COMPONENTS: COMBINED LOADING

Transcription

1 LECTURE COMONENTS: COMBINED LODING Third Edition. J. Clark School of Engineering Department of Civil and Environmental Engineering 5 Chapter 8. by Dr. Ibrahim. ssakkaf SRING 00 ENES 0 Mechanics of Materials Department of Civil and Environmental Engineering University of Maryland, College ark LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 1 Introduction In many industrial cases, structural or machine members are subjected to combinations of three general types of loads: ial Torsional Fleural

2 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. Introduction ial Loading E 1 E E L 1 L L Figure 8. Structural Member under ial Loads LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. Introduction Torsional Loading Cylindrical members Figure 9. Structural Member under Torsional Loads

3 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. Introduction Fleural Loading Figure 0. Beams under Combinations of Loads LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 5 Introduction When a fleural load is combined with torsional and aial loads, it is often difficult to locate the points where most severe stresses (imum) occur. Eample of structural member that is subjected to combined aial, torsional, and fleural loadings is shown in Figure 1

4 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 6 y Introduction a 1 w 0 b M 0 N. y B N. a b C D Figure 1. Beam under a Combination of Loads LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 7 Maimum Stress in a Member due to Combined Loads When a machine or a structural element is subjected to combinations of loads, it is usually difficult to locate the point or points that gives (give) the absolute largest or smallest stress (tension or compression) For eample, the imum absolute normal stress in the cross section of the beam of Fig. 1 that is subjected to the

5 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 8 Maimum Stress in a Member due to Combined Loads combinations of loads shown, could be located at any point lying in the cross section. Furthermore, the location of the critical section along the beam length that provides the critical stress needs to be identified. LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 9 Maimum Stress in a Member due to Combined Loads Depending on the shear and bending moment variations (shear and moment diagrams) along the beam, the critical cross section of the beam of Fig. 1 could be Section a-a, Section b-b, or any other section located a distance from the origin.

6 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 10 Some Helpful Remarks for Identifying the Maimum Stresses 1. The longitudinal and transverse shearing stresses in a beam are imum where Q is imum; usually at the centroidal ais of the section where V is imum. QV τ = (7) It LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 11 Some Helpful Remarks for Identifying the Maimum Stresses. The fleural stress is imum at the greatest distance (fiber) from the centroidal ais on the section where M is imum M y = (8) I

7 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 1 y c c Some Helpful Remarks for Identifying the Maimum Stresses a b N. a b Figure. Beam under In-plane Load N. y t C B d d LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 1 Some Helpful Remarks for Identifying the Maimum Stresses Eample The imum V for the cantilever beam of Fig. is at the fied support. Therefore, the imum shearing stress is represented by point C on the cross section at the support (Section c-c). τ = Q V It, Q td =, td I =

8 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 1 Some Helpful Remarks for Identifying the Maimum Stresses Eample B The imum M for the cantilever beam of Fig. is at the fied support. Therefore, the imum fleural normal stress is represented by points or B on the cross section at the support (Section c-c). = M y I, y = d, td I = LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 15 Some Helpful Remarks for Identifying the Maimum Stresses. The torsional shearing stress is imum at the surface of a shaft at the section where T is imum T = ρ J Tc J τ = (9)

9 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 16 Some Helpful Remarks for Identifying the Maimum Stresses. With reference to remarks 1,, and, it is possible to locate one or more possible point of high stress. Even so, it may be necessary to compute the various stresses at more than one point of a member before locating the most critically stressed point. LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 17 Some Helpful Remarks for Identifying the Maimum Stresses 5. The superposition method can be used to combine stresses on any given plane at any specific point of a loaded member provided that the stresses are below the proportional limit of the material.

