8.1 Internal Forces in Structural Members

Transcription

1 8.1 Internal Forces in Structural Members

2 8.1 Internal Forces in Structural Members xample 1, page 1 of 4 1. etermine the normal force, shear force, and moment at sections passing through a) and b). 4 kn 2 m 2 m 4 m 4 m

3 + 8.1 Internal Forces in Structural Members xample 1, page 2 of 4 Part a): Internal forces and moment at 4 kn 1 Pass a section through point. 2 Free-body diagram of part of beam to left of (This is a much better choice of free body than the part of the beam to the right, since we won't have to calculate the reactions at and.) 4 Solving qs. 1, 2, and 3 gives N = 0 ns. 4 kn V = 4 kn ns. V 2 m N M 3 quations of equilibrium for part of beam to left of : + Fx = 0: N = 0 (1) + F y = 0: V 4 kn = 0 (2) M = 0: (4 kn)(2 m) + M = 0 (3) 5 M = 8 kn m 4 kn 2 m V = 4 kn M = 8 kn m ns. Free-body diagram showing correct senses of internal forces and moment at : N = 0

4 + 8.1 Internal Forces in Structural Members xample 1, page 3 of 4 Part b): Internal forces and moment at 6 Pass a section through point. 4 kn 7 Free-body diagram of part of beam to right of N M V x y 4 m 8 + quations of equilibrium for part of beam to right of : Fx = 0: N + x = 0 (4) 9 Three equations but five unkowns. nother free body is needed. + F y = 0: V y = 0 (5) M = 0: M + y (4 m) = 0 (6)

5 + 8.1 Internal Forces in Structural Members xample 1, page 4 of 4 10 Free-body diagram of entire beam 4 kn x y y 4 m 8 m 11 quilibrium equations for entire beam: + Fx = 0: x = 0 (7) M = 0: (4 kn)(4 m) + y (8 m) = 0 (8) Solving qs. 7 and 8 gives x = 0 and y = 2 kn 12 Using these results in qs. 4, 5, and 6 and solving gives 13 Free-body diagram showing correct senses of internal forces and moment at N = 0 M = 8 kn m V = 2 kn 2 kn N = 0 V = 2 kn M = 8 kn m ns. ns. ns.

6 8.1 Internal Forces in Structural Members xample 2, page 1 of etermine the normal force, shear force, and moment at a section passing through. 300 lb/ft 6 ft 6 ft 3 Note that the resultant of the entire distributed load, (300 lb/ft) (12 ft) = 3,600 lb, does not act at. The entire distributed load acts on the entire 12-ft span; the free body shown below has 300 lb/ft acting over only a 6-ft span, and the resultant of this distributed load acts halfway between and. 1 Pass a section through point. 300 lb/ft 4 quations of equilibrium for part of beam to right of : + Fx = 0: N = 0 (1) Therefore 2 Free-body diagram of part of beam to right of 300 lb/ft N V M 6 ft y + N = 0 ns. F y = 0: V + y 300 lb/ft)(6 ft) = 0 (2) M = 0: M (300 lb/ft)(6 ft)( 6 ft ) + y (6 ft) = 0 2 (3) qs. 2 and 3 involve three unknowns. n additional free-body diagram is needed.

7 + 8.1 Internal Forces in Structural Members xample 2, page 2 of 2 5 Free-body diagram of entire beam. 300 lb/ft x y y 12 ft 6 quation of equilibrium for entire beam: M = 0: y (12 ft) (300 lb/ft)(12 ft)( 12 ft ) = 0 (4) 2 Solving q. 4 gives y = 1800 lb. Using y = 1800 lb in qs. 2 and 3 and then solving gives V = 0 M = 5,400 lb ft ns. ns.

