Lecture 2: Behavior of Beams in Flexure (Bending) PDF

Lec Design Reinforced Concrete Structures Misan University Lec Third Year Enginee...

Lec Design Reinforced Concrete Structures Misan University Lec Third Year Engineering College 2 Asst.Prof.Dr.Abbas Oda Dawood Civil Department 2 Lecture 2: Behavior of Beams in Flexure (Bending) Behavior of Beams in Flexure (Bending) 1. Design Assumptions Reinforced concrete sections are heterogeneous (nonhomogeneous), because they are made of two different materials, concrete and steel. Therefore, proportioning structural members by strength design approach is based on the following assumptions: 1. Cross sections continue that plane before loading remain plane under loading → The strain distribution is l inear. 2. The strain of reinforcement equal to the strain of concrete at the same level → perfect bond. 3. Concrete is assumed not resist any tension if: Applied tensile stress > modulus of rupture of the concrete ft < fr ft > fr 4. The stresses in the concrete and reinforcement can be computed from the strains by using stress–strain curves for concrete and steel. See figure below Lecture 2....... Page 1 Lec Design Reinforced Concrete Structures Misan University Lec Third Year Engineering College 2 Asst.Prof.Dr.Abbas Oda Dawood Civil Department 2 Lecture 2: Behavior of Beams in Flexure (Bending) 2.Behavior of Reinforced Concrete Beams in Flexure Loaded to Failure If it is assumed that a small transverse load is placed on a concrete beam with tensile reinforcing and that the load is gradually increased in magnitude until the beam fails. As this takes place, the beam will go through three distinct stages before collapse occurs. These are: 1. The uncracked concrete stage 2. The concrete cracked–elastic stresses stage 3. The ultimate-strength stage (Failure) 2.1 Uncracked Elastic Section At small loads when the tensile stresses are less than the modulus of rupture (the bending tensile stress at which the concrete begins to crack), the entire cross section of the beam resists bending. Thus Lecture 2....... Page 2 Lec Design Reinforced Concrete Structures Misan University Lec Third Year Engineering College 2 Asst.Prof.Dr.Abbas Oda Dawood Civil Department 2 Lecture 2: Behavior of Beams in Flexure (Bending) ft  fr  Uncracked Section 1 fc  f c  Elastic Section 2 fs  f y N.A. εt ft < fr Mc Stress  I Use transform section method for calculations. Modular Ratio The modular ratio n is the ratio of the modulus of elasticity of steel to the modulus of Es elasticity of concrete: n  Ec Lecture 2....... Page 3 Lec Design Reinforced Concrete Structures Misan University Lec Third Year Engineering College 2 Asst.Prof.Dr.Abbas Oda Dawood Civil Department 2 Lecture 2: Behavior of Beams in Flexure (Bending) 2.2 Cracked Elastic Section As the load is increased after the modulus of rupture of the concrete is exceeded, cracks begin to develop in the bottom of the beam. The moment at which these cracks begin to form—that is, when the tensile stress in the bottom of the beam equals the modulus of rupture—is referred to as the cracking moment, Mcr. As the load is further increased, these cracks quickly spread up to the vicinity of the neutral axis, and then the neutral axis begins to move upward. The cracks occur at those places along the beam where the actual moment is greater than the cracking moment, as shown in Figure below. This stage