Lecture-03 Design of Reinforced Concrete Members for Flexure and Axial Loads By: Prof. Dr. Qaisar Ali Civil Engineering Department UET Peshawar drqaisarali@uetpeshawar.edu.pk www.drqaisarali.com Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 1 Topics Addressed General Reinforced Concrete Members Subjected to Flexure Load only Reinforced Concrete Members Subjected to Axial Compressive Load only Reinforced Concrete Members Subjected to Axial Compressive Load with Uniaxial Bending Reinforced Concrete Members Subjected to Axial Compressive Load with Biaxial Bending Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 2 1
General While transmitting load from floors and roof to the foundations, frame members (beams and columns) of a RC frame structure are subjected to one or more of the following load effects : Axial Load (compression or tension), Flexure, Shear and Torsion If all of these effects exist together in a RC frame member, Axial and Flexure loads are considered as one set of effects in the design process; whereas Shear and Torsion are considered as another set of load effects. It means that the design for Axial+ Flexure is not affected by Shear + Torsion and vice versa. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 3 When frame members are designed for the effects of Axial and Flexure loads (with or without shear+ torsion), following cases are possible Members subjected to Flexure Load only In this case normal beam design procedures are followed. Members subjected to Axial Load only Pure compression member design procedures are used Members subjected to Combined Axial and Flexure Loads Interaction diagram procedures, considering Axial and Flexure effects together, are used. The Provisions of Chapter 10 shall apply for design of members subjected to flexure or axial loads or to combined flexure and axial loads. General These cases will be discussed one by one in the next slides Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 4 2
Reinforced Concrete Members Subjected to Flexure Load only Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 5 Contents Loading Stages Before Collapse Design of Solid Rectangular Members Design of Solid T Members Design of Hollow Rectangular Members Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 6 3
Loading Stages Before Collapse Beam Test In order to clearly understand the behavior of RC members subjected to flexure load only, the response of such members at three different loading stages is discussed. BEAM TEST VIDEO Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 7 Loading Stages Before Collapse 1. Un-cracked Concrete Elastic Stage: At loads much lower than the ultimate, concrete remains uncracked in compression as well as tension and the behavior of steel and concrete both is elastic. 2. Cracked Concrete (tension zone) Elastic Stage With increase in load, concrete cracks in tension but remains uncracked in compression. Concrete in compression and steel in tension both behave in elastic manner. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 8 4
Loading Stages Before Collapse 3. Cracked Concrete (tension zone) Inelastic (Ultimate Strength) Stage Concrete is cracked in tension. Concrete in compression and steel in tension both enters into inelastic range. At collapse, steel yields and concrete in compression crushes. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 9 Loading Stages Before Collapse Stage-1: Behavior Compression zone f c h d f t = f r M = M cr f c = f t << f c ' Tension Zone b Strain Diagram f t Stress Diagram f c ' Compressive Stress This is a stage where concrete is at the verge of failure in tension Tensile Stress f t = f r = 7.5 f c ' Concrete stress-strain diagram Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 10 5
Loading Stages Before Collapse Stage-1: Calculation of Forces Compression zone h d 1/2 h f c C= 0.5f c (b 0.5h) 2/3 h M 1/2 h T=0.5f t (b 0.5h) b C = T ; f c = f t M = 0.5f c (b 0.5h) (2/3 h) = 1/6 f c b h 2 f c = f t = 6M/(bh 2 ) OR At f t = f r, where modulus of rupture, f r = 7.5 f c f c = f t = Mc/I g where c = 0.5h I g = bh 3 /12 f c = f t = 6M/(bh 2 ) Cracking Moment Capacity, M cr = f r I g /(0.5h) = (f r b h 2 )/6 f t The contribution of steel is ignored for simplification. If there is no reinforcement, member will fail in tension. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 11 Loading Stages Before Collapse Stage-2: Behavior Compression zone ε c < 0.003 f c = 0.45f c ' h d f t > f r M > M cr f c = 0.45f c ' f s =0.5 f y ε s = f s /E s f s = 0.5 f y Tension Zone Concrete Cracked b Strain Diagram Stress Diagram f c ' Compressive Stress 0.5f y f y 0.45f c ' ε c 0.003 E s ε t Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 12 6
