SHEAR AND DIAGONAL TENSION IN BEAMS

Transcription

1 CHAPTER REINFORCED CONCRETE Reinorced Concrete Design Fith Edition A Fundamental Approach - Fith Edition SHEAR AND DIAGONAL TENSION IN BEAMS A. J. Clark School o Engineering Department o Civil and Environmental Engineering 6a SPRING 004 By Dr. Ibrahim. Assakka ENCE 454 Design o Concrete Structures Department o Civil and Environmental Engineering University o Maryland, College Park CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 1 The previous chapter dealt with the lexural strength o beams. Beams must also have an adequate saety margin against other types o ailure such as shear, which may be more dangerous than lexural ailure. The shear orces create additional tensile stresses that must be considered.

2 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. Shear Failure Shear ailure o reinorced concrete beam, more properly called diagonal tension ailure, is diicult to predict accurately. In spite o many years o experimental research and the use o highly sophisticated computational tools, it is not ully understood. I a beam without properly designed or shear reinorcement is overloaded to ailure, shear collapse is likely to occur suddenly. CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 3 Figure 1. Shear Failure (Nilson, 1997) (a) Overall view, (b) detail near right support.

3 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 4 Shear Failure (cont d) Figure 1 shows a shear-critical beam tested under point loading. With no shear reinorcement provided, the member ailed immediately upon ormation o the critical crack in the high-shear region near the right support. CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 5 Shear Failure (cont d) When are the shearing eects so large that they cannot be ignored as a design consideration? It is somehow diicult to answer this question. Probably the best way to begin answering this question is to try to approximate the shear stresses on the cross section o the beam.

4 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 6 Shear Failure (cont d) Suppose that a beam is constructed by stacking several slabs or planks on top o another without astening them together. Also suppose this beam is loaded in a direction normal to the surace o these slabs. When a bending load is applied, the stack will deorm as shown in Figure a. Since the slabs were ree to slide on one another, the ends do not remain even but staggered. CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 7 Shear Failure (cont d) P (a) Unloaded Stack o Slabs (b) Unglued Slabs loaded Figure a

5 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 8 Shear Failure (cont d) Each o the slabs behaves as independent beam, and the total resistance to bending o n slabs is approximately n times the resistance o one slab alone. I the slabs o Figure b is astened or glued, then the staggering or relative longitudinal movement o slabs would disappear under the action o the orce. However, shear orces will develop between the slabs. In this case, the stack o slabs will act as a solid beam. CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 9 Shear Failure (cont d) P (c) Glued Slabs Unloaded (d) Glued Slabs loaded Figure b

6 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 10 Shear Failure (cont d) The act that this solid beam does not exhibit this relative movement o longitudinal elements ater the slabs are glued indicates the presence o shearing stresses on longitudinal planes. Evaluation o these shearing stresses will be discussed in the next set o viewgraphs. CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 11 Theoretical Background The concept o stresses acting in homogeneous beams are usually covered in various textbooks o mechanics o materials (strength o materials). It can be shown that when the material is elastic, shear stresses can be computed by VQ v = Ib v = shear stress V = external shear orce I = moment o inertia about neutral axis Q = statical moment o area about N.A. b = width o the cross section (1)

7 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 1 Theoretical Background (cont d) Also, when the material is elastic, bending stresses can be computed rom = Mc I () = bending stress M = external or applied moment c = the distance rom the neutral axis to out iber o the cross section I = moment o inertia o the cross section about N.A. CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 13 Theoretical Background (cont d) All points along the length o the beam, where the shear and bending moment are not zero, and at locations other than the extreme iber or neutral axis, are subject to both shearing stresses and bending stresses. The combination o these stresses produces maximum normal and shearing stresses in a speciic plane inclined with respect to the axis o the beam.

8 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 14 Theoretical Background (cont d) y The distributions o the bending and shear stresses acting individually are shown in Figs. 3, 4, 5, and 6. Mc P = I w Neutral axis Centroidal axis F C c x F T da y C c c y dy R V r Figure 3. Bending Stress CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 15 Theoretical Background (cont d) Figure 4. Bending Stress = Mc I

9 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 16 Theoretical Background (cont d) VQ v = Ib Figure 5. Vertical Shearing Stress N.A V Max Stress CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 17 Theoretical Background (cont d) Figure 6. Vertical Shearing Stress VQ v = Ib

