CE 201 (STATICS) DR. SHAMSHAD AHMAD CIVIL ENGINEERING ENGINEERING MECHANICS-STATICS

Transcription

1 COURSE: CE 201 (STATICS) LECTURE NO.: 28 to 30 FACULTY: DR. SHAMSHAD AHMAD DEPARTMENT: CIVIL ENGINEERING UNIVERSITY: KING FAHD UNIVERSITY OF PETROLEUM & MINERALS, DHAHRAN, SAUDI ARABIA TEXT BOOK: ENGINEERING MECHANICS-STATICS STATICS by R.C. HIBBELER, PRENTICE HALL

2 LECTURE NO. 28 to 30 FRAMES AND MACHINES Objectives: To analyze the forces acting on the members of frames and machines composed of pin-connected members

3 FRAMES AND MACHINES Frames and machines are two common types of structures which are often composed of pin-connected multi-force members Frames are generally stationary and are used to support loads Machines contain moving parts and are designed to transmit and alter the effect of forces Following are the examples of frames and machines: P 2 P 3 P 2 P 1 C D C B P 1 M B E A D R Ax R Ay R Dy A G R Ax R Gx R Ay R Gy

4 STEPS FOR THE ANALYSIS OF FRAMES AND MACHINES Frames and machines can be analyzed using the following steps: I. Free-Body Diagram Draw the FBD of either the entire structure, or a portion of the structure, or each of the members of the structure The choice should be made so that it leads to the most direct solution of the problem Forces common to two members which are in contact act with equal magnitude but opposite sense on the respective FBD of the members II. Equations of Equilibrium Apply the equations of equilibrium (i.e., ΣF x = 0, ΣF y = 0, ΣM o = 0) to each of the FBDs considered for the analysis of the structure There will be one extra equation of equilibrium for each of the hinge joints in the FBD (i.e., ΣM M about the hinge joint = 0)

5 PROBLEM SOLVING: Example # 1 Determine the force P needed to hold the 20-lb block in equilibrium.

6 PROBLEM SOLVING: Example # 1 ree-body diagram of the pulleys at C nd A are shown below: T C P T P A The force P can be determined by the above free-body diagrams, as applying equilibrium conditions to follows: At C Σ F y = 0 T P P = 0 T = 2P (1) At A Σ F y = 0 P + T + P 20 = 0 T + 2P = 20 (2) Substituting T = 2P in Eq. (2) from Eq. (1) 4P = 20 P = 5 lb Ans. P F.B.D. at C P 20lb F.B.D. at A

7 PROBLEM SOLVING: Example # 2 The compound beam is supported by a rocker at B and is fixed to the wall at A. If it is hinged (pinned) together at C, determine the reactions at the supports.

8 PROBLEM SOLVING: Example # 2 B y = 1600 sin = lb Ans. Free-body diagram of the portion CB of the compound beam is shown below: C x Hinge C C y 8ft 200lb 60 o (F.B.D. of BC portion) 4ft 4000lb-ft Applying equilibrium conditions to the above free-body diagram, as follows: ΣM about C = B y + (200 sin 60 ) 8 = 0 B y Free-body diagram of the entire compound beam is shown below: MA lb 200lb 60 o A x 4ft 4ft 8ft 4ft A y (F.B.D. of entire structure) 4000lb-ft Applying equilibrium conditions to the above free-body diagram, as follows: ΣF x = 0 A x cos 60 = 0 13 A x = lb Ans. B y

9 PROBLEM SOLVING: Example # 2 MA lb 200lb 60 o A x 4ft 4ft 8ft 4ft A y (F.B.D. of entire structure) 4000lb-ft ΣF y = 0 A y sin 60 + B y = 0 A y + B y = lb Substituting the value of B y calculated earlier, A y = = lb Ans. B y ΣM about B = 0 20 A y M A sin = 0 20 A y M A = Substituting the value of A y calculated earlier, M A = = = lb-ft ( ) Ans.

10 PROBLEM SOLVING: Example # 3 Determine the horizontal and vertical components of force that pins A and C exert on the two member arch.

11 PROBLEM SOLVING: Example # 3 ree-body diagram of the entire arch s shown below: A x A y 0.5m 2kN B 3m (F.B.D. of entire structure) 1m C y 1.5kN Applying equilibrium conditions to the above free-body diagram, as follows: C x ΣM about A = C y = 0 C y = kn Ans. ΣF y = 0 A y + C y 2 = 0 A y + C y = 2 Substituting the value of C y calculated earlier, A y = = kn Ans. ΣF x = 0 A x C x = 0 C x A x = 1.5 Free-body diagram of the portion AB of the arch is shown below: A x A y 2kN 1m 1.5m (hinge) B B y (F.B.D. of portion AB) B x 1.5m

12 PROBLEM SOLVING: Example # 3 2kN 1m (hinge) B B y B x 1.5m Applying equilibrium conditions to the above free-body diagram, as follows: ΣM about B = A x A y 2 1 = 0 A x A y 1.5m (F.B.D. of portion AB) A A x = y = = kn Ans. Substituting the value of A x in the equation: C x A x = 1.5, obtained earlier C x = A x = ( 0.166) = kn Ans.

