1. CE 201 - STATICS Dr. Mustafa Y. Al-Mandil Department of Civil Engineering LECTURE 18 1 34 EQUILIBRIUM OF RIGHD BODY “ It is necessary and sufficient for the equilibrium of rigid body that the resultant of all forces acting on the body is zero and that the resultant of all moments taken about any point inside or outside the body is zero”. O x FRFR MRMR MCMC MFMF O MCMC &         0 0 0 z y x F F F... ... ... ... ... ... 

2. CE 201 - STATICS Support for Rigid Bodies Subjected to Two-Dimensional Forces System. Table 5 - 1 Type of ConnectionReactionNumbers of Unknowns One unknown. The reaction is the tension force which acts away from the member in the direction of the cable One unknown. The reaction is a force which acts along the axis of the link. One unknown. The reaction is a force which acts perpendicular to the surface at the point of contact. One unknown. The reaction is a force which acts perpendicular to the slot. One unknown. The reaction is a force which acts perpendicular to the surface at the point of contact. Rocker Roller or pin in confined smooth slot Roller Weightless Link Cable

3. CE 201 - STATICS Member fixed connected to collar on smooth rod Support for Rigid Bodies Subjected to Two-Dimensional Forces System. Table 5 - 1 Type of ConnectionReactionNumbers of Unknowns One unknown. The reaction is a force which acts perpendicular to the rod. Two unknown. The reaction are two components of force, or the magnitude and its direction  of the resultant force. Note that  and  are not necessarily equal. (usually not, unless the rod shown is a link as in (2)). Two unknowns. The reactions are the couple moment and the force which acts perpendicular to the rod. Three unknowns. The reaction are the couple and the two forces components, or the couple and the magnitude and direction  of the resultant force. One unknown. The reaction is a force which acts perpendicular to the surface at the point of contact. Fixed support Smooth pin or hinge Member pin connected to collar on smooth rod Smooth contacting surface

4. CE 201 - STATICS Dr. Mustafa Y. Al-Mandil Department of Civil Engineering LECTURE 18 3 36 FREE BODY DIAGRAM EQUILIBRIUM IN 2-DIMENIONS F By F Bx FBFB B A D C FcFc O MoMo FyFy FxFx FOR EQUILIBRIUM OM OF OF z y x    ……..(1) ……..(2) ……..(3) R Ax R Ay F cy R Dy

5. CE 201 - STATICS Dr. Mustafa Y. Al-Mandil Department of Civil Engineering LECTURE 19 1 37 EXAMPLE Determine the support reaction in the beam (ABCD). Consider beam weight equals 50N. Determine the support reaction in the beam (ABCD). Consider beam weight equals 50N. A C  4 3 5 100 N B 3m3m 2m2m 5m5m D X A B C D R Ax R Ay 60 N 80 N 50 N R Dy SOLUTION First we draw FBD for beam (ABCD) + ……… (1) ……… (2) Solve (1) & (2) R Dy = 49 N (  ) R Ay = 81 N (  ) + ++

6. CE 201 - STATICS Dr. Mustafa Y. Al-Mandil Department of Civil Engineering LECTURE 19 2 38 A B C D R Ax R Ay 60 N 80 N 50 N R Dy ALTERNATE SOLUTION I SOLUTION             N Ay D N Dy A N Ax x 81 10 560250 R 010R780550 0ΣM )(49 10 240250 R 038055010R 0ΣM )(60R0 R 0ΣF + ++ CONCLUSION One force equation and two moment equations on a line not  to the force direction.

7. CE 201 - STATICS Dr. Mustafa Y. Al-Mandil Department of Civil Engineering LECTURE 19 3 39 A B C D R Ax R Ay 100 N 50 N R Dy + 4m4m E 4 3 5 3m3m 2m2m 5m5m ALTERNATE SOLUTION II CONCLUSION Three moment equations taken at three points not on a straight line. + + +

8. CE 201 - STATICS Given: k A = k B = 15 kN/m Required:  c = ? 40  9.8 = 392 N A B C 40 kg 392 N 1m1m 3m3m R A =K A  A R B =K B  B CC 0.1045 m 0.0784 m + ++ AA BB

9. CE 201 - STATICS LECTURE 20 2 41 TWO FORCE MEMBERS: OR THREE FORCE MEMBERS: A B B A     B A    O B A C Dr. Mustafa Y. Al-Mandil Department of Civil Engineering

10. CE 201 - STATICS Dr. Mustafa Y. Al-Mandil Department of Civil Engineering DISTRIBUTED LOADING A) Uniform Distributed Loading: B) Linear Distributed Loading: C) Trapezoidal Distributed Loading: 6m6m 3m3m 6m6m 4m4m 5 N/m 30 N 45 N 15 N/m 10 N/m 60 N 30 N 20 N/m 6m6m 4m4m 3m3m