10 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 18 Principal Planes The combination o bending moment and shearing stresses is o such a nature that maximum normal and shearing shearing stresses at a point in a beam exist on planes that are inclined with the axis o the beam. These planes are commonly called principal planes, and the stresses that act on them are reerred to as principal stresses. CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 19 Principal Planes General Plane State o Stress y y v yz v xy v yx v xy v zy v z v zx v xz x x v xy x v yx y

11 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 0 Principal Planes General Plane State o Stress (cont d) Components: Normal Stress x Normal Stress y Shearing Stress v xy Shearing Stress v yx v xy = v yx x A θ v xy v yx y y v yx v xy A x CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 1 Principal Planes when pr pr General Plane State o Stress (cont d) = = x x Principal Stresses y y ± = 0, y ± x ( ) x 4 =, ν xy ( x y ) 4 xy y = ν + v + v xy x A θ v xy v yx y y v yx v xy A x

12 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. Principal Stresses The principal stresses in a beam subjected to shear and bending may be computed using the ollowing equation: pr + v 4 = ± (3) pr = principal stress = bending stress computed rom Eq. v = shearing stress computed rom Eq. 1 A θ v v A v v CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 3 Orientation Principal Planes The orientation o the principal planes may be calculated using the ollowing equation: 1 1 = tan v α (4) Note that at the neutral axis o the beam, the principal stresses will occur at a 45 0 A v angle. v θ v v A

13 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 4 State o Stress at the Neutral Axis o a Homogeneous Beam w v v N.A. v v (a) Beam under Uniorm Loading (b) Stresses on Unit Element Figure 7. Shear Stress Relationship CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 5 State o Stress at the Neutral Axis o a Homogeneous Beam C Diagonal Tension Figure 8 v v v v B This plane is subject to compression v v D A v This plane is subject to tension v

14 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 6 State o Stress at the Neutral Axis o a Homogeneous Beam Diagonal Tension Plane A-B is subjected to tension While Plane C-D is subjected to compression. The tension in Plane A-B is historically called diagonal tension. Note that concrete is strong in compression but weak in tension, and there is a tendency or concrete to crack on the plane subject to tension, Plane A-B. When the tensile stresses are so high, it is necessary to provide reinorcement. CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 7 Diagonal Tension Failure In the beams with which we are concerned, where the length over which a shear ailure could occur (the shear span) is in excess o approximately three times the eective depth, the diagonal tension ailure would be the mode o ailure in shear. Such a ailure is shown in Figures 1 and 8. For longer shear spans in plain concrete beams, cracks due to lexural tensile stresses would occur long beore cracks due to diagonal tension.

15 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 8 Diagonal Tension Failure Shear Span Portion o span in which Shear stress is high Figure 8. Typical Diagonal Tension Failure CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 9 Figure 1. Shear Failure (Nilson, 1997) (a) Overall view, (b) detail near right support.

16 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 30 Diagonal Tension Failure The principal stress in tension acts at an approximately 45 0 plane to the normal at sections close to support, as seen in Figure 9. Because o the low tensile strength o concrete, diagonal cracking develops along planes perpendicular to the planes o principal tensile stress; hence the term diagonal tension cracks. To prevent such cracks, special diagonal tension reinorcement has to be provided. CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 31 Diagonal Tension Failure Tensile trajectories compressive trajectories Figure 9. Trajectories o principal stresses in a homogeneous Isotropic beam

17 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 3 Beams without Diagonal Tension Reinorcement Modes o Failure o Beams without Diagonal Tension Reinorcement The slenderness o the beam or its shear span/depth determines the ailure o the plain concrete beam. Figure 10 demonstrates schematically the ailure patters. The shear span a or concentrated load is the distance between the point o application o the load and the ace o the support. CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 33 Beams without Diagonal Tension Reinorcement Modes o Failure o Beams without Diagonal Tension Reinorcement (cont d) Figure 10. Failure Patterns as a unction o beam slenderness; (a) lexure ailure; (b) diagonal tension ailure; (c) shear compression ailure.

18 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 34 Beams without Diagonal Tension Reinorcement Modes o Failure o Beams without Diagonal Tension Reinorcement (cont d) Basically, three modes o ailure or their combinations are o interest: 1. Flexure Failure. Diagonal Tension Failure 3. Shear Compression Failure The more slender the beam, the stronger the tendency toward lexure behavior. CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 35 Beams without Diagonal Tension Reinorcement Modes o Failure o Beams without Diagonal Tension Reinorcement (cont d) The shear span l c or distributed loads is the clear beam span. Table 1 gives expected range values o the shear span/depth ratio or the dierent modes o ailure and types o beams.