Loading Stages Before Collapse Stage-2: Calculation of Forces Compression zone f c h d c C = 0.5f c (bc) l a = d c/3 M T= A s f s b Stress Diagram In terms of moment couple ( M = 0) M = Tl a = A s f s (d c/3) A s = M/f s (d c/3) C = T ( F x = 0) (½)f c bc = A s f s c = 2A s f s / f c b {where f s = nf c and n =E s /E c } c = 2A s n/b Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 13 Loading Stages Before Collapse Stage-3: Behavior Compression zone ε c = 0.003 f c h d f t > >f r M > >M cr f s = f y f c = αfc, where α < 1 ε s = f y /E s T = A s f y b Tension Zone Concrete Cracked Strain Diagram f y Stress Diagram f c ' Compressive Stress E s ε c 0.003 ε t Stress-Strain Diagram for Reinforcing Steel in Tension Stress-Strain Diagram for Concrete in Compression Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 14 7
Loading Stages Before Collapse Stage-3: Calculation of Forces h d ε c = 0.003 f c M a = β 1 c 0.85f c C = 0.85f c ab l a = d a/2 ε s = f y /E s T = A s f y T = A s f y b Stress Diagram Equivalent Stress Diagram In terms of moment couple ( M = 0) M = Tl a = A s f y (d a/2) A s = M/f y (d a/2) C = T ( F x = 0) 0.85f c ab = A s f y a = A s f y / 0.85f c b Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 15 Loading Stages Before Collapse Stage-3: Calculation of Forces According to the strength design method (ACI 10.3.3), the nominal flexural capacity of RC Members shall be calculated from the conditions corresponding to stage 3. ACI code, R10.3.3 The Nominal Flexural Strength (M n ) of a RC member is reached when the strain in the extreme compression fiber reaches the assumed strain limit of 0.003, (i.e. strains at stage 3.) In other words, the member finally fails by crushing of concrete, even if steel in tension has yielded well before crushing of concrete. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 16 8
Loading Stages Before Collapse Stage-3: Calculation of Forces When concrete crushes at ε c =0.003, depending on the amount of steel (A s ) present as tension reinforcement, following conditions are possible for steel strain (ε s ) 1. ε s =ε y Balanced Failure Condition, Brittle Failure 2. ε s <ε y Over reinforced condition, brittle failure 3. ε s >ε y Under Reinforced Condition, Ductile Failure For relative high amount of tension reinforcement, failure may occur under conditions 1 & 2, causing brittle failure. It is for this reason that ACI code restricts maximum amount of reinforcement in member subjected to flexural load only. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 17 Loading Stages Before Collapse Stage-3: Calculation of Forces To ensure ductile failure & hence to restrict the maximum amount of reinforcement, the ACI code recommends that for tension controlled sections (Beams)ε s =ε t = 0.005 Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 18 9
Design of Solid Rectangular Members Singly Reinforced: Flexural Capacity M n = A s f y (d a/2) ΦM n =ΦA s f y (d a/2) [Nominal capacity] [Design capacity] To avoid failure, ΦM n M u For ΦM n = M u; ΦA s f y (d a/2) =M u ; A s = M u / {Φf y (d a/2)} and a = A s f y /0.85f c b h d ε c = 0.003 f c M a = β 1 c 0.85f c C = 0.85f c ab l a = d a/2 ε s = f y /E s T = A s f y T = A s f y b Stress Diagram Equivalent Stress Diagram Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 19 Design of Solid Rectangular Members Singly Reinforced: Maximum reinforcement (A smax ): From equilibrium of internal forces, F x = 0 C = T 0.85f c ab = A s f y (a) From similarity of triangles, in strain diagram at failure condition, c/ε u = (d c)/ε s c = dε u /(ε u +ε s ) substituting a =β 1 c, A s =ρ max b d andε s =ε t, in equation (a) yields; ρ max = 0.85β 1 (f c /f y )ε u / (ε u +ε t ) Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 20 10