13 PROBLEM SOLVING: Example # 4 Determine the support reactions of the frame shown in the following figure. Also calculate the force in member GB.

14 PROBLEM SOLVING: Example # 4 Applying equilibrium conditions to the entire frame: ΣM about A = 0 R Hy = 0 R Hy = = 115 kn Ans. ΣF y = R Ay = 0 R Ay = 35 kn = 35 kn ( ) Ans. ΣF x = R Ax + R Hx = 0 R Ax + R Hx = 40

15 PROBLEM SOLVING: Example # 4 Applying equilibrium conditions to the free-body diagram of the portion CDE, as follows: R Cx C 4 m D 80kN E R Ex 8 m R Cy R Ey ΣM about C = 0 8 R Ey = 0 R Ey = 40 kn

16 PROBLEM SOLVING: Example # 4 J m F GB R Hx R Ey = 40 kn E G H R Hy = 115 kn R Ex 6 m 9 m Applying equilibrium conditions to the free-body diagram of the portion EGH, as follows: ΣM about J = R Hx 15 = 0 R Hx = 80 kn = 80 kn ( ) Ans. ΣF y = F GB 40 = 0 F GB = kn Ans. Substituting the value of R Hx in equation: R Ax + R Hx = 40, obtained earlier: R Ax = R Hx 40 = ( 80) 40 = 40 kn ( ) Ans.

17 PROBLEM SOLVING: Example # 5 The two-member structure is connected at C by a pin, which is fixed to BDE and passes through the smooth slot in member AC. Determine the horizontal and vertical components of reaction at the supports.

18 PROBLEM SOLVING: Example # 5 Applying equilibrium conditions to the above free-body diagram: ΣM about A = 0 5 C = 0 Free-body diagram of the portion AC of the structure is as follows: 4 tan θ = 3 4 θ = tan = θ = tan A x A y 600lb-ft C 90-θ 5ft 5ft 5ft 5ft θ C C = 120 lb ΣF x = 0 A x 120 cos (90 - θ) = 0 A x = = 96 lb Ans. ΣF y = 0 A y sin (90 - θ) = 0 A y = = 72 lb = 72 lb ( ) Ans. (F.B.D. of the portion AC)

19 PROBLEM SOLVING: Example # 5 ΣF x = cos (90 θ) E x = 0 E x = = 96 lb Ans. Free-body diagram of the portion BCDE of the structure is as follows: 500lb C=120 B 90-θ C 3ft 3ft 2ft D y (F.B.D. of the portion B-C-D-E) Applying equilibrium conditions to the above free-body diagram: E E y E x ΣM about E = D y = 0 D y = 2180 lb Ans. ΣF y = 0 D y + E y = 0 D y + E y = 572 E y = = 1608 lb = 1608 lb ( ) Ans.

20 PROBLEM SOLVING: Example # 6 Analyze the frame shown in the following figure: R Ex E 2 m D 2 m G C H 4 m 100 N 3 m B 2 6 m m 5 m R Ax A R Ay

21 PROBLEM SOLVING: Example # 6 R Ex E Applying equilibrium conditions to the above free-body diagram: 2 m 2 m G D C H ΣM about A = 0 R Ex = 0 R Ex = N = N ( )Ans. 4 m 5 m 3 m R Ax B A 6 m 2 m 100 N ΣF x = 0 R Ax +R Ex = 0 R Ax = R Ex = ( 61.54) = N ( ) Ans. R Ay ΣF y = 0 R Ay 100 = 0 R Ay = 100 N ( )

22 2 m 2 m R Ex D PROBLEM SOLVING: Example # 6 E Free-body diagram of the portion GCH of the structure is as follows: G C H 4 m 100 N 5 m 3 m R Ax 100 N 100 N B A R Ay 100 N 2 6 m m Reactions at center, H, of the pulley are shown below: H 100 N Applying equilibrium conditions to the above free-body diagram: ΣM about C = T GB = 0 T GB = 250 N (T) ΣF x = R Cx 100 = 0 R Cx = 50 N = 50 N ( ) Ans. Ans. ΣF y = R 5 Cy 100 = 0 R Cy = 300 N ( ) Ans.

23 PROBLEM SOLVING: Example # 7 If each of the three uniform links of the mechanism has a length L and weight W, determine the angle θ for equilibrium. The spring which always remains vertical, is unstretched when θ = 0.

