Simple calculations in mechanics of materials

1. SIMPLE CALCULATIONS
IN MECHANICS OF
MATERIALS
Andi cakra

2. OBJECTIVES
The main objective of this basic mechanics course
should be to develop in the participant the ability to
analyze a founded problem and to find its solution
according to fundamental and well understood
principles.

3. GENERAL APPROACH
In this course the study of mechanics of material base
understanding of a few basic concepts and the use of
simplified models. This approach makes it possible to
determine appropriate formula to solve the founded
problems.

4. GENERAL CONSIDERATION
1. Type of load and Stress caused by the load.
2.Selection of materials.
3.Form and size of the parts.

5. Example
Determine the shear and bending moment formula for the beam loading
shown

6. Courtessy of Mechanics of Material 3rd Edition Solution Manual Problem 5.1

7. From the solution above the formula to determine the reaction at
support shear and the bending moment are,
M =
Pab
L
, Bending moment
RA =
Pb
L
, Reaction at support A
RC =
Pa
L
, Reaction at support C
V =
Px
L
, Shear stress at any given poin

8. If P = 300 kg, L = 400 cm, a = 200 cm, b= 200 cm, Thus
Reaction at support, R = 300.200/400 = 150 kgf
Shear Force, V = 300.200/400 = 150 kgf
Bending Moment, M = 300.2002/400 = 30000 kgf.cm

9. Step 2
Determine the properties of the beam.
yG = Sa/A = 50760/1500 = 33.8 cm
Ia = Ai.yi2 + Ixi = 2248560+69840 = 2318400 cm4
Ix = Ia – A.yG
2= 2318400 – 1500(33.8)2 = 604740 cm4
Z = Ix/yG = 604740/33.8 =17892 cm3
Caonica,Lucio “Memahami Mekanika Teknik 2” penerbit Angkasa Bandung
i Ai yi Ai.yi Ai.yi
𝐈 𝐱𝐢 =
𝐛𝐡 𝟑
𝟏𝟐
1 600 54 32400 1749600 7200
2 540 30 16200 486000 58320
3 360 6 2160 12960 4320
A = 1500 Sa = 50760 Σ = 2248560 69840

10. Step 3
Determine the Bending Stress and shear stress.
Bending Stress , σ =
M
Z
=
30000
17892
= 1.68 kgf/cm2
Shear stress, τ =
V
A
=
150
1500
= 0.1 kgf/cm2
The result obtained by these formula shall be less then
allowable stress. The allowable stress prices on vary depend
on standard that is used

11. Alternative to Solve similiar Case
1. Find the appropriate formula in text book, for examples
Megyesy, Eugene F “Pressure Vessel Handbook” 11th edition

12. Or
Property of Machinery Handbooks 26th Edition

13. If P = 300 kg, L = 400 cm, a = 200 cm, b= 200 cm, Thus from the table
above
Bending Moment, M =
ML
4
= 300.400/4 = 30000 kgf.cm
Reaction at support & shear, R = V =
P
2
= 300/2 = 150 kgf

14. 2. Determine the properties of the beam
Tabel 5.1 Properties of common cross section “Textbook of Machine Design”
Since the profile to be used not the I-Section aforementioned, we shall proceed with
the formula from previous step 2.

15. 3. Determine the Bending Stress and shear stress.
Bending Stress , σ =
M
Z
=
30000
17892
= 1.68 kgf/cm2
Shear stress, τ =
V
A
=
150
1500
= 0.1 kgf/cm2

16. If the profile is a standard Profile, it is easier to find it’s mechanical
properties.
Example: The same load to be applied to profile I 300 x 150 JIS G3101
find the shear and bending stress
Solution
1. Bending Moment, M =
ML
4
= 300.400/4 = 30000 kgf.cm
Reaction at support & shear, R = V =
P
2
= 300/2 = 150 kgf

17. 2. Find the properties

18. 3. Determine the shear and bending stress
from the table above
A = 46.78 cm2
Ix =7210 cm4
Shear stress, τ =
V
A
=
150
46.78
= 3.2 kgf/cm2
Bending Stress , σ =
M
Z
=
30000
7210
= 4.16 kgf/cm2

19. Reference
1. Beer, Ferdinand “Mechanics of Materials” 6th Edition,McGraw Hill
2012.
2. Khurmi,R.S,Ghupta, J.K “A Textbook of Machine Design” Eurasia
Publishing House(pvt.) Ltd. 2004
3. Canonica,Luca “Memahami Mekanika Teknik 2” Penerbit Angkasa
Bandung.