ENGINEERING MECHANICS, EQUILIBRIUM, STATICS

1. DEVAPRAKASAM DEIVASAGAYAM
Professor of Mechanical Engineering
Room:11, LW, 2nd Floor
School of Mechanical and Building Sciences
Email: devaprakasam.d@vit.ac.in, dr.devaprakasam@gmail.com
MEE1002: Engineering Mechanics (2:1:0:0:3)
Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933

2. • Express the 2D and 3D equilibrium equations
for particle resulting from the application of
Newton’s 1st Law
0
0
0






Y
X
F
F
F
3D
0
0
0
0








Z
Y
X
F
F
F
F2D
0
0
0
|| 






F
F
F
2 Independent Eqns & 3 Independent Eqns &
Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
2D and 3D Equilibrium

3. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
3D Force rectangular Components

4. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
3D Force rectangular Components

5. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
3D Force rectangular Components

6. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
3D Force rectangular Components

7. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
3D Force rectangular Components

8. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
3D Force rectangular Components

9. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
3D Force rectangular Components

10. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
3D Force rectangular Components

11. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
3D Force rectangular Components

12. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
3D Force rectangular Components

13. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
Examples

14. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
Examples

15. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
Solution
The forces at A are,
TAB, TAC, TAD and P
P= P j. Express all the forces in unit vectors I, j, k
AB= - 4.2 i- 5.6j ; |AB|= 7.00 m
AC = 2.4 i+ -5.6j+ 4.2k; |AC|=7.4m
AD=-5.6 j - 3.3 k; |AD|= 6.5m
TAB = TAB λAB = [-0.6i-0.8j] TAB
TAC = TAC λAC =[0.3240i-756j+0.567k] TAC
TAD = TAD λAD = [-0.861j-0.5076k] TAD

16. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
Solution
Equilibrium condition
  0F TAB + TAC + TAD +Pj =0
Substituting the values of TAB ,TAC , TAD
Equating the I, j, k components to zero
We will get TAB =259 (known) , TAC =479.15N , TAD =535.66 N, P=1031N

17. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
Moment

18. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
Moment (M)

19. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
Moment (M)

20. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
Moment (M)

21. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
Moment (M)

22. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
Moment (M)

23. Devaprakasam D, Email: devaprakasam.d@vit.ac.in, Ph: +91 9786553933
Moment (M)