10 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 18 Some Helpful Remarks for Identifying the Maimum Stresses 6. fter the stresses on a pair of mutually perpendicular planes at a specific point are determined, plane-stress transformation equations and Mohr s circle can be used to compute the principal stresses as well as the imum shearing stress at the point. LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 19 Eample The dimensions of a T-shaped beam and the loads on it are shown in Figs. a and b. The horizontal 5000-lb tensile force acts through the centroid of the crosssectional area. Determine the imum principal normal stress and the imum shearing stress in the beam, assuming elastic behavior.

11 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 0 Eample (cont d) Figure 6 in in 6 in N.. 5,000 lb 100 lb/ft y C 10 ft in (a) (b) LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 1 in 6 in Eample (cont d) First, we need to locate the neutral ais. Let s make our reference from the bottom edge. 6 in y C N.. y y C ten ( )( 6) + ( 6 + 1)( 6) = = 8 5 =.0 in y M r y Ma. Stress = I com = 5.0 in = 5.0 in = y in

12 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. in 6 in Eample (cont d) Net find the moment of inertia about the neutral ais: ( 5) 6( ) ( 1) I = + = 16 in 6 in Both the imum shearing force V and imum bending moment M occur at N.. the fied support of the beam 5 in V = = 1000 lb in M ( ) ( )(5) = 5,000 ft lb = LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. in 6 in Eample (cont d) Maimum shearing stress: The imum value of Q occurs at the neutral ais. Since in this cross section the 6 in in 5 in width t is minimum at the neutral ais, the imum shearing stress will occur there. Choosing the area below neutral ais, we have N.. 5 Q = V τ = ( )( 5) Q It = 5 in 1000 = 16 ( 5) ( ) = 91.9 psi

13 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. Eample (cont d) The longitudinal normal stress on the uppermost fibers is M yup 5000 ( )( ) up = + I = + 16 = = 1,51.8 psi (T) While the normal stress on the lowest fibers is M yup 5000 ( )( 5) up = I = 16 = 08.,05.9 = 1,998 psi (C) LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 5 Eample (cont d) t any point between either outer surface and centroid, there will be both a normal stress ans a shear stress. However, a comparison of their orders of magnitude suggests that the normal stresses at the outside far eceed the transverse shear stresses at the centroid. Hence we can conclude that the critical fibers will be those at the outside surface, where the imum principal normal stresses will be.

14 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 6 Eample (cont d) Therefore, the principal normal stresses are = psi (T) and = 1998 psi (C) nd the imum shearing stress will occur at the lower fiber, that is min τ = = = 999 psi 0 and will act 5 to the longitudinal ais of beam. LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 7 Eccentric ial Loading in a lane of Symmetry When the line of action of the aial load passes through the centriod of the cross section, it can be assumed that the distribution of normal stress is uniform throughout the section. Such a loading is said to be centric, as shown in Fig.

15 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 8 Eccentric ial Loading in a lane of Symmetry Fig.. Centric Loading LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 9 Eccentric ial Loading in a lane of Symmetry When the line of action of the concentrated load dose not pass through the centroid of the cross section, the distribution of normal stress is no longer uniform. Such loading is said to eccentric, as shown in Fig 5.

16 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 0 Eccentric ial Loading in a lane of Symmetry Fig. 5. Eccentric Loading LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 1 Eccentric ial Loading in a lane of Symmetry Consider the member of Fig. 6a that subjected to an aial load. The internal force acting on a given cross section may then be represented by a force F applied at the centroid y C of the section and a couple M acting in the plane of symmetry of the member (Fig. 6.b).

17 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. Eccentric ial Loading in a lane of Symmetry D C E B D M F Fig. 6. Eccentric Loading e (a) (b) LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. Eccentric ial Loading in a lane of Symmetry The conditions of equilibrium of the free body C require that the force F be equal and opposite to and that the moment of the couple M be equal and opposite to the moment of about C. Denoting by e the distance from the centroid y C to line of action B of the forces

18 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. Eccentric ial Loading in a lane of Symmetry we have F = and M = e (50) The internal forces in the section would have been represented by he same force and couple if the straight portion DE of member B had been detached from B and subjected simultaneously to the LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 5 Eccentric ial Loading in a lane of Symmetry centric loads and the bending couple M of Fig. 7 M D E M C Figure 7 M C M F = M = e

