyl Tiles 0.05
Jan
uar
y 20
15
85
![Page 84: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/84.jpg)
• LIVE LOADS
Live Loads are not permanent and can
change in magnitude. They include
items found within a building such as
furniture, pianos, safes, people, books,
cars, computers, machinery, or stored
materials, as well as environmental
effects such as loads due to the sun,
earth or weather.
• WIND LOADS
Wind and earthquakes loads are put
into the special category of lateral live
loads due to the severity of their action
upon a building and their potential to
cause failure.
Jan
uar
y 20
15
86
![Page 85: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/85.jpg)
Secondary loads
Structures can be subjected to secondary loads from
temperature changes, shrinkage of members and settlement of
supports.
Jan
uar
y 20
15
87
![Page 86: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/86.jpg)
Example 1
Figure shows a precast concrete
Beam which is 10.5 m long.
a) Calculate the weight of the
beam per unit length in kN/m
b) Calculate the total weight of
the beam
Jan
uar
y 20
15
88
![Page 87: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/87.jpg)
Solution
a) Cross sectional area of the
beam
= (0.6 x 0.25) – (0.4 x 0.15)
= 0.09 m2
Unit weight of concrete = 24 kN/m3
Weight per unit length = 0.09 x 24
= 2.16 kN/m
b) Total weight of the beam
= 2.16 x 10.5 = 22.68 kN
Jan
uar
y 20
15
89
![Page 88: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/88.jpg)
Example 2 The floor in a multi-storey office. Building consists of the following:
• Vinyl tiles
• 40 mm sand/cement screed
• 125 mm reinforced concrete
slab
• Acoustic tile suspended ceiling
Determine the dead load in kN/m2
Jan
uar
y 20
15
90
![Page 89: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/89.jpg)
Solution
From the table given
• Vinyl tiles = 0.05
• 40 mm sand/cement screed = 0.6 x (40/25) = 0.96
• 125 mm reinforced concrete = 0.125 x 24 = 3.00
slab
• Acoustic tile suspended ceiling = 0.10
The dead load = 4.11 kN/m2
Jan
uar
y 20
15
91
![Page 90: Chapter 2 basic structure concepts](https://reader030.vdocuments.us/reader030/viewer/2022021508/5a64a6767f8b9a31568b5d45/html5/thumbnails/90.jpg)
Exercise 1
Figure shows the outer wall of a multi-storey
building which is supported on a beam at
each floor level. The wall consist of a 1.2 m
height of cavity wall supporting 1.3 m high
double glazing. The cavity wall construction
is 102.5 mm of brickwork, a 75 mm cavity
and 100 mm of plastered lightweight concrete
blockwork.
Determine the dead load on one beam
in kN/m of beam.
(Ans: 4.77 kN/m)
Jan
uar
y 20
15
92
TUTORIAL 4 (LOADS)