11. CE 201 - STATICS Given: Beam ABC Required: Find Reactions Given: Beam ABC Required: Find Reactions ++ (    +  50 kN 3 4 5m5m 5m5m 3m3m 4m4m 5m5m A B D C B C A 5m5m 3 4 RBRB R Ay R Ax 40 kN 30 kN RBRB RDRD 3m3m 4m4m Two Force Member RBRB = R D d D/2 W a o

12. CE 201 - STATICS Dr. Mustafa Y. Al-Mandil Department of Civil Engineering LECTURE 21 1 42 Use vector algebra in the two vector equations :....  Three scaler equations Solve for (6) unknown components. Type of Constraints (Support): A - STATICALLY DETERMINATE No. of Unknowns = No. of Equations B - STATICALLY INDETERMINATE No. of Unknowns > No. of Equations C - STATICALLY UNSTABLE No. of Unknowns < No. of Equations Also check for Improper Constraints M F

13. CE 201 - STATICS Q. Dr. Mustafa Y. Al-Mandil Department of Civil Engineering D etermine the support reaction of beam ABCD ? 3m3m 4m4m B D C A 2m2m w = 15 N/m 3 4 50 N 3m3m 4m4m 2m2m R Ax R By R Ay A B C 60 N 40 N 30 N 60 N·m Free body Diagram A. Draw F.B.D. Transfer the (50N) force from point (D) to point (C) +

14. CE 201 - STATICS Support for Rigid Bodies Subjected to Three-Dimensional Forces System. Table 5 - 2 Type of Connection ReactionNumbers of Unknowns One unknown. The reaction is a force which acts away from the member in the direction of the cable. One unknown. The reaction is a force which acts perpendicular to the surface at the point of contact. One unknown. The reaction is a force which acts perpendicular to the surface at the point of contact. Three unknowns. The reactions are three rectangular force components. Four unknowns. The reactions are two force and two couple components which act perpendicular to the shaft. Cable smooth surface support Roller on a smooth surface Ball and socket Single journal bearing

15. CE 201 - STATICS Support for Rigid Bodies Subjected to Three-Dimensional Forces System. Table 5 - 2 Type of Connection ReactionNumbers of Unknowns Five unknowns. The reactions are three force and two couple components. Five unknowns. The reactions are three force and two couple components. Five unknowns. The reactions are three force and two couple components. Six unknowns. The reactions are three force and three couple components. Single thrust bearing Single smooth pin Single hinge Fixed support

16. CE 201 - STATICS Dr. Mustafa Y. Al-Mandil Department of Civil Engineering 1m 1.5m 4m D A C B y x z

17. CE 201 - STATICS Dr. Mustafa Y. Al-Mandil Department of Civil Engineering A 0.3m 0.2m 0.6m 0.5m 1m 0.5m 100 kg C B E D x y

18. CE 201 - STATICS Dr. Mustafa Y. Al-Mandil Department of Civil Engineering z x y A C B 3ft 4ft 6ft D E R Az R Ax 8ft 4ft R Ay T CD T CE

19. CE 201 - STATICS Dr. Mustafa Y. Al-Mandil Department of Civil Engineering TRUSSES Chep. 6 A truss is an assembly of prismatic “members” connected at their ends by smooth pin.Forces can only be applied at such connections (joints) A truss is an assembly of prismatic “members” connected at their ends by smooth pin.Forces can only be applied at such connections (joints) Simple truss: Assumptions for Analysis & Design: 1- Loading and reactions are applied at joints (nodes) only. Member weight is distributed equally to the adjacent nodes. 2- Members are loaded at their ends only (i.e. two-force members). So, they undergo axial forces only (tension or compression). 1- Loading and reactions are applied at joints (nodes) only. Member weight is distributed equally to the adjacent nodes. 2- Members are loaded at their ends only (i.e. two-force members). So, they undergo axial forces only (tension or compression). R xy R Ay R Ax B A C E D G H K J LECTURE 23 1 51 LECTURE 23 1 51

20. CE 201 - STATICS LECTUR E 23 2 52 Dr. Mustafa Y. Al-Mandil Department of Civil Engineering Analyze the truss shown below: — F.B.D. For Whole Truss : — Take Joint (E) : + ++ 4m4m R Ay R Ax R Bx A C B D E 3m3m 3m3m 20 kN F EC F ED E

21. CE 201 - STATICS LECTUR E 23 Analyze the truss shown below: 2 52 Dr. Mustafa Y. Al-Mandil Department of Civil Engineering — Take Joint (C) : — Take Joint (A) : ++ F CA F CD 25 kN C 40 kN 45 kN 15 kN A F AD F AB