Design of Solid Rectangular Members Singly Reinforced: Maximum reinforcement (A smax ): For ductility in Tension Controlled sections (Beams) ε s =ε t = 0.005 (ACI 10.3.5) and at failureε u = 0.003 (ACI R10.3.3), c = dε u /(ε u +ε s ) c = 0.375d and, a =β 1 c =β 1 0.375d Therefore, when a =β 1 0.375d, A s = A smax in equation (a). Hence equation (a) becomes, 0.85f c β 1 0.375db = A smax f y A smax = 0.31875β 1 bd f c /f y (b) 10.2.7.3 Factorβ 1 shall be taken as 0.85 for concrete strengths f c up to and including 4000 psi. For strengths above 4000 psi,β 1 shall be reduced continuously at a rate of 0.05 for each1000 psi of strength in excess of 4000 psi, butβ 1 shall not be taken less than 0.65. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 21 Design of Solid Rectangular Members Singly Reinforced: Maximum reinforcement (A smax ): A smax = 0.31875β 1 bd f c /f y (b) Forβ 1 = 0.85; f c = 3 ksi ; and f y = 40 ksi A smax = 0.0203 bd; which means 2 % of effective area of concrete β 1 = 0.85; f c = 3 ksi ; and f y = 60 ksi A smax = 0. 0135 bd; which means 1.35 % of gross area of concrete Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 22 11
Design of Solid Rectangular Members Singly Reinforced: Maximum flexural capacity ( M nmax ): Table 1: Maximum factored flexural capacity (M n in in-kips) of singly reinforced RC rectangular beam for specified material strength and dimensions f c = 3 ksi b (in) f y = 40 ksi 12 15 18 12 740 (2.32) 925(2.90) 1110(3.47) h (in) Assuming distance from Centre of the main bar to outer tension fiber=2.5 18 1970(3.78) 2462(4.72) 2955(5.67) 20 2511(4.27) 3139(5.33) 3767(6.40) 24 3790(5.24) 4738(6.55) 5685(7.86) 30 6201(6.71) 7751(8.38) 9301(10.06) Note: The values in brackets represents A smax in in 2. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 23 Design of Solid Rectangular Members Singly Reinforced: Flexural capacity at other strains We know that the ductility requirement of ACI code does not allow us to utilize the beam flexural capacity beyond ΦM nmax. The code wants to ensure that steel in tension yield before concrete crushes in compression. However, if we ignore ACI code restriction, let see what happens. We know that c = dε u /(ε u +ε s ) ; a= 0.85c ; A s = 0.85f c ab/ f s ; M n = A s f s (d a/2) ; f s = Eε s f y; Forε u = 0.003 and assuming various values ofε s, we can determine A s and M n Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 24 12
Design of Solid Rectangular Members Singly Reinforced: Flexural capacity at other strains Table 2: Flexural Capacity (Mn) of 12 x 24 inch [d=21.5 ] RC beam at different tensile strain condition ε s (in/in) 0.0005 0.001 0.00137 * 0.0021 0.003 0.004 0.005 ** 0.007 c (in) A s (in 2 ) f s (ksi) M n (in-kips) 18.43 16.13 14.76 12.65 10.75 9.21 8.06 6.46 33.06 14.46 9.66 8.22 6.99 5.99 5.24 4.19 14.5 29 39.73 40 40 40 40 40 6551 6143 5846 5304 4734 4214 3790 3147 *Yield strain for grade 40 steel **ACI limit Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 25 Design of Solid Rectangular Members Singly Reinforced: Flexural nominal capacity at other strains Conclusions At balance condition, Yield strain = 0.00137, M = 5856; we see no substantial increase in capacity beyond this point i.e. with further increase in steel reinforcement, or decrease in strain there is no appreciable increase in flexural capacity. At ACI code limit of strain = 0.005, M = 3790; we see that there is considerable gap between moment capacity at balance and moment capacity at ACI limit. Therefore if ductility is not required, beam capacity can be further increased up to capacity at balanced point. However if ductility is also required, we can increase moment capacity (without changing dimensions) only if we provide reinforcement in compression (doubly reinforced). Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 26 13
Design of Solid Rectangular Members Singly Reinforced: Minimum reinforcement (A smin ): According to ACI 10.5.1, at every section of a flexural member where tensile reinforcement is required by analysis, the area A s provided shall not be less than that given byρ min b w d where,ρ min is equal to 3 (f c )/f y and not less than 200/f y. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 27 Design of Solid Rectangular Members Doubly Reinforced Background We have seen that we can not provide tensile reinforcement in excess of A smax = 0.31875β 1 bd f c /f y, so there is a bar on maximum flexural capacity. We can increase A smax if we increase b, d, f c or decrease f y. If we can t do either of these and provide reinforcement in excess of A smax, concrete in compression may crush before steel in tension yields. However if we provide this excess reinforcement also on compression side so that the compression capacity of concrete also increases, we would be able to increase the flexural capacity of the member. In this case the member is called doubly reinforced. In other words the range of A smax is increased. In such a case A smax = 0.31875β 1 bd f c /f y + compression steel. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 28 14