will continue as long as the compression stress in the top fibers is less than 0.5f'c , and as long as the steel stress is less than its yield stress. Thus ft  fr  Cracked Section 1 fc  f c  Elastic Section 2 fs  f y Mc Stress  I cr N.A. The cracking moment, Mcr is calculated from the equation: f r Ig M cr  yt Lecture 2....... Page 4 Lec Design Reinforced Concrete Structures Misan University Lec Third Year Engineering College 2 Asst.Prof.Dr.Abbas Oda Dawood Civil Department 2 Lecture 2: Behavior of Beams in Flexure (Bending) 2.2.1 Permissible service load stresses According to ACI Code 1995,Appendix A In working stress method (elastic method) the allowable stresses according to ACI code 1995 are: Extreme fiber stress in compression in concrete , fcall  0.45 f c Allowable Tensile stress in reinforcement=140 MPa → fy=275-350 MPa =170 MPa → fy=420 MPa (or 400) and greater Lecture 2....... Page 5 Lec Design Reinforced Concrete Structures Misan University Lec Third Year Engineering College 2 Asst.Prof.Dr.Abbas Oda Dawood Civil Department 2 Lecture 2: Behavior of Beams in Flexure (Bending) 2.2.2 Interior Couple Method In working stress method the stresses can be found by internal couple method in additional to transform section method discussed previously. kd N.A. T=As fs  Fx  0 C  T As fy  2 fc. kd. b 1            (1) c kd From strain diagram →             (2)  s d  kd As Solve Eqs. (1) and (2) yields k    n  ( n) 2  2 n , where   bd kd k jd  d  j 1  3 3 M  C. jd  T. jd  M fs   fs all As. jd 2M fc   fc all b.kd. jd Lecture 2....... Page 6 Lec Design Reinforced Concrete Structures Misan University Lec Third Year Engineering College 2 Asst.Prof.Dr.Abbas Oda Dawood Civil Department 2 Lecture 2: Behavior of Beams in Flexure (Bending) 2.3 Failure State (Ultimate-Strength Stage) As the load is increased further so that the compressive stresses are greater than 0.50f'c ,the tensile cracks move farther upward, as does the neutral axis, and the concrete compression stresses begin to change appreciably from a straight line. 3.Types Of Flexural Failure And Strain Limits Three types of flexural failure of a structural member can be expected depending on the percentage of steel used in the section. 1. Tension Failure (Secondary Compression Failure):  tension-controlled Section or under-reinforced section c  u  s   y Lecture 2....... Page 7 Lec Design Reinforced Concrete Structures Misan University Lec Third Year Engineering College 2 Asst.Prof.Dr.Abbas Oda Dawood Civil Department 2 Lecture 2: Behavior of Beams in Flexure (Bending)  Steel may reach its yield strength before the concrete reaches its maximum strength. In this case, the failure is due to the yielding of steel reaching a high strain equal to or greater than 0.005.  when steel reach yield point, at this stress, the reinforcement yields suddenly and stretches large amount, and the tension cracks in concrete widen visibly and propagate upward, with simultaneous significant deflection of the beam. When this happens, the strains in the remaining compression zone of the concrete increase to such a degree that crushing of the concrete, the secondary compression failure. Such yield failure is gradual and is preceded by visible signs of distress, such as the widening and lengthening of cracks and the market increase in deflection. 