19 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 6 Eccentric ial Loading in a lane of Symmetry Thus, the stress distribution due to the original eccentric loading may be obtained by superposing the uniform stress distribution corresponding to the centric load and the linear distribution corresponding to the bending moment M, as shown in Fig. 8. LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 7 Eccentric ial Loading in a lane of Symmetry y y Figure 8 y C = C + C ( ) centric ( ) bending = +

20 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 8 Eccentric ial Loading in lane of Symmetry The stress due to eccentric loading on a beam cross section is given by D C E B = ± My I (51) LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 9 Equivalent Force System for Eccentric Loading When there are eccentric loadings, that is loadings that their line of action does not pass through the centroid of the cross section, it is necessary to provide an equivalent-force system, consisting of and M for use in Eq. 51. Eample of such system is shown in Fig 9.

21 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 0 =.8 kn Equivalent Force System for Eccentric Loading y M M z =.8 =.8 ( 0) = 19 kn m ( 60 5) = 10 kn m =.8 kn 10 mm 80 mm M z = 10 kn m M = 19 kn m y 5 mm Figure 9 z LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 1 Eample The T-section shown in Fig. 50 is used as a short post to support a compressive load of 150 kip. The load is applied on centerline of the stem at a distance e = in. from the centroid of the cross section. Determine the normal stresses at points and B on a transverse plane C-C near the base of the post.

22 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. Eample (cont d) e 6 in Figure 50 in in 6 in C C Section C-C LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. Eample (cont d) Computing the cross-sectional properties: 6 in rea = C = = [ 6 ] = in ( 6 ) + ( 6 + 1)( 6 ) in = 5 in.from point in 6 in I y = ( 5) 6( ) ( 1) + = 16 in C = 5 in N..

23 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. Eample (cont d) Equivalent force system: = 150 kip acts through centroid M = e = ( 150)( ) 1 =,600 kip in Computations of normal stresses: B = = + My I ( 5) = + 16 () My = I 16 =.78 ksi (T) = ksi (C) LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 5 Eample The beam shown in Fig. 51 has a 8-in rectangular cross section and is loaded in a plane of symmetry. Determine and show on a sketch the principal and imum shearing stresses at point.

24 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 6 Eample (cont d) 8 80,000 lb 16 Section C-C C 10 C 8 Figure 51 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 7 Eample (cont d) 80 kip On Section C-C, we have 8 kip 6 kip M 16 6 kip α 8 = 80,000 cosθ = 80,000 = 6 kip 5 = 80,000sinθ = 80,000 = 8 kip 5 C 8 kip θ C 16 csinα = 16 = cosα = 16 = 9.6 5

25 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 8 Eample (cont d) ( ) + 8( 9.6) = 1,90 kip in M = 6 80 kip On Section C-C, we have 8 kip 6 kip 1,90 kip in 16 6 kip α 8 10 C 8 kip θ C 16 csinα = 16 = cosα = 16 = LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 9 Eample (cont d) On Section C-C through point, we have = 8 kip = ( 8) = in V = 6 kip Q = ()() = 18 in () 8 M = 190 kip in Therefore, I = My y = + = + I 18 VQ 6( 18) τ = = =.0 ksi It 18() ( ) 1 = 18 in =.0 ksi (T)

26 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 50 Eample (cont d) rincipal Stresses: = ± + ( ) = =. ksi (T) p1, p θ p1 p p p = = = 0.79 ksi (C) = 0 1 tan 1 τ ( ) 0 = = τ = 16.8 ksi p LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 51 Eample (cont d) Sketch of the principal and imum shearing stresses at point : y ksi ksi θ p 0 θ p = ksi ksi. ksi

