Ho1

PN. IRINAH BINTI ABDULLAH013 - 7443474

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%

JOHOR BAHRU POLYTECHNICDEPARTMENT OF

MECHANICAL ENGINEERINGKAJIDAYA BAHAN 1 / J3009

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.

6 CHAPTER IN STRENGHT OF MATERIAL 1

1. FORCES ON MATERIALS2. COMPOUND BAR AND THERMAL STRESSES3. SHEAR FORCE AND BENDING MOMENT4. BENDING STRESS5. DEFLECTION6. TORSION

CHAPTER 1

FORCES ON MATERIALS

Unit 1

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

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

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.

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

Unit 2

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.

Kawasan anjal. Tegasan berkadar

terus dengan terikan

P

Graf 2.1

A = Yeild pointB = Elastic pointC = Upper yield point

D = Lower yield pointE = Maximum forceF = Rupture

Max. force

yeildyeild Strain hardeningStrain hardeningElongation,Elongation,LL

Force,Force,PP

i. Yield stress= force on point C cross sectional area

O

Yy A

P

o

mm A

P

%100%

L

LL

ii. Max. stress = max.force on point E cross sectional area

iii. Percent of elongation= change on length original length X 100%

iv. Percent of area = change on area original area X 100%

%100%

oA

AA

example 1 :The tensile tests were conducted on a substance, the data acquired is recorded as follows:

Gauge length = 260mmFinal gauge length = 295mmOriginal diameter = 30mmEnd 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

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

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 loadb. Minimum load at yield levels Ac. minimum lengthd. maximum diameter

UNIT 3

3.1 Safety Factor Formula, n

nm allowed

ny

allowed

work

y

P

Pn

work

m

P

Pn

3.2 safety margin

n -1Note: the ultimate tensile or yield stress

must be two times then allowable stress ,and FK should be larger than 1

example :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 stressb. cross-sectional areac. outside diameter of the cylinder

3.6 Strain Energy

Force Force

Stain energyStain energy

elongationelongation

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

200kN

40mm

520mm

figure 3.1figure 3.1

3.7 Poisson Ratio

AA

EEDD CC

BB

HH

GGFF

A tensile force P is act on square ABCD and it will change A tensile force P is act on square ABCD and it will change shape into EFGH, the action of this force, the strain that shape into EFGH, the action of this force, the strain that occurs towards x is known as LONGITUDINAL strain (occurs towards x is known as LONGITUDINAL strain (xx), ), while the square is also a shortage in the y-direction width while the square is also a shortage in the y-direction width measurement known as LATERAL strain (measurement known as LATERAL strain ( y). y).

LONGITUDINAL LONGITUDINAL strain strain

LATERAL LATERAL strain strain

EL

Lx

d

dy

dd

dd

LL

LL

example :A load of 45kN is act on figure 3.2 below. Given E = 200GN/m2, Ʋ = 0.4 calculate:a. stressb. longitudinal strainc. lateral strain

45kN45kN

45kN45kN

250mm30mm30mm

50mm50mm

figure3.2figure3.2

3.8 Shear stress

3.8.1 shear stress is the frictional between two surfaces abed and EFGH as shown in Figure 3.3

aa

ee

ccdd

bb

hh ggff

Shear force,PShear force,P


Figure 3.3Figure 3.3

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

8mmpunch D=20mm


Shear act on the surface

Three plates are connected using two ribet as figure 3.5. If shear stress is 35N/m2. calculate the diameter of ribet.

Example 2 :

8kN8kN


3.8.2 Shear strain will lead to distortions on one surface or structural blocks as in Figure 3.4


xx

yy


Example :

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.

V=120kNV=120kN

hh

dd

h=30mm

b=25mm

a=50mmbb

Rajah 3.5Rajah 3.5

Ho1

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