8 + 8.1 Internal Forces in Structural Members xample 3, page 1 of 2 3. etermine the normal force, shear force, and moment acting at a section passing through point on the quarter-circular rod shown. 70 lb 30 lb 2 Free-body diagram of portion above 70 lb 30 lb V 4 ft M 4 ft 40 O N 40 O 1 Pass a section through point 70 lb 30 lb quilibrium equations for part of rod to right and above section at : Fx = 0: N sin V cos lb = 0 (1) F y = 0: N cos + V sin lb = 0 (2) M O = 0: M + N (4 ft) (30 lb)(4 ft) = 0 (3) hoosing point O for summing moments eliminates both the 70-lb force and the shear V from the moment equation because their lines of action pass through O.

9 8.1 Internal Forces in Structural Members xample 3, page 2 of 2 4 Geometry = = O 5 Substituting = 40 into qs. 1, 2, and 3 and solving gives N = 72.9 lb V =.0 lb M = lb ns. ns. ns.

10 8.1 Internal Forces in Structural Members xample 4, page 1 of 6 4. etermine the internal torque at sections passing through points a) and b) of the shaft. 2 kip 2 ft 2 ft 6 kip ft (Torque applied at end) 2 kip

11 8.1 Internal Forces in Structural Members xample 4, page 2 of 6 Part a) point 1 Pass a section through point. 2 kip 2 ft 2 ft 6 kip ft (Torque applied at end) 2 kip

12 8.1 Internal Forces in Structural Members xample 4, page 3 of 6 2 Free-body diagram of part to right of. This is a better choice than using the part to the left because the reaction at the support would have to be calculated, if we used the part to the left. 2 kip 2 ft M Torque from part of the shaft to the left of point acting on the part to the right. The choice of sense of the torque is arbitrary. 2 kip 2 ft kip ft (Torque applied at end) x quilibrium equation for part of shaft: Mx = 0: M (2 kip)(2 ft) (2 kip)(2 ft) + 6 kip ft = 0 Solving gives M = 2 kip ft ns.

13 8.1 Internal Forces in Structural Members xample 4, page 4 of 6 4 Free-body diagram of part to right of showing correct sense of torque at 2 kip 2 ft M = 2 kip ft 2 ft 6 kip ft (Torque applied at end) x 2 kip

14 8.1 Internal Forces in Structural Members xample 4, page 5 of 6 Part b) point 5 Pass a section through point. 2 kip 2 ft 2 ft 6 kip ft 2 kip

15 8.1 Internal Forces in Structural Members xample 4, page 6 of 6 6 Free-body diagram of part to right of M 6 kip ft x 7 + quilibrium equation for part of shaft: Mx = 0: M + 6 kip ft = 0 Solving gives M = 6 kip ft ns.

16 8.1 Internal Forces in Structural Members xample 5, page 1 of 6 5. etermine the axial force, shear force, and moment at sections passing through a) and b) immediately to the left of roller, and c) immediately to the right of roller. 4 kn 5 kn m Hinge (pin connection) F 4 m 2 m 1 m 5 m 5 m 1 Part a): point Pass a section through point. Hinge 5 kn m 4 kn F 4 m 2 m 1 m 5 m 5 m

17 8.1 Internal Forces in Structural Members xample 5, page 2 of 6 quilibrium equations for part of beam: + 2 Free-body diagram of part of beam to right of section. This is a better choice than the part to the left because only one unknown reaction (the vertical force at ) occurs on the right while three (two force components and a moment) occur at. V Hinge 4 kn N M y 1 m 5 m 5 m 3 + Fx = 0: N = 0 (1) Therefore, N = 0 ns. + F y = 0: V + y 4 kn = 0 (2) M = 0: M + y (1 m + 5 m) (4 kn)(1 m + 5 m +5 m) = 0 (3) 4 Three equations but four unknowns so an additional equation is needed.