19 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 36 Beams without Diagonal Tension Reinorcement Table 1. Beam Slenderness Eect on Mode o Failure Beam Category Failure Mode Shear Span/Depth Ratio as a Measure o Slenderness Concentrated load, a/d Distributed Load, l c /d Slender Flexure (F) Exceeds 5.5 Exceeds 16 Intermediate Diagonal Tension (DT).5 to to 16 Deep Shear Compression (SC) 1 to.5 1 to 5 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 37 ACI Philosophy and Rationale Basis o ACI Design or Shear The ACI provides design guidelines or shear reinorcement based on the vertical shear orce V u that develops at any given cross section o a member. Although it is really the diagonal tension or which shear reinorcing must be provided, diagonal tensile orces (or stresses) are not calculated. Traditionally, vertical shear orce has been taken to be good indicator o diagonal tension present.

20 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 38 Shear or Web Reinorcement The basic rationale or the design o the shear reinorcement, or web reinorcement as it usually called in beams, is to provide steel to cross the diagonal tension cracks and subsequently keep them rom opening. In reerence to Figure 8, it is seen that the web reinorcement may take several orms such as: CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 39 Shear or Web Reinorcement (cont d) 1. Vertical stirrups (see Fig. 11). Inclined or diagonal stirrups 3. The main reinorcement bent at ends to act as inclined stirrups (see Fig. 1). The most common orm o web reinorcement used is the vertical stirrup. This reinorcement appreciably increases the ultimate shear capacity o a bending member.

21 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 40 Shear or Web Reinorcement (cont d) Vertical Stirrups Vertical Stirrups L L Figure 11. Types o Web Reinorcement CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 41 Shear or Web Reinorcement (cont d) Bent-up Longitudinal Bars Bent-up bar L L Figure 1. Type o Web Reinorcement

22 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 4 ACI Code Provisions or Shear Reinorcement For member that are subject to shear and lexure only, the amount o shear orce that the concrete (unreinorced or shear)can resist is Vc = λ cb Note, or rectangular beam b w = b λ= 1.0 or normal concrete λ = 0.85 or sand-lightweight concrete λ = 0.75 or all light weight concrete w d (5) CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 43 ACI Code Provisions or Shear Reinorcement When axial compression also exists, V n in Eq. 5 becomes N u Vc = λ 1 + cb A 000 g Note, or rectangular beam b w = b, and N u /A g is expressed in psi N u = axial load (positive in compression) A g = gross area o the section w d (6)

23 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 44 ACI Code Provisions or Shear Reinorcement When signiicant axial tension also exists, V n in Eq. 5 becomes N u Vc = λ 1 + cb A 500 g Note, or rectangular beam b w = b, and N u /A g is expressed in psi N u = axial load (negative in compression) A g = gross area o the section w d (7) CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 45 ACI Code Provisions or Shear Reinorcement The ACI also permits the use o more reined equation in replacement o Eq. 5, that is V u c = 1.9λ cbwd + 500ρ w bwd 3. 5 M u Note, or rectangular beam b w = b, and V u d/m u 1 V u = φv n = actored shear orce, taken d distance rom FOS M u = φm n = actored bending moment, taken d distance rom FOS V d b w d c (8)

24 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 46 ACI Code Provisions or Shear Reinorcement The design shear orce V u results rom the application o actored loads. Values o V u are most conveniently determined using a typical shear orce diagram. Theoretically, no web reinorcement is required i V φ (9) u V c CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 47 Table. Resistance or Strength Reduction Factors Structural Element Factor φ Beam or slab; bending or lexure 0.90 Columns with ties 0.65 Columns with spirals 0.75 Columns carrying very small axial load (reer to Chapter 9 or more details) or Beam: shear and torsion 0.75

25 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 48 ACI Code Provisions or Shear Reinorcement However, the code requires that a minimum area o shear reinorcement be provided in all reinorced concrete lexural members when V u > ½ φv c, except as ollows: In slabs and ootings In concrete joist construction as deined in the code. In beams with a total depth o less than 10 in., ½ times the lange thickness, or one-hal the width o the web, whichever is greater. CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 49 ACI Code Provisions or Shear Reinorcement In cases where shear reinorcement is required or strength or because V u > ½ φv c, the minimum area o shear reinorcement shall be computed rom A v b ws 50bws = max 0.75 c, (10) y y

