EQUILIBRIUM OF RIGHD BODY Department of Civil Engineering LECTURE 18 1 34 EQUILIBRIUM OF RIGHD BODY MF MC MR FR O O x MC “ 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”. ï î í ì = S z y x F ... ... ... ... & ... ... M. HUSSAIN KHAN

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

Support for Rigid Bodies Subjected to Two-Dimensional Forces System. Table 5 - 1 Support for Rigid Bodies Subjected to Two-Dimensional Forces System. Type of Connection Reaction Numbers of Unknowns One unknown. The reaction is a force which acts perpendicular to the surface at the point of contact. Smooth contacting surface One unknown. The reaction is a force which acts perpendicular to the rod. Member pin connected to collar on smooth 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)). Smooth pin or hinge Member fixed connected to collar on smooth rod 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. M. HUSSAIN KHAN Fixed support

Department of Civil Engineering FREE BODY DIAGRAM LECTURE 18 3 36 FREE BODY DIAGRAM FB FBx FBy B D A RAx C Fc RAy Fcy RDy EQUILIBRIUM IN 2-DIMENIONS FOR EQUILIBRIUM Fy S F = O ……..(1) x S = O F O ……..(2) Fx Mo y S M = O ……..(3) z M. HUSSAIN KHAN

Department of Civil Engineering EXAMPLE SOLUTION LECTURE 19 1 37 EXAMPLE Determine the support reaction in the beam (ABCD). Consider beam weight equals 50N. SOLUTION 100N First we draw FBD for beam (ABCD) 4 5 3 B C D +  X A 3m 2m 5m 80N + 60N A C D RAx B RAy 50N RDy ……… (1) + Solve (1) & (2) RDy = 49N (  ) RAy = 81N (  ) ……… (2) M. HUSSAIN KHAN

Department of Civil Engineering ALTERNATE SOLUTION I SOLUTION LECTURE 19 2 38 ALTERNATE SOLUTION I 80N 60N A C D RAx B SOLUTION RAy 50N RDy + ΣF = ® x R + 60 = Þ R = - 60 N ( ¬ ) CONCLUSION Ax Ax ΣM = + A R ´ 10 - 50 ´ 5 - 80 ´ 3 = One force equation and two moment equations on a line not  to the force direction. Dy 250 + 240 R = = 49 N ( ­ ) Dy 10 ΣM = + D 50 ´ 5 + 80 ´ 7 - R ´ 10 = Ay 250 + 560 M. HUSSAIN KHAN \ R = = 81 N ­ Ay 10

Department of Civil Engineering SOLUTION ALTERNATE SOLUTION II LECTURE 19 3 39 + E ALTERNATE SOLUTION II 100N 4m 4 5 3 SOLUTION A B C D RAx 3m 2m 5m + RAy 50N RDy CONCLUSION + Three moment equations taken at three points not on a straight line. + M. HUSSAIN KHAN

PROBLEM 5.48 Solution: + Given: kA = kB = 15kN/m Required: c = ? 40kg Given: kA = kB = 15kN/m Required: c = ? A B C Solution: 40  9.8 = 392N + 1m 3m 392N + RA=KAA RB=KBB 0.0784m A 0.1045m B C M. HUSSAIN KHAN

Department of Civil Engineering LECTURE 20 2 41 TWO FORCE MEMBERS: THREE FORCE MEMBERS: A   A A   A B O B B  B  C  OR M. HUSSAIN KHAN

Department of Civil Engineering A) Uniform Distributed Loading: 5N/m 30N A) Uniform Distributed Loading: 6m 3m 15N/m 45N B) Linear Distributed Loading: 6m 4m 20N/m 10N/m 60N 30N C) Trapezoidal Distributed Loading: 3m 6m 4m M. HUSSAIN KHAN

Department of Example: Civil Engineering Solution:   (   Given: Beam ABC Required: Find Reactions + Solution: + 50kN 4 A 5m B 5m 3 C 4m D (   3m 40kN  + 30kN a W RAx A 5m B 5m C 3 RAy 4 RB o RB  D/2 D/2 Two Force Member 4m RB = RD d 3m M. HUSSAIN KHAN RD

Department of Civil Engineering EQUILIBRIUM IN 3-D LECTURE 21 1 42 EQUILIBRIUM IN 3-D Use vector algebra in the two vector equations : . . . .  Three scaler equations Also check for Improper Constraints Solve for (6) unknown components. Type of Constraints (Support): M 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 F M. HUSSAIN KHAN

Q. A. Department of Civil Engineering 4 w = 15N/m 3 Q. D Determine the support reaction of beam ABCD ? B 2m A C 4m 3m A. Draw F.B.D. Transfer the (50N) force from point (D) to point (C) 4m 3m 2m 60N 40N RAx A B C 30N RAy RBy 60N·m + Free body Diagram M. HUSSAIN KHAN

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

Department of Civil Engineering Table 5 - 2 Support for Rigid Bodies Subjected to Three-Dimensional Forces System. Type of Connection Reaction Numbers of Unknowns Five unknowns. The reactions are three force and two couple components. Single thrust bearing Five unknowns. The reactions are three force and two couple components. Single smooth pin Five unknowns. The reactions are three force and two couple components. Single hinge Six unknowns. The reactions are three force and three couple components. M. HUSSAIN KHAN Fixed support