Design of Solid Rectangular Members Doubly Reinforced Flexural Capacity Consider figure d and e, the flexural capacity of doubly reinforced beam consists of two couples: The forces A s f y and 0.85f c ab provides the couple with lever arm (d a/2). M n1 = A s f y (d a/2).. (c) The forces A s f y and A s f s provide another couple with lever arm (d d ). M n2 = A s f s (d d ).. (d) Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 29 Design of Solid Rectangular Members Doubly Reinforced Flexural Capacity The total nominal capacity of doubly reinforced beam is therefore, M n = M n1 + M n2 = A s f y (d a/2) + A s f s (d d ) Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 30 15
Design of Solid Rectangular Members Doubly Reinforced Flexural Capacity Factored flexural capacity is given as, ΦM n =ΦA s f y (d a/2) +ΦA s f s (d d ).. (e) To avoid failure,φm n M u. ForΦM n = M u, we have from equation (e), M u =ΦA s f y (d a/2) +ΦA s f s (d d ).. (f) Where,ΦA s f y (d a/2) is equal toφm nmax (singly) for A s = A smax Therefore, M u =ΦM nmax (singly) +ΦA s f s (d d ) {M u ΦM nmax (singly) } =ΦA s f s (d d ) A s = {M u ΦM nmax (singly) }/ {Φf s (d d )}.... (g) ; where, f s f y Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 31 Design of Solid Rectangular Members Doubly Reinforced Maximum reinforcement C c + C s = T [ F x = 0 ] 0.85f c ab + A s f s = A st f y For A max ; a = β 1 c = 0.85 0.375d A st will become A stmax C c = Compression force due to concrete in compression region, C s = Compression force in steel in compression region needed to balance the tension force in addition to the tension force provided by A smax (singly). 0.85f c β 1 0.375db + A s f s = A stmax f y A stmax = β 1 0.31875bdf c /f y + A s f s /f y A stmax = A smax (singly) + A s f s /f y Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 32 16
Design of Solid Rectangular Members Doubly Reinforced Maximum reinforcement A stmax = A smax (singly) + A s f s /f y The total steel area actually provided A st as tension reinforcement must be less than A stmax in above equation i.e. A st A stmax A stmax (singly ) is a fixed number, whereas A s is steel area actually placed on compression side. (For more clarification, see example) Note that Compression steel in the above equation may or may not yield when tension steel yields. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 33 Design of Solid Rectangular Members Doubly Reinforced Conditions at which f s = f y when tension steel yields. By similarity of triangle (fig b), compression steel strain (ε s ) is, ε s =ε u (c d )/ c.. (h) For tensile steel strain (ε s ) =ε t = 0.005 (for under reinforced behavior): c = 0.375d Substituting the value of c in eqn. (h), we get, ε s =ε u (0.375d d )/ 0.375d = (0.003 0.008d /d)... (i) Equation (i) gives the value of ε s for the condition at which reinforcement on tension side is at strain of 0.005 ensuring ductility. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 34 17
Design of Solid Rectangular Members Doubly Reinforced Conditions at which f s = f y when tension steel yields. ε s = {0.003 0.008d /d}.... (i) OR d /d = (0.003 -ε s )/0.008. (j) Substitutingε s =ε y,in equation (j). d /d = (0.003 -ε y )/0.008..... (k) Equation (k) gives the value of d /d that ensures that when tension steel is at a strain of 0.005 (ensuring ductility), the compression steel will also be at yield. Therefore for compression to yield, d /d should be less than the value given by equation (k). Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 35 Design of Solid Rectangular Members Doubly Reinforced Conditions at which f s = f y when tension steel yields. Table 3 gives the ratios (d /d) and minimum beam effective depths (d) for compression reinforcement to yield. For grade 40 steel, the minimum depth of beam to ensure that compression steel will also yields at failure is 12.3 inch. Table 3: Minimum beam depths for compression reinforcement to yield f y, psi Maximum d'/d Minimum d for d' = 2.5" (in.) 40000 0.2 12.3 60000 0.12 21.5 75000 0.05 48.8 Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 36 18
