1. PN. IRINAH BINTI ABDULLAH
013 - 7443474

2. JOHOR BAHRU POLYTECHNIC
DEPARTMENT OF
MECHANICAL ENGINEERINGKAJIDAYA BAHAN 1 / J3009
CONTINUOUS ASSESSMENT (CA)
a. Quiz (2) = 20%
b. Assignments (2) = 30%
c. Case Study (1) = 20%
d. Test (2) = 30%
FINAL ASSESSMENT (FA)
a. Final Exam = 50%

3. REFERENCES
• 1. Ahmad Zafri Bin Zainudin & Yazid Bin Yahya
(1998) Mekanik Pepejal 1. Universiti
Teknologi Malaysia.
• 2. Ahmad Zafri Bin Zainudin (1998) Mekanik Bahan 1.
Universiti Teknologi Malaysia.
• 3. Beer & Johnston (1992) Mechanic Of Materials. McGraw
Hill.
• 4. G.H.Ryder (1983) Kekuatan Bahan, Edisi Bahasa
Malaysia. UPM.
• 5. I.Granet (1980) Strength Of Materials For Engineering
Technology. Reston.

4. 6 CHAPTER IN
STRENGHT OF MATERIAL 1
1. FORCES ON MATERIALS
2. COMPOUND BAR AND THERMAL STRESSES
3. SHEAR FORCE AND BENDING MOMENT
4. BENDING STRESS
5. DEFLECTION
6. TORSION

5. CHAPTER 1
FORCES ON MATERIALS

6. Unit 1

7. example 1 :
Figure 1.0 shows a steel bar has a rectangular
cross section measuring 30mm x 25mm. 40kN
tensile force is imposed on the bar. Calculate
the resulting stress.
40kN
40kN
Figure 1.0

8. Example 2 :
Figure 1.1 shows a block cylinder 30cm in
length, have a diameter of 10cm. This block is
a compression load of 70kN and shrink
0.02cm. Find a compressive stress and
compressive strain.
70kN
70kN
figure 1.1

9. example 3 :
A glass bar that has a 2m long and cross
section of 25mm x 20mm. The bar is forced by
52kN load. Determine the elongation of the
glass given modulus of elasticity 60GPa.

10. practice
1. A brick that has a cross
section of 15mm x 20mm
and length 30cm as
shown in Figure 1.2 bear
Abu weight of 68kg. Given
that the Young modulus
30GPa. Calculate stresses
and length changes that
occur on the bricks.
figure 1.2

11. Unit 2

12. 2.0 Hooke’s Law
elastic material is a material that changes
shape easily if the burden imposed against
him. It expenses or not subject to the
condition and would return to real condition.

13. ∆P
Kawasan anjal.
Tegasan berkadar
terus dengan terikan
Graf 2.1

14. Max. force
Force,
P
Elongation,∆ L
yeild Strain hardening
A = Yeild point D = Lower yield point
B = Elastic point E = Maximum force
C = Upper yield point F = Rupture

15. i. Yield stress= force on point C
cross sectional area
PY
σy =
AO
ii. Max. stress = max.force on point E
cross sectional area
Pm
σm =
Ao
iii. Percent of elongation= change on length
original length X 100%
∆L
%L = ×100%
L

16. iv. Percent of area = change on area
original area X 100%
∆A
%A = ×100%
Ao

17. example 1 :
The tensile tests were conducted on a
substance, the data acquired is recorded as
follows:
Gauge length = 260mm
Final gauge length = 295mm
Original diameter = 30mm
End of Diameter = 19mm
Force , kN 20 60 100 140 160 170 172 176 178
elongation x 10-3 50 160 260 360 410 440 470 550 720
Force , kN 180 190 220 240 257 261 242 229 180
elongation x 10-3 760 900 1460 1990 3120 4500 5800 5850 760

18. Draw a graph of load versus elongation and determine:
a.Young's modulus
b. The yield stress
c. The maximum stress
d. Percentage of area
e. Percent of elongation

19. Example 2 :
The tensile tests were conducted on a rod with a diameter
of 6.4mm and length of 30mm, the determinant of the
material are as follows:
Tensile strength = 500MN/m2
Yield stress = 210MN/m2
% elongation = 20%
% reduction in area = 60%
calculate:
a. maximum load
b. Minimum load at yield levels A
c. minimum length
d. maximum diameter

20. UNIT 3

21. 3.1 Safety Factor Formula, n
σy σm
σ allowed = σ allowed =
n n
Py Pm
n= n=
P work Pwork

22. 3.2 safety margin
n -1
Note: the ultimate tensile or yield stress
must be two times then allowable
stress ,and FK should be larger than 1

23. example :
A hollow cylinder used to support a
compressive load of 13kN. Ultimate stress is
350kN/m2, the cylinder is 1 cm thick and
safety factor is 3.0. calculate:
a. allowable stress
b. cross-sectional area
c. outside diameter of the cylinder

24. 3.6 Strain Energy
Force
Stain energy
elongation

25. Example :
A bar is loaded by a force as in Figure 3.1.
calculate the strain energy stored in the bar.
Given E = 200GN/m2.
200kN
∅40mm
0mm
52
200kN
figure 3.1

26. 3.7 Poisson Ratio
G A B
F
H E
D C
A tensile force P is act on square ABCD and it will change
shape into EFGH, the action of this force, the strain that
occurs towards x is known as LONGITUDINAL strain (ε x),
while the square is also a shortage in the y-direction width
measurement known as LATERAL strain (ε y).

27. LONGITUDINAL LATERAL
strain strain
∆L σ ∆d
εx = = εy =
L E d
∆L d
∆d
L

28. example :
A load of 45kN is act on figure 3.2 below. Given
E = 200GN/m2, Ʋ = 0.4 calculate:
a. stress
b. longitudinal strain
c. lateral strain
250mm
30mm
45kN
45kN
50mm
figure3.2

29. 3.8 Shear stress
3.8.1 shear stress is the frictional between two
surfaces abed and EFGH as shown in Figure 3.3
Shear force,P d c
a b
e f
Shear force,P g
h
Figure 3.3

30. Example 1 :
Figure 3.4 shows a punch of diameter 20 mm is used to drill a
hole on the plate of 8mm thick. Force 120kN is loaded at
punch, determine the shear stresses that occur on the
plate.
120kN
punch D=20mm
8mm
Shear act
on the
surface Figure 3.4

31. Example 2 :
Three plates are connected using two ribet as
figure 3.5. If shear stress is 35N/m2. calculate
the diameter of ribet.
8kN
Figure 3.5

32. 3.8.2 Shear strain will lead to distortions on one
surface or structural blocks as in Figure 3.4
x
Shear force,P γ
y
Figure 3.4

33. Example :
V=120kN
d
b=25mm
γ
h=30mm h
b
a=50mm
Rajah 3.5
Figure 3.5 above shows a pure copper plates the elastic
material located above its subjected to horizontal forces V.
Determine the shear stress, shear strain and horizontal
displacement of the plate. Given the rigidity modulus G =
45GPa.