2. Balanced Failure:  Steel may reach its yield strength at the same time as concrete reaches its ultimate strength. c  u  s   y 3. compression Failure  Compression-controlled Section or Over-reinforced section c  u  s   y   Concrete may fail before the yield of steel due to the presence of a high percentage of steel in the section. In this case, the concrete strength and its maximum strain of 0.003 are reached, but the steel stress is less than the yield strength, that is, fs is less than fy. The strain in the steel is equal to or less than 0.002.  Compression failure due to crushing of concrete is sudden, of an almost explosive nature, and occurs without warning. For this reason beam design to failed by yielding of the steel rather than by crushing of the concrete Lecture 2....... Page 8 Lec Design Reinforced Concrete Structures Misan University Lec Third Year Engineering College 2 Asst.Prof.Dr.Abbas Oda Dawood Civil Department 2 Lecture 2: Behavior of Beams in Flexure (Bending) 4. Strain Limits The design provisions for both reinforced and prestressed concrete members are based on the concept of tension or compression-controlled sections, ACI Code, Section 21.2. Both are defined in terms of net tensile strain (NTS), εt, in the extreme tension steel at nominal strength. In addition to tension and compression-controlled sections, two other conditions may develop: (1) the balanced strain condition and (2) the transition region condition. These four conditions are defined as follows:  Compression-controlled sections are those sections in which the net tensile strain, NTS, in the extreme tension steel at nominal strength is equal to or less than the compression-controlled strain limit (For grade 60 steel, (f = 60 ksi), the compression- controlled strain limit may be taken as a net strain of 0.002) at the time when concrete in compression reaches its assumed strain limit of 0.003, ( c = 0.003). This case occurs mainly in columns subjected to axial forces and moments.  Tension-controlled sections are those sections in which the NTS, t , is equal to or greater than 0.005 just as the concrete in the compression reaches its assumed strain limit of 0.003.  Transition region condition : sections in which the NTS in the extreme tension steel lies between the compression controlled strain limit (0.002 for fy= 60 ksi) and the tension-controlled strain limit of 0.005 constitute the transition region.  The balanced strain condition develops in the section when the tension steel, with the first yield, reaches a strain corresponding to its yield strength, just as the maximum strain in concrete at the extreme compression fibers reaches 0.003. ACI Code, Section 9.3.3.1 indicates that the net tensile strain, εt, at nominal strength, within the transition region, shall not be less than 0.004 for reinforced concrete flexural members without or with an axial load less than 0.10 f'c Ag,where Ag=gross area of the concrete section. Lecture 2....... Page 9 Lec Design Reinforced Concrete Structures Misan University Lec Third Year Engineering College 2 Asst.Prof.Dr.Abbas Oda Dawood Civil Department 2 Lecture 2: Behavior of Beams in Flexure (Bending)  Example 1: 1-Uncracked Section A simply supported beam having rectangular cross section, is loaded as shown in figure below. Investigate Es 200,000 200,000 the state of beam during load increase from zero till n    8.5 E c 4700 f c 4700 25 failure load. f c  25 MPa f y  400 MPa.  