Design of Solid Rectangular Members Doubly Reinforced Example Design a doubly reinforced concrete beam for an ultimate flexural demand of 4500 in-kip. The beam sectional dimensions are restricted. Material strengths are f c = 3 ksi and f y = 40 ksi. d = 20 b = 12 Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 37 Design of Solid Rectangular Members Doubly Reinforced Solution: Step No. 01: Calculation ofφm nmax (singly) ρ max (singly) = 0.0203 A smax (singly) =ρ max (singly) bd = 4.87 in 2 ΦM nmax (singly) = 2948.88 in-kip Step No. 02: Moment to be carried by compression steel M u (extra) = M u ΦM nmax (singly) = 4500 2948.88 = 1551.12 in-kip Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 38 19
Design of Solid Rectangular Members Doubly Reinforced Solution: Step No. 03: Findε s and f s From table 2, d = 20 > 12.3, and for d = 2.5, d /d is 0.125 < 0.20 for grade 40 steel. So compression steel will yield. Stress in compression steel f s = f y Alternatively, ε s = (0.003 0.008d /d).. (i) ε s = (0.003 0.008 2.5/20) = 0.002 >ε y = 40/29000 = 0.00137 Asε s is greater thanε y, so the compression steel will yield. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 39 Design of Solid Rectangular Members Doubly Reinforced Solution: Step No. 04: Calculation of A s and A st. A s = M u(extra) /{Φf s (d d )}=1551.12/{0.90 40 (20 2.5)}= 2.46 in 2 Total amount of tension reinforcement (A st ) is, A st = A smax (singly) + A s = 4.87 + 2.46 = 7.33 in 2 Using #8 bar, with bar area A b = 0.79 in 2 No. of bars to be provided on tension side = A st / A b = 7.33/ 0.79 = 9.28 No. of bars to be provided on compression side = A s / A b =2.46/ 0.79 = 3.11 Provide 10 #8 (9.7 in 2 in 3 layers) on tension side and 4 #8 (3.16 in 2 in 1 layer) on compression side. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 40 20
Design of Solid Rectangular Members Doubly Reinforced Solution: Step No. 05: Ensure that d /d < 0.2 (for grade 40) so that selection of bars does not create compressive stresses lower than yield. With tensile reinforcement of 10 #8 bars in 3 layers and compression reinforcement of 4 #8 bars in single layer, d = 19.625 and d = 2.375 d /d = 2.375/ 19.625 = 0.12 < 0.2, OK Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 41 Design of Solid Rectangular Members Doubly Reinforced Solution: Step No. 06: Ductility requirements: A st A stmax A st, which is the total steel area actually provided as tension reinforcement must be less than A stmax. A stmax = A smax (singly) + A s f s /f y A stmax (singly ) is a fixed number for the case under consideration and A s is steel area actually placed on compression side. A smax (singly) = 4.87 in 2 ; A s = 4 0.79 = 3.16 in 2 ; A stmax = 4.87 + 3.16 = 8.036 in 2 A st = 7.9 in 2 Therefore A st = 7.9 in 2 < A stmax OK. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 42 21
Design of Solid T Members Difference between T-beam and T-beam Behavior Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 43 Design of Solid T Members Flexural Capacity Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 44 22
Design of Solid T Members Flexural Capacity A sf =0.85f c (b b w )h f /f y A sf, is the steel area which when stressed to f y, is required to balance the longitudinal compressive force in the overhanging portions of the flange that are stressed uniformly at 0.85f c. ΦM n1 =ΦA sf f y (d h f /2) A s =ΦM n2 /Φ f y (d a/2) = (M u ΦM n1 )/Φ f y (d a/2) a = A s f y / (0.85f c b w ) A s represents the steel area which when stressed to f y, is required to balance the longitudinal compressive force in the rectangular portion of the beam. Total steel area required (A st ) = A sf + A s Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 45 Design of Solid T Members Flexural Capacity (Alternate Formulae) ΦM n = M u = ΦA st f y (d x) A st = M u / {Φf y (d x)} x = {b w a 2 /2 + (b b w )h f2 /2}/ {b w a + (b b w )h f } a = {A st f y 0.85f c (b b w )h f }/0.85f c b w Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 46 23