26 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 50 ACI Code Provisions or Shear Reinorcement Where A v = total cross-sectional area o web reinorcement within a distance s, or single loop stirrups, A v = A s A s = cross-sectional area o the stirrup bar (in ) b w = web width = b or rectangular section (in.) s = center-to-center spacing o shear reinorcement in a direction parallel to the longitudinal reinorcement (in.) y = yield strength o web reinorcement steel (psi) = compressive strength o concrete (psi) c CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 51 ACI Code Provisions or Shear Reinorcement Figure 13. Isometric section showing stirrups partially exposed

27 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 5 Example 1 A reinorced concrete beam o rectangular cross section shown in the igure is reinorced or moment only (no shear reinorcement). Beam width b = 18 in., d = 10.5 in., and the reinorcing is ive No. 4 bars. Calculate the maximum actored shear orce V u permitted on the member by the ACI Code. Use c = 4,000 psi, and y = 60,000 psi, and normal concrete. CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 53 Example 1 (cont d) Since no shear reinorcement is provided, the ACI Code requires that maximumv u 1 = φv 1 = φ 1 = c 5 #4 ( λ b d ) c w Eq ( 0.75)( )( 4000)( 18)( 10.5) = 8,75 lb 10.5

28 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 54 ACI Code Provisions or Shear Design According to the ACI Code, the design o beams or shear is based on the ollowing relation: φvn V u (11) Where φ = strength reduction actor (= 0.75 or shear) V n = V c + V s V s = nominal shear strength provided by reinorcement A v y d = or vertical stirrups (1) s CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 55 ACI Code Provisions or Shear Design For inclined stirrups, the expression or nominal shear strength provided by reinorcement is Av yd( sin α + cosα ) (13) Vs = s For α = 450, the expression takes the orm V s 1.414A = s v y d (14)

29 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 56 ACI Code Provisions or Shear Design The design or stirrup spacing can be determined rom Av yd required s = V and 1.414Av yd required s = V where V s s s V = (or vertical stirrups) u V φ (or 45 φ c 0 stirrups) (15) (16) (17) CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 57 ACI Code Provisions or Shear Design According to the ACI Code, the maximum spacing o stirrups is the smallest value o Av y smax = 50b s s max max = = w d 4 in. (18) I V s exceeds 4 c bwd, the maximum spacing must not exceed d/4 or 1 in.

30 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 58 ACI Code Provisions or Shear Design It is not usually good practice to space vertical stirrups closer than 4 in. It is generally economical and practical to compute spacing required at several sections and to place stirrups accordingly in groups o varying spacing. Spacing values should be made to not less than 1- in. increments. CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 59 ACI Code Provisions or Shear Design Critical Section The maximum shear usually occurs in this section near the support. For stirrup design, the section located a distance d rom the ace o the support is called the critical section Sections located less than a distance d rom the ace o the support may be designed or the same V u as that o the critical section.

31 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 60 ACI Code Provisions or Shear Design Critical Section Critical Section d d L Figure 14 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 61 ACI Code Provisions or Shear Design Critical Section (cont d) The stirrup spacing should be constant rom the critical section back to the ace o the support based on the spacing requirements at the critical section. The irst stirrup should be placed at a maximum distance o s/ rom the ace o the support, where s equals the immediately adjacent required spacing (a distance o in. is commonly used.

32 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 6 Shear Analysis Procedure The shear analysis procedure involves the ollowing: Checking the shear strength in an existing member Veriying that the various ACI code requirements have been satisied and met. Note that the member may be reinorced concrete section or plain concrete section. CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 63 Shear Analysis Procedure Example A reinorced concrete beam o rectangular cross section shown is reinorced with seven No. 6 bars in a single layer. Beam width b = 18 in., d = 33 in., single-loop No. 3 stirrups are placed 1 in. on center, and typical cover is 1 ½ in. Find V c, V s, and the maximum actored shear orce permitted on this member. Use c = 4,000 psi and y = 60,000 psi. Assume normal concrete.

33 CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 64 Shear Analysis Procedure Example (cont d) 18 # stirrup 1 1 COV #6 bars CHAPTER 6a. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 65 Shear Analysis Procedure Example (cont d) The orce that can be resisted by concrete alone is (1) 4,000( 18)( 33) V λ c = cbwd = = 75.1 kips 1000 The nominal shear orce provided by the steel is Av yd ( 0.11)( 60)( 33) V s = s = = 36.3 kips The maximum actored shear orce is maximumv φv + φv = 0.75( ) 1 u = c s = 83.6 kips