Department of Civil Engineering z B D 1m A y 1.5m 4m C x M. HUSSAIN KHAN

Department of Civil Engineering 0.3m 0.2m A 1m C x 0.6m D y B E 0.5m 100 kg M. HUSSAIN KHAN

Department of Prob. 5 - 81 Civil Engineering z 3ft 3ft D 4ft E RAz 6ft TCD RAy RAx A TCD 4ft B x TCE C 8ft y M. HUSSAIN KHAN

Department of Civil Engineering A truss is an assembly of prismatic LECTURE 23 LECTURE 23 1 1 51 51 Chep. 6 Simple truss: TRUSSES A truss is an assembly of prismatic “members” connected at their ends by smooth pin.Forces can only be applied at such connections (joints) 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). B D G J RAx K A C E H RAy M. HUSSAIN KHAN Rxy

Department of Civil Engineering Example: LECTURE 23 2 52 Example: Analyze the truss shown below: RAy — F.B.D. For Whole Truss : A C RAx + 4m RBx E 3m B D 3m 20kN 20kN — Take Joint (E) : + FEC FED E M. HUSSAIN KHAN

Department of Civil Engineering Example: LECTURE 23 2 Example: 52 Analyze the truss shown below: — Take Joint (C) : FCA C — Take Joint (A) : 40kN FCD 25kN A 45kN 15kN + FAD FAB M. HUSSAIN KHAN

Department of Civil Engineering Example: Example: e.g. LECTURE 24 1 ZERO-FORCE MEMBERS 53 “ Members which carry no forces in a truss either because of particular loading pattern or because they enhance geometric stability:. Example: F3 EQUAL-FORCE MEMBERS F1 = F2 F3 = 0 Example: F1 F1 = F3 F2 = F4 F2 F1 F4 F2 e.g. F1 F3 40kN 30kN ZERO-FORCE MEMBERS BC, BE KM, KN PQ, PN, LN 30kN D I L B G P K F2 A T C E H J M N Q M. HUSSAIN KHAN 25kN 35kN

Department of Civil Engineering e.g. LECTURE 24 2 METHOD OF SECTIONS 54 By sectioning the truss into two or more parts (each part containing more then one joint), then we can find forces in certain numbers directly. Find FGH ? e.g. 80kN Use Section A-A 50kN 3m 80kN 50kN 3m A 3m A G J F3 F2 F1 F1 =  3m FGH 3m + 4m M. HUSSAIN KHAN

Department of Civil Engineering Example: THEORY OF PLANE TRUSSES LECTURE 25 2 THEORY OF PLANE TRUSSES 55 Example: 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 = 8 R = 3 J = 6 M + R < 2J UNSTABLE M = 9 R = 3 J = 6 M + R = 2J Determinate B D M = 10 R = 3 J = 6 M + R > 2J Indeterminate A G M. HUSSAIN KHAN C E

Department of Civil Engineering e.g. FRAMES: LECTURE 27 1 58 P2 e.g. C FRAMES: Frames are structural systems composed of members connected and loaded at unrestricted locations along their axis. B RAx RGx RAy RGy • P1 P3 P2 C G E A B D • P1 M • A D • • RAx RAy 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. RDy M. HUSSAIN KHAN

Department of Civil Engineering Example: Analyze Frame (ABCDE) LECTURE 27 2 Analyze Frame (ABCDE) 58 Example: RAy A E C 40kN • 60kN·m REy 4m D B 3m — Take whole Frame — Take Part (ABC) As F.B.D. + RCy + C 40kN RCx B A RAx RAy 45kN — Take Whole Frame Again M. HUSSAIN KHAN

Department of Civil Engineering GENERAL EXAMPLE LECTURE 25 3 56 (b) Howe Truss Pratt Truss (a) Howe Truss (c) (d) Pratt Truss (b) Warren Truss Warren Truss with verticals (c) Fan Truss (e) (f) K Truss Sub-divided Warren Truss Fink Truss (d) (g) (h) Sub-divided Pratt Truss or Baltimore Truss Baltimore Truss with inclined chord or Petit Truss Conventional Roof Trusses M. HUSSAIN KHAN Conventional Bridge Trusses

Department of Civil Engineering GENERAL EXAMPLE LECTURE 25 4 57 M. HUSSAIN KHAN

Department of Civil Engineering EXAMPLE LECTURE 28 1 59a RAx RHx RAy RHy • 80kN C H G A B E 40kN D 12m 3m 6m 9m 4m — Take whole Frame : + Analyze the Frame — Take ( C.D.E. ) : + 80kN RCx C D E REx M. HUSSAIN KHAN RCy REy

Department of Civil Engineering EXAMPLE Analyze the Frame LECTURE 28 1 59b Analyze the Frame — Take whole Frame : RHx H G 6m 9m 115kN FGB J REx 40kN 16m E 3 8 + M. HUSSAIN KHAN

Department of Civil Engineering Analyze the Frame LECTURE 28 2 60a REx E 2m D — Take whole Frame : 2m • G C H 4m 100N + B 3m 6m 2m 5m RAx A RAy  Take Pulley as F.B.D. : 100N M. HUSSAIN KHAN

Department of Civil Engineering Analyze the Frame LECTURE 28 2 60b  Take GCH as F·B·D : + RCy G C H 100N RCx TGB 100N M. HUSSAIN KHAN