As  3 * ( * 22 2 )  1140.4 mm 2 P 4 600 550 As (n  1) 1140.4 * (8.5  1)  8553 3  22 3m 3m 600 Solution 300 600 * 300 *  As (n  1) * 550 y 2  311.3 mm 300 * 600  As (n  1) P 3m 3m 1/2 P y  311. 3 S.F.D + 600 - 1/2P y t  288. 7 1.5 P 300 B.M.D + Lecture 2....... Page 10 Lec Design Reinforced Concrete Structures Misan University Lec Third Year Engineering College 2 Asst.Prof.Dr.Abbas Oda Dawood Civil Department 2 Lecture 2: Behavior of Beams in Flexure (Bending) Example 1: : cont. kd 300 * kd *  n As * (d  kd) 3 3 2 311.3 * 300 288.7 * 300 I    As (n  1) * N.A 3 3 150 kd 2  9693 * (550  kd) (550  311.3) 2  5.91*10 9 mm 4 150 kd 2  9693 kd  5331150  0 Solve kd  158.96  159 mm f t all  f r  0.62  f ' c  0.62 *1.0 * 25  3.1MPa 300 *1593 M yt M * 288.7 Icr   nAs * (550  159) 2  1.88*109 mm4 f t all  3.1  63.46 *106 N.mm 3 I 5.91*109  63.46 KN.m f call  0.45 f c 11.25  M *c  M *159 Icr 1.88 *109 M 63.46 M 1.5 P P    42.31 KN 1.5 1.5  M 133 KN.m M 133 M * 311.3 P   88.66 KN f c all  0.45 f c  11.25   M  213.6 KN.m 1.5 1.5 5.91*109 M 213.6 For f y  400  fs, all 170 MPa P   142.4 KN 1.5 1.5 M.c f s _ all  n.  Icr For f y  400  fs _ all 170 MPa M * (550  159) M.c M * (550  311.3) 170  8.5 *  M  96.16 KN.m f s _ all  n. 170  8.5. 1.88 *10 9 I 5.91*109 M 96.16 M 495.2 P    64 KN  M  495.2 KN.m P    330.12 KN 1.5 1.5 1.5 1.5 Use smaller value P  64 KN Use smaller value P  42.13 KN Thus at this force P=64 KN the stresses in steel and Thus for uncracked section maximum allowable force concrete are P=42.13 KN and the stresses at this force is: f s 170 MPa 6 (1.5 * 42.13 *10 ) * 311.3 fc   3.33 MPa 11.25 (1.5 * 64 *106 ) *159 5.91*109 fc   8.12 MPa 11.25 1.88*109 (1.5 * 42.13 *106 ) * (550  311.3) f s  8.5 *  21.70 MPa 5.91*109 2-Cracked Section n As  8.5 *1140.4  9693.0 300 kd d n As d  kd Lecture 2....... Page 11 Lec Design Reinforced Concrete Structures Misan University Lec Third Year Engineering College 2 Asst.Prof.Dr.Abbas Oda Dawood Civil Department 2 Lecture 2: Behavior of Beams in Flexure (Bending) 5. Failure State The distribution of compressive concrete stresses at failure may be assumed to be a rectangle, trapezoid, parabola, or any other shape that is in good agreement with test results. 5.1 Actual Stress Distribution h d As b C   f c b c , T  A s f s ,  Fx  0 C  T  f c b c  As f s          (1) M n  C * z   f c b c (d  c) or M n  T * z  A s f s (d  c)      (2) For failure initiated by yielding of the tension steel, fs=fy, thus from Eq.(1) As f y As  fy d c , but    c           (3)  f c b bd  f c As Substitute Eq. (3) into Eq. (2) with   and fs=fy yields bd fy  M n   f y b d 2 (1  ) , where Mn is the nominal strength of the section or nominal-  f c section moment capacity. Substitute values of  and  yields a nominal strength of fy M n   b d 2 f y (1  0.59  ) f c Lecture 2....... Page 12 Lec Design Reinforced Concrete Structures Misan University Lec Third Year Engineering College 2 Asst.Prof.Dr.Abbas Oda Dawood Civil Department 2 Lecture 2: Behavior of Beams in Flexure (Bending) 5.2 Equivalent Rectangular Stress Distribution ( Whitney Block )  This model is adopted by ACI Code h d As b a  1 c c= location of neutral axis N.A. at failure C  0.85 f c b c According to ACI section 10.2.7.3 1  0.85 for f c  28 MPa f   28 1  [ 0.85  0.05 * c ]  0.65 for f c  28 MPa 7 Thus 0.65  1  0.85 6. Balanced Failure State (Balanced Steel Ratio b) Lecture 2....... Page 13 Lec Design Reinforced Concrete Structures Misan University Lec Third Year Engineering College 2 Asst.Prof.Dr.Abbas Oda Dawood Civil Department 2 Lecture 2: Behavior of Beams in Flexure (Bending) From equilibrium equation  Fx  0 C  T 0.85 f c b a b  A sb f y          (1) From the strain diagram u cb u   cb  dt ,  y d t  cb u  y u a b  1 c b a b  1 d t          (2) u   y Thus from Eqs.