27 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 5 Eample 5 The cantilever beam shown in Fig. 5a has the cross section shown in Fig.5b. If the allowable stresses are 10,000 psi shear and 16,000 psi tension and compression at point (just below the flange), determine the imum allowable load. LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 5 Eample 5 (cont d) Figure 5 0 in 7.5 in 5 1 in 1 in 1 in 10 in 1 in (a) (b)

28 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 5 Eample 5 (cont d) On a section through point, we have N = 5 = ( 10)( 1) = 0 in V = Q = 10()( 1 7) = 70 in 10( 15) 9.5( 1) M = 0 Therefore, I = 1 1 ( 0)( 6.5) N My 5 = + = + I VQ ( 70) τ = = = It 107(0.5) = 107 in = 0.10 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 55 Eample 5 (cont d) τ rincipal Stresses and Ma Shear Stress = τ = < 10,000 psi p Therefore, =, lb. kip = p1 = + + ( 0.105) = = 0.67 < 16,000 psi <, lb

29 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 56 Eample mm-diameter steel rod is loaded and supported as shown in Fig. 5. Determine the principal stresses and imum shearing stress (a) t point on a section adjacent to the support. (b) t point B on a section adjacent to the support. LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 57 Eample 6 (cont d) Figure 5 9 kn m 8 kn B 750 mm 10 kn 00 kn

30 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 58 Eample 6 (cont d) On a section adjacent to the support, π π 1 = d = ( 100) = 785 mm QN = ( 100) = mm 1 π π 6 I = d = ( 100) = mm M = ( 0.750) = 7500 kn m 6 6 π π 6 J = d = ( 100) = mm M B = 8 10 ( 0.750) = 6000 kn m Note for QN : area πd d d QN = distance from N to Centroid = = 8 π 1 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 59 Commonly Used Second Moments of lane reas (For Eample 6) Figure 5 or d I = π (for d = R) 6 d d = = π π (for d = R) πr J = πd =

31 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 60 Eample 6 (cont d) (a) For point : ( 0.1) ( 0.05) Mc = + = + = N/m 6 6 I VQ Tc 8 10 ( ) 9 10 τ y = = =.8 10 N/m 6 6 It J = ± + (.8) = = 11. Ma (T) p1, p p1 = 0 p θ = p 1 tan 1 τ (.8) 0 = = τ = 77.7 Ma p p ( 0.05) 6 = = 1 Ma (C) LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 61 Eample 6 (cont d) (b) For point B: ( 0.1) ( 0.05) Mc = + = + = N/m 6 6 I VQ Tc ( ) 9 10 τ y = = = N/m 6 6 It J = ± + ( 7.5) = = 19.5 Ma (T) p1, p p1 = 0 p θ = p 1 tan 1 τ ( 7.5) 0 = = τ = 7.5 Ma p p ( 0.05) 6 = = 17.5 Ma (C)

32 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 6 Eample 7 -in-diameter solid circular steel shaft is loaded and supported as shown in Fig. 55. If the imum normal and shearing stresses at point must be limited to 7500 psi tension and 5000 psi, respectively, determine the imum permissible value for the the transverse load V. LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 6 Eample 7 (cont d) Figure 55 7,500 ft lb -in Diameter 1,500 ft lb V ft ft 10,000 lb

33 LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 6 Eample 7 (cont d) On a section adjacent to the support, π π = d = ( ) = in T = 1,500 7,500 = π I = d 6 π J = d π = 6 π = ( ) = in M = V ( 6) ( ) = 5.1 in = 10,000 lb ( ) Mc 10,000 6V 1 = + = + I Tc ( ) τ y = = = 775 psi J 5.1 = V = 6V ft lb 5000 ft lb LECTURE 5. COMONENTS: COMBINED LODING (8.) Slide No. 65 Eample 7 (cont d) rincipal Stresses and Ma Shearing Stress p1 = + τ Therefore, and = τ = p + + ( 775) ( 775) = V <,966 V, = = 189. lb < 7,500 psi <,60 psi < 5,000 psi <,966 psi