22. CE 201 - STATICS LECTURE 24 1 53 Dr. Mustafa Y. Al-Mandil Department of Civil Engineering ZERO-FORCE MEMBERS “ Members which carry no forces in a truss either because of particular loading pattern or because they enhance geometric stability:. EQUAL-FORCE MEMBERS F3F3 F1F1 F2F2 F2F2 F1F1 F 1 = F 2 F 3 = 0 F3F3 F1F1 F2F2 F4F4 F 1 = F 3 F 2 = F 4 30 kN 35 kN 25 kN 30 kN 40 kN A B D IL P T G K C E H J M N Q Z ERO-FORCE MEMBERS BC, BE KM, KN PQ, PN, LN

23. CE 201 - STATICS B y sectioning the truss into two or more parts (each part containing more then one joint), then we can find forces in certain numbers directly. LECTURE 24 2 54 Dr. Mustafa Y. Al-Mandil Department of Civil Engineering METHOD OF SECTIONS Use Section A-A Find F GH ? 80 kN 50 kN JG F GH + 4m4m 50 kN 80 kN 3m3m 3m3m 3m3m 3m3m 3m3m A A F3F3 F2F2 F1F1 0 F1 = 

24. CE 201 - STATICS LECTURE 25 2 55 Dr. Mustafa Y. Al-Mandil Department of Civil Engineering THEORY OF PLANE TRUSSES M = Number of members R = Number of Unknown Reactions J = Number of joints if M + R > 2J Statically Indeterminate if M + R = 2J Statically Determinate if M + R < 2J Statically Unstable M = 9 R = 3 J = 6 M + R = 2J Determinate M + R > 2J Indeterminate M = 10 R = 3 J = 6 M + R < 2J UNSTABLE M = 8 R = 3 J = 6 B C E D A G

25. CE 201 - STATICS F rames are structural systems composed of members connected and loaded at unrestricted locations along their axis. LECTURE 27 1 58 Dr. Mustafa Y. Al-Mandil Department of Civil Engineering ANALYSIS & DESIGN 1- SOLVE FOR EXTERNAL REACTIONS ( partially or totally ). 2- DIS-ASSEMBLE & TREAT EACH COMPONENT AS RIGID BODY UNDER EQUILIBRIUM. 3- FORMULATE SETS OF SIMULTANEOUS EQUATIONS & SOLVE FOR UNKNOWNS. R Ax R Ay R Dy A D B C P2P2 P1P1 M R Ax R Gx R Ay R Gy P1P1 P3P3 P2P2 C G E A B D

26. CE 201 - STATICS Analyze Frame (ABCDE) LECTURE 27 2 58 Dr. Mustafa Y. Al-Mandil Department of Civil Engineering — Take whole Frame — Take Part (ABC) As F.B.D. — Take Whole Frame Again R Ay A E C 40 kN 60 kN·m R Ey 4m4m 4m4m D B 3m3m 3m3m 3m3m 3m3m R Ax C R Cy R Ay R Cx A B 45 kN 40 kN + +

27. CE 201 - STATICS LECTURE 25 3 56 Dr. Mustafa Y. Al-Mandil Department of Civil Engineering G E NERAL EXAMP L E (a)(b) (c) (d) (e) (g) (h) (f) Howe Truss Pratt Truss Warren Truss Warren Truss with verticals K Truss Sub-divided Warren Truss Sub-divided Pratt Truss or Baltimore Truss Baltimore Truss with inclined chord or Petit Truss (a) (d) (c) (b) Howe Truss Pratt Truss Fan Truss Fink Truss Conventional Roof Trusses Conventional Bridge Trusses

28. CE 201 - STATICS LECTURE 25 4 57 Dr. Mustafa Y. Al-Mandil Department of Civil Engineering G E NERAL EXAMP L E

29. CE 201 - STATICS LECTURE 28 1 59a Dr. Mustafa Y. Al-Mandil Department of Civil Engineering E X AMP L E Analyze the Frame — Take whole Frame : + — Take ( C.D.E. ) : + R Ax R Hx R Ay R Hy 80 kN C H G A B E 40 kN D 12 m 3m3m 6m6m 9m9m 4m4m 4m4m R Cx C R Cy E D 80 kN R Ey R Ex

30. CE 201 - STATICS LECTURE 28 1 59b Dr. Mustafa Y. Al-Mandil Department of Civil Engineering E X AMP L E — Take whole Frame : + Analyze the Frame R Hx H G 6m6m 9m9m 115 kN F GB J R Ex 40 kN 16 m E 3 8

31. CE 201 - STATICS LECTURE 28 2 60a Dr. Mustafa Y. Al-Mandil Department of Civil Engineering Analyze the Frame — Take whole Frame : +  Take Pulley as F.B.D. : 100 N 100 N 2m2m 6m6m H E D C B 2m2m 4m4m 5m5m 2m2m R Ex R Ax R Ay 3m3m G A

32. CE 201 - STATICS LECTURE 28 2 60b Dr. Mustafa Y. Al-Mandil Department of Civil Engineering  Take GCH as F·B·D : Analyze the Frame G C H 100 N TGB R Cy R Cx +