(1) and (2) u 0.85 f c b 1 d t  A sb f y u   y 0.003 0.85 f c b 1 d t  A sb f y ( b d) 0.003  f y / E s dt 0.003 A A 0.85 f c 1  sb f y but  b  sb d 0.003  f y / E s bd bd dt 0.003 0.85 f c 1  b f y divide by fy d 0.003  f y / E s f c d 0.003  b  0.85 1 t substitute Es=200,000 MPa fy d 0.003  f y / E s f c d 600 dt  b  0.85 1 t , If one layer reinforcement d=dt and  1.0 fy d 600  f y d  If  = b → Balanced Failure  If  > b → Compression failure  If  < b → Tension Failure (Secondary Compression Failure)  In actual practice the upper limit on  should be below b.  Previous codes (1963–1999) limited flexural members to 75% of the balanced steel ratio, ρb (ρmax = 0.75 ρb. However, this approach was changed in the 2002 code to the new philosophy which based on strain limits as explained in section 4 of this lecture, Lecture 2....... Page 14 Lec Design Reinforced Concrete Structures Misan University Lec Third Year Engineering College 2 Asst.Prof.Dr.Abbas Oda Dawood Civil Department 2 Lecture 2: Behavior of Beams in Flexure (Bending) whereby the member capacity is penalized by reducing the φ factor when the strain in the reinforcing steel at ultimate is less than 0.005.  To ensure tension failure or under-reinforced section, ACI Code 9.3.3.1 establish a minimum tensile net strain εt at the nominal member strength of 0.004 for reinforced concrete flexural members without or with an axial load less than 0.10 f'c Ag, where Ag=gross area of the concrete section.  Thus the upper limit or the maximum steel percentage, ρmax , that can be used in a singly reinforced concrete section in bending is based on the net tensile strain in the tension steel εt of 0.004. 7. Nominal Strength of Under-Reinforced Rectangular Section If    max Tension Failure (S.C.F.) f s  f y  Fx  0 C  T As f y 0.85 f c b a  A s f y a  0.85 f c b M  0 a M n  C * (d  ) or 2 a a M n  T * (d  )  A s f y (d  ) 2 2 Lecture 2....... Page 15 Lec Design Reinforced Concrete Structures Misan University Lec Third Year Engineering College 2 Asst.Prof.Dr.Abbas Oda Dawood Civil Department 2 Lecture 2: Behavior of Beams in Flexure (Bending) Example 1: Cont. dt  c 550  84.18 t  u  0.003  0.017 Find nominal strength and failure load of the beam in c 84.18 example 1. t  0.017  0.005 Tension controlled fs  f y Solution a M n  A s f y (d  ) As 1140.4 mm 2 2 71.55  1140.4 * 400 * (550  )  234.57 KN.m  Fx  0 C  T 2 As f y 0.85 f c b a  A s f y a  Thus  Pn  M n 234.57   156.38 KN 0.85 f c b 1.5 1.5 1140.4 * 400 a  71.55 mm 0.85 * 25 * 300  f c  25 MPa  28 MPa 1  0.85 a 71.55 Location of N.A. = c    84.18 m 1 0.85 dt d Summary of Example Section Pmax, KN Location of I, mm4 fc, MPa fs, MPa State N.A., mm Uncracked 42.13 311.3 5.91*109 3.33 21.70 Cracked 64 159 1.88*109 8.12 170 Failure 156.38 84.18 ------- 25 400 P=42.13KN P=64KN P=156.38KN P=0 Uncracked Cracked Cracked Elastic Elastic Inelastic Failure Stages of beam in example 1 from zero load till failure load Lecture 2....... Page 16 Lec Design Reinforced Concrete Structures Misan University Lec Third Year Engineering