24 PROBLEM SOLVING: Example # 7 Free-body diagram of the portion AB of the structure is as follows: A x A A y θ L/2 KL F s = Sin θ 2 W L/2 B Using the angle θ, the stretch s in the spring can be determined as follows: A θ L/2 s (F.B.D. of portion AB) F B Applying equilibrium conditions to the above free-body diagram: ΣM about A = 0 (W KL sin 2 L 2 θ ) cos θ FB Lcosθ = 0 F B = 0.5 W 0.25 KL sinθ (1) s = 2 L sinθ

25 PROBLEM SOLVING: Example # 7 F B = 0.5 W 0.25 KL sinθ (1) Applying equilibrium conditions to the above free-body diagram: ΣM about D = 0 (F B + W) Lcosθ + W cos θ = 0 F B + 1.5W = 0 (2) L 2 Free-body diagram of the portion BCD of the structure is as follows: D x D D y θ F B B W Substituting F B from Eq. (1) in Eq. (2): 0.5 W 0.25 KL sinθ +1.5 W = 0 2 W = 0.25 KL sinθ sinθ = 8W KL θ = sin -1 8W KL Ans. W C (F.B.D. of portion BCD)

26 PROBLEM SOLVING: Example # 8 The structure is subjected to the loading shown. Member AD is supported by a cable AB and roller at C and fits through a smooth circular hole at D. Member ED is supported by a roller at D and a pole that fits in a smooth snug circular hole at E. Determine the x, y, z components of reaction at E and the tension in cable AB.

27 PROBLEM SOLVING: Example # 8 The tension in cable AB can be determined by taking ΣM about C = 0 using the free-body diagram of the portion AC, as follows: A F AB G F={-2.5k}kN (F.B.D. of portion AC) C Coordinates of points A, B, C, and G are as follows: A (0.6, 0.9, 0) m; B (0, 0.9, 0.8) m; C (0, 0.9, 0) m; and G (0.3, 0.9, 0) m r AB = { 0.6 i k}m r = = 1 m AB F AB = AB r FAB r = { 0.6 i k} AB 1 F AB = F AB {- 6 i k} kn r CA = {0.6 i}m and r CG = {0.3 i}m ΣM about C = 0 r CA F AB + r CG F = 0 {0.6 i} F AB {0.6 i k} + {0.3 i} {- 2.5 k} = j F AB j = 0 F AB = = kn Ans.

28 PROBLEM SOLVING: Example # 8 M Ez k=0 E M Ex i E x j E z k=0 M Ey j E y j D C x i D z k C G F AB A {-2-5k} The x, y, z components of reactions at E can be determined by applying equilibrium conditions to the free-body diagram of the entire structure, as follows: ΣF = 0 E x i + E y j + D z k + C x i 2.5 k i k = 0 ( E x + C x )i + (E y )j + (D z ) = 0 Equating the coefficients of the unit vectors i, j, and k to zero, E x + C x = 0 (1) E y = 0 (2) D z = 0 (3) D z = 1.25 kn

29 PROBLEM SOLVING: Example # 8 M Ez k=0 E M Ex i E x j E z k=0 M Ey j E y j D D z k C C x i ΣM about y axis = 0 M Ey = 0 M Ey = 0 F AB A G {-2-5k} ΣM about z axis = C x = 0 C x = (4) Substituting the value of C x in Eq. (1): E x = 0 Σ M about x axis = 0 M Ex + D z = 0 M Ex = = 0.5 kn-m Summary: FAB = KN Ex = 0 Ey = 0 Ans. EZ = 0 M Ex = 0.5 KN-m M Ey = 0 M Ez = 0

30 Multiple Choice Problems 1. The support reactions A y and D y of the frame shown below are Ans: (a) Feedback: (a) 16.2 kn ( ) and 13.8 kn ( ) (b) 16.2 kn ( ) and 13.8 kn ( ) (c) 15.0 kn ( ) and 15.0 kn ( ) (d) 15.0 kn ( ) and 15.0 kn ( ) ΣM about D = Ay (8 5) = 0 8Ay = 130 Ay = = 16.2 kn = 16.2 kn ( 8 F = 0 A D = 0 D = A 30 = ( 16.2) 30 = 13.8 kn =13.8 kn ( ) y y y y y

31 Multiple Choice Problems 2. The support reactions C x and C y of the frame shown below are Ans: (b) Feedback: (a) 577 kn ( ) and 1000 N ( ) (b) 577 N ( ) and 1000 N ( ) (c) 500 kn ( ) and 1000 N ( ) (d) 500 N ( ) and 1000 N ( ) Free-body diagram of the portion BC of the frame is shown below: Applying equilibrium condition to the above free-body diagram, Mabout C = 0 ( FAB sin 60 ) = 0 FAB = N F = 0 F cos 60 C = 0 C = cos 60 = 577 N ( ) x AB x x F = 0 F sin C = 0 C = sin 60 = 1000 N ( ) y AB y y