18 8.1 Internal Forces in Structural Members xample 5, page 3 of 6 5 To calculate the reaction at, pass a section immediately to the left of hinge. Hinge 5 kn m 4 kn F 4 m 2 m 1 m 5 m 5 m 7 The moment is zero at a hinge (a "hinge" is just another name for a pin connection between two parts of the beam; no moment is transmitted at a pin connection). We will make use of this fact by summing moments about. 6 N Free-body diagram of part of beam to right of section M = 0 V Hinge y 4 kn 5 m 5 m

19 + 8.1 Internal Forces in Structural Members xample 5, page 4 of 6 8 quilibrium equation for part of beam: M = 0: y (5 m) 4(5 m +5 m) = 0 (4) Note that V does not appear in this equation because its moment arm about is infinitesimal. Solving q. 4 gives y = 8 kn Substituting y = 8 kn in qs. 2 and 3 and solving gives V = 4 kn M = 4 kn m ns. ns. 9 Free-body diagram of part of beam showing correct senses of forces and moment at 4 kn V = 4kN Hinge N = 0 M = 4 kn m y = 8 kn 1 m 5 m 5 m

20 8.1 Internal Forces in Structural Members xample 5, page 5 of 6 Part b): point immediately to left of. 11 Free-body diagram of part of beam to right of section 4 kn 10 Pass a section through the beam immediately to the left of. Hinge 5 kn m 4 kn F N M V y = 8 kn 5 m 4 m 2 m 1 m 5 m 5 m 12 quilibrium equations for part to right of section: + Fx = 0: N = 0 (5) Therefore, N = 0 ns. 13 Free-body diagram of part of beam showing correct senses of forces and moment at section to left of M = 20 kn m N = 0 V = 4 kn y = 8 kn 5 m 4 kn + + F y = 0: V + 8kN 4 kn = 0 (6) Solving gives V = 4 kn ns. M = 0: M (4 kn)(5 m) = 0 (7) Solving gives M = 20 kn m ns.

21 8.1 Internal Forces in Structural Members xample 5, page 6 of 6 14 Part c): point immediately to right of roller. Pass a section through the beam immediately to the right of. 5 kn m Hinge 4 kn F 15 Free-body diagram of part of beam to right of section N ' M ' V ' 5 m 4 kn 4 m 2 m 1 m 5 m 5 m 16 quilibrium equations for part to right of section: + Fx = 0: N ' = 0 (8) 17 Free-body diagram of part of beam showing correct senses of forces and moment at section to right of 4 kn M ' = 20 kn m N ' = 0 V ' = 4 kn 5 m + + Therefore, N ' = 0 ns. F y = 0: V ' 4 kn = 0 (9) Solving gives V ' = 4 kn ns. M = 0: M ' (4 kn)(5 m) = 0 (10) Solving gives M ' = 20 kn m ns.

22 8.1 Internal Forces in Structural Members xample 6, page 1 of 2 6. Two wooden blocks have been glued together and a compressive force of 80 N is applied by the clamp to press the blocks together as the glue dries. etermine the normal force, shear force, and moment at a vertical section through point. 35 mm 1 Pass a section through point. 35 mm

23 8.1 Internal Forces in Structural Members xample 6, page 2 of Free-body diagram of part of clamp to right of section. The block is not included in the free body. V N M 35 mm 80 N 4 + quilibrium equations for part of clamp: Fx = 0: N + 80 N = 0 (1) 3 The force from the block points towards the clamp because the block is in compression. + F y = 0: V = 0 (2) M = 0: M + (80 N)(35 mm) = 0 (3) Solving gives N = 80 N V = 0 M = 2800 N mm = 2.8 N m ns. ns. ns.

24 8.1 Internal Forces in Structural Members xample 7, page 1 of 3 7. etermine the axial force, shear force, and moment at. 100 lb 5 ft 5 ft 1 Pass a section through point. 100 lb 5 ft 5 ft

25 8.1 Internal Forces in Structural Members xample 7, page 2 of 3 2 Free-body diagram of part of member above section at 100 lb V F action of F is known to be vertical. 3 is a two-force member so the line of M N 5 ft 5 ft 4 quations of equilibrium for part of member: + Fx = 0: V = 0 (1) Therefore V = 0 ns. + + F y = 0: N + F 100 lb = 0 (2) M = 0: M + F (5ft) (100 lb)(5 ft +5 ft) = 0 (3) 5 Four unknowns but only three equations so an additional equation is needed.