College 2 Asst.Prof.Dr.Abbas Oda Dawood Civil Department 2 Lecture 2: Behavior of Beams in Flexure (Bending) 8. Analysis of Non- Rectangular Sections " Irregular Sections" Failure Tension Failure Balanced Failure Compression Failure Rectangular Section  < b  = b  > b ( < max) ( > max) Non-Rectangular Section As < Asb As = Asb As > Asb (As > Asmax) (As < Asmax) yo C = 0.85 f'c Ac The analysis steps are summarized in table below. STEP 1 Find Ac  Fx  0 C  T 0.85 fc Ac  As f y STEP 2 Find a Calculate "a"= the depth of Ac from compression fiber STEP 3 Check Tension dt  c d c a  t  0.004,  t   u  0.003 t , c controlled state c c 1 STEP 4 Find yo yo = centroid of Ac, for rectangular section , yo = a/2 STEP 5 Find Mn M n  C * (d  yo ) or Mn  T * (d  yo )  As f y (d  yo ) Lecture 2....... Page 17 Lec Design Reinforced Concrete Structures Misan University Lec Third Year Engineering College 2 Asst.Prof.Dr.Abbas Oda Dawood Civil Department 2 Lecture 2: Behavior of Beams in Flexure (Bending) Example 2: Example 2: Cont. For the beam shown, determine the flexural behavior: f t all  f r  0.62  f ' c  0.62 *1.0 * 25  3.1 MPa 1. Uncracked elastic moment capacity 2. Cracked elastic moment capacity M yt M * 258.74 f t all  3.1  M  71.29 KN.m 3. Cracked Moment I 5.95 *109 4. Nominal moment capacity Hence uncracked elastic moment capacity = 71.29 KN.m f c  25 MPa f y  400 MPa 2- Cracked Elastic Moment Capacity 0.5 kd 740 kd 800 d 319 n As d  kd 200 200 n As  8.5 * 850.59  7230 Solution 1 1 ( * 0.5 kd * kd) * ( * kd)  n As * (d  kd) 1- Uncracked Elastic Moment Capacity 2 3 1 kd 3  7230 * (740  kd) Es 2 105 2 105 12 n    8.5 Ec 4700 f c 4700 25 1 kd 3  7230 kd  5350200  0 12  Solve kd  329.1 mm As  3* ( *19 2 )  850.59 mm2 4 164.55 * 329.13 I cr   nAs * (740  329.1) 2 12  1.71*109 mm4 y  541.26 740 800 M * c M * 329.1 x  270.63 f call  0.45 f c  11.25   I cr 1.71 *109 y t  258.74  M  58.45 KN.m 200 200 For f y  400  fs, all 170 MPa M.c 1 2 f s _ all  n.  ( 800 * 400) * ( * 800)  7.5 * 850.59 * 740 Icr y 2 3  541.26 mm 1 ( 800 * 400)  7.5 * 850.59 M * (740  329.1) 2 170  8.5*  M  83.23 KN.m 1.71*109 541.26 3 * 270.63 258.74 3 * 400 Use smaller value M  58.45 KN.m I N.A  [ ]   12 3  258.74 3 * 64.69    7.5* 850.59 * (740  541.26) 2 2  12   5.95 *10 9 mm 2 Lecture 2....... Page 18 Lec Design Reinforced Concrete Structures Misan University Lec Third Year Engineering College 2 Asst.Prof.Dr.Abbas Oda Dawood Civil Department 2 Lecture 2: Behavior of Beams in Flexure (Bending) Example 2: Cont. 4-Nominal moment capacity 3.Cracked Moment a 2 f r Ig M cr  yt c a 2 y * 800  533.33 mm 3 533.333 * 266.67 266.67 3 * 400 Ig [ ]    f c  25 MPa  28 MPa 12 3 1  0.85  266.67 3 * 66.67    5.88 *10 mm 9 4 2  12  d t  d  740 f r  0.62  f ' c  3.1 MPa  As  3* ( *19 2 )  850.59 mm2 4 y t  266.67 T  C 0.85 f c. Ac  As f y 850.59 * 400 f r Ig 3.1* 5.88 *109  Ac   16011.3 mm2 M cr    68.35 KN.m 0.85 * 25 yt 266.67 1 a  Ac  16011.3  * a * a  253.1mm 2 2 a 253.1 c   297.76 mm 1 0.85 dt  c 740  297.76  t  0.003  0.003 c 297.76  t  0.0045  0.004 Tension controlled 2 2 y o  a  * 253.1 168.71mm 3 3 M n  A s f y (d  y o )  850.59 * 400 * (740  168.71) 194.37 KN.m Lecture 2....... Page 19 Lec Design Reinforced Concrete Structures Misan University Lec Third Year Engineering College 2 Asst.Prof.Dr.Abbas Oda Dawood Civil Department 2 Lecture 2: Behavior of Beams in Flexure (Bending) H.W. Lecture 2....... Page 20