26 + 8.1 Internal Forces in Structural Members xample 7, page 3 of 3 6 Free-body diagram of entire structure 5 ft 5 ft 100 lb 8 quations of equilibrium for entire structure: x M = 0: F (5ft) (100 lb)(5 ft +5 ft) = 0 (4) Solving gives y F 7 ecause is a two-force member, the line of action of F is known to be vertical. F = 200 lb Substituting F = 200 lb in qs. 2 and 3 and solving gives N = 100 lb ns. M = 0 ns.

27 8.1 Internal Forces in Structural Members xample 8, page 1 of 3 8. The spreader bar is used to spread the load acting on the 2,000 lb beam FGH while it is being lifted. etermine the normal force, shear force, and moment at the midpoint of the spreader bar. Neglect the weight of the bar. 2,000 lb 6 ft 1 Pass a section through point. 2,000 lb F G H 5 ft 5 ft 5 ft 5 ft F G H

28 8.1 Internal Forces in Structural Members xample 8, page 2 of 3 2 Free-body diagram of part of spreader bar to left of + + T 3 T and T F are the tensions in the cables at. M N quilibrium equations for part of spreader bar to left of : Fx = 0: N + T cos = 0 (1) F y = 0: T sin T F V = 0 (2) M = 0: T F (5 ft) T (sin )(5 ft) + M = 0 (3) T F 5 ft V 5 Three equations but five unknowns, so at least one more free body must be used. 7 quilibrium equations for beam: 6 Free-body diagram of beam M G = 0: T F (5 ft+5 ft) + (2,000 lb)(5 ft) = 0 (4) T F F 2,000 lb G T G H Solving gives T F = 1,000 lb 5 ft 5 ft 5 ft 5 ft nother free body is needed, if we are to calculate the value of T in qs. 1, 2, and 3.

29 8.1 Internal Forces in Structural Members xample 8, page 3 of 3 8 Free-body diagram of connection 2,000 lb quilibrium equations for connection : Fx = 0: T cos + T cos = 0 (5) F y = 0: 2,000 lb T sin T sin = 0 (6) T T 11 Substituting = in qs. 5 and 6 and solving gives T = T = 1,302 lb 10 Geometry Substituting T = 1,302 lb and T F = 1,000 lb into qs. 1, 2, and 3 and solving gives 5 ft 6 ft = tan -1 ( ) = ft 5 ft N = 833 lb V = 0 M = 0 ns. ns. ns.

30 8.1 Internal Forces in Structural Members xample 9, page 1 of 5 9. The frame shown is pin-connected at and rests on smooth surfaces at and G. etermine the normal force, shear force, and moment acting at a section passing through point. 10 m 100 kg F G 1 Pass a section through. 1 m 2 m 3 m 2 m 3 m 1 m 10 m 100 kg F G

31 8.1 Internal Forces in Structural Members xample 9, page 2 of 5 2 Free-body diagram of part quilibrium equations for part : + 4 N + Fx = 0: T + N cos + V cos = 0 (1) M + F y = 0: F + N sin V sin = 0 (2) L T V L M = 0: T(L ) V ( L ) + M = 0 (3) Three equations and five unknowns, so at least one more free body is needed. ut first, the angles and and distances L and L must be determined. F 1 m 2 m 3 Since the floor is smooth, only a normal force is present; friction is absent.

32 + 8.1 Internal Forces in Structural Members xample 9, page 3 of Geometry L L 1 m 2 m 3 m 10 m quilibrium equation for entire frame: = tan m ( 1 m + 2 m + 3 m ) = = 90 = L = (1 m) tan = m (1 m + 2 m) L = cos = m 6 Free-body diagram of entire frame. This free body will enable us to calculate F. Weight = (100 kg)(9.81 m/s 2 ) = 981 N F G 10 m M G = 0: F (8 m + 4 m) + (981 N)(4 m) = 0 (4) Solving this equation gives F 8 m 4 m F G F = N One more free body is needed, since we now have four equations but five unknowns.

33 + 8.1 Internal Forces in Structural Members xample 9, page 4 of 5 8 Free-body diagram of member. This free body will give us an equation relating F and T. y 9 x Forces from the pin connection at 10 m T L = m F = N 6 m 10 quilibrium equations for member : M = 0: (327.0 N)(6 m) + T(10 m m) = 0 (5) Solving gives T = N

34 8.1 Internal Forces in Structural Members xample 9, page 5 of 5 11 Substituting T = N 12 Free-body diagram showing correct sense of internal moment and forces at F = N N = 402 N = = L = m M = N m V = 33.6 N T = N L = m into qs. 1, 2, and 3 and solving gives N = 402 N V = 33.6 N M = N m ns. ns ns F = 327 N

35 8.1 Internal Forces in Structural Members xample 10, page 1 of In the floor-beam girder system shown, the four floor panels at the top are simply supported at their ends by floor beams. The beams in turn transmit forces to the girder I. etermine the axial force, shear force, and moment in the girder at sections passing through a) point and b) point J. Floor panels 2 kip F G H Floor beam (end view) J I Girder (side view) 10 ft 3 ft 7 ft 10 ft 6 ft 2 ft 2 ft

36 8.1 Internal Forces in Structural Members xample 10, page 2 of 8 Part a) point 1 Pass a section through the girder at. Floor panels 2 kip F G H Floor beam (end view) J I Girder (side view)

37 8.1 Internal Forces in Structural Members xample 10, page 3 of 8 2 Free-body diagram of part of structure to left of point, including floor panels and F F M N F F 3 Force from floor beam F acting on floor panel F ("simply supported" so no moment acts at F, only a force) F V 13 ft 7 ft 4 quilibrium equations for part of structure: + Fx = 0: N = 0 (1) + + F y = 0: F V + F F = 0 (2) M = 0: F (13 ft) + M + F F (7 ft) = 0 (3) 5 Three equations but five unknowns so at least one more free body is needed.

38 8.1 Internal Forces in Structural Members xample 10, page 4 of 8 6 Free-body diagram of floor panel F: F F F F 7 Obviously, F F = 0 (Just consider the sum of moments about ). 8 Free-body diagram of entire structure (This free body will enable us to calculate F ): 2 kip F G H x I F y 10 ft 7 ft 10 ft 6 ft 3 ft 2 ft 2 ft

39 + 8.1 Internal Forces in Structural Members xample 10, page 5 of 8 9 quilibrium equation for entire structure: M = 0: F (10 ft + 3 ft + 7 ft + 10 ft) (2 kip)(6 ft + 2ft) = 0 (4) Solving gives F = kip Substituting F = kip in qs. 2 and 3 and solving gives V = 0.53 kip M = 6.93 kip ft ns. ns. 10 Free-body diagram showing correct senses of internal reactions at F V = 0.53 kip N = 0 F F = 0 M = 6.93 kip ft F 13 ft 7 ft

40 8.1 Internal Forces in Structural Members xample 10, page 6 of 8 Part b) Internal reactions at section J 11 Pass a section through the girder at J. 2 kip F G H J I

41 Internal Forces in Structural Members xample 10, page 7 of 8 12 Free-body diagram of part of structure to right of section at J, including floor beam H F H 13 Force from panel GH acting on floor beam H M J H N J V J J 4 ft I 14 quilibrium equations for part of structure: + Fx = 0: N J = 0 (5) Therefore N J = 0 ns. F y = 0: V J F H = 0 (6) M J = 0: M J F H (4 ft) = 0 (7) 15 Three equations but four unknowns so another free body is needed.

42 + 8.1 Internal Forces in Structural Members xample 10, page 8 of 8 16 Free-body diagram of floor panel GH. This free body will enable us to calculate F H. 2 kip G H 17 quilibrium equation for floor panel GH: M G = 0: (2 kip)(8 ft) + F H (8 ft + 2 ft) = 0 Solving gives F H = 1.60 kip Substituting F H = 1.60 kip in qs. 6 and 7 and solving gives F G F H V J = 1.60 kip ns. 8 ft 2 ft M J = 6.40 kip ft ns. 18 Free-body diagram of part of beam to right of J showing internal forces and moment with correct senses. F H = 1.6 kip V J = 1.60 kip H N J = 0 M J = 6.40 kip ft J I 4 ft

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