Design of Solid T Members Ductility Requirements T = C 1 + C 2 [ F x = 0 ] A st f y = 0.85f c ab w + 0.85f c (b b w )h f A st f y = 0.85f c ab w + A sf f y For ductility ε s = ε t = 0.005 (ACI 10.3.5), a = a max = β 1 c = β 1 0.375d, and A st will become A stmax, Therefore, A stmax f y = 0.85f c β 1 0.375db w + A sf f y A stmax f y = 0.85f c β 1 0.375db w + A sf A stmax = 0.31875 β 1 (f c /f y )db w + A sf OR A stmax = A smax (singly) + A sf So, for T-beam to behave in a ductile manner A st, provided A stmax Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 47 Design of Solid T Members Effective Flange width for T and L beam (ACI 8.10) Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 48 24
Design of Solid T Members Example 03 Design a beam to resist a factored moment equal to 6500 in-kip. The beam is 12 wide, with 20 effective depth and supports a 3 slab. The beam is 25 long and its c/c distance to next beam is 4 ft. Material strengths are f c = 3 ksi and f y = 40 ksi. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 49 Design of Solid T Members Example Solution: Span length (l) = 25 d = 20 ; b w = 12 ; h f = 3 Effective flange width (b) is minimum of, l/4 = 25 12/4 = 75 16h f + b w = 16 3 + 12 = 60 c/c distance to next beam = 4 12 = 48 Therefore, b = 48 Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 50 25
Design of Solid T Members Example Solution: Check if the beam behaviour is T or rectangular. Let a = h f = 3 A s = M u /Φf y (d a/2) = 6500/{0.90 40 (20 3/2)} = 9.76 in 2 a = A s f y /(0.85f c b) = 9.76 40/ (0.85 3 48) = 3.20 > h f Therefore, design as T beam. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 51 Design of Solid T Members Example Solution: Design: We first calculate A sf, the steel area which, when stressed to f y, is required to balance the longitudinal compressive force in the overhanging portions of the flange that are stressed uniformly at 0.85f c. A sf = 0.85f c (b b w ) h f /f y = 0.85 3 (48 12) 3/40 = 6.885 in 2 The nominal moment resistance (ФM n1 ), provided by A sf is, ФM n1 =ФA sf f y {d h f /2} = 0.9 6.885 40 {20 3/2} = 4585.41 in-kip Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 52 26
Design of Solid T Members Example Solution: Design: The nominal moment resistance (ФM n2 ), provided by remaining steel A s is, ФM n2 = M u ФM n1 = 6500 4585.41 = 1914.45 in-kip Let a = 0.2d = 0.2 20 = 4 A s =ФM n2 / {Фf y (d a/2)} = 1914.45/ {0.9 40 (20 4/2)}= 2.95 in 2 a = A s f y /(0.85f c b w ) = 2.95 40/(0.85 3 12) = 3.90 This value is close to the assumed value of a. Therefore, A st = A sf + A s = 6.885 + 2.95 = 9.84 in 2 (13 #8 Bars) Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 53 Design of Solid T Members Example Solution: Ductility requirements, (A st = A s + A sf ) A stmax A stmax = A smax (singly) + A sf = 4.87 + 6.885 = 11.76 in 2 A st = A s + A sf = 13 0.79 = 10.27 in 2 < 11.76 O.K. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 54 27
Design of Solid T Members Example Solution: Ensure that A st > A smin A st = 10.27 in 2 A smin =ρ min b w d ρ min = 3 (f c )/f y 200/f y 3 (f c )/f y = 3 (3000)/60000 = 0.004 200/f y = 200/40000 = 0.005 ρ min = 0.005 ; A smin = 0.005 12 20 = 1.2 in 2 < A st, O.K. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 55 Design of Solid T Members Example Solution: Design: We design the same beam by alternate method. Trial 01: Assume a = h f = 3 x = {b w a 2 /2 + (b b w )h f2 /2}/ {b w a + (b b w )h f } = {12 3 2 /2+(48 12) 3 2 /2}/ {12 3+ (48 12) 3} = 1.5 A st = M u / {Φf y (d x)} = 6500/ {0.90 40 (20 1.5) = 9.76 in 2 Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 56 28
Design of Solid T Members Example Solution: Design: Trial 02: a = {A st f y 0.85f c (b b w )h f }/0.85f c b w = {9.76 40 0.85 3 (48 12) 3}/ (0.85 3 12)= 3.76 x = {12 3.76 2 /2+(48 12) 3 2 /2}/ {12 3.76+ (48 12) 3} = 1.61 A st = 6500/ {0.90 40 (20 1.61)} = 9.81 in 2 Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 57 Design of Solid T Members Example Solution: Design: Trial 03: a = {9.81 40 0.85 3 (48 12) 3}/ (0.85 3 12)= 3.83 x = {12 3.83 2 /2+(48 12) 3 2 /2}/ {12 3.83+ (48 12) 3} = 1.62 A st = 6500/ {0.90 40 (20 1.62)} = 9.83 in 2, O.K. This is same as calculated previously for T-beam. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 58 29
Design of Hollow Rectangular Members Flexural Capacity Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 59 Design of Hollow Rectangular Members Flexural Capacity A s1 =0.85f c b o h f /f y A s1 represents the steel area which when stressed to f y, is required to balance the longitudinal compressive force in the rectangular portion of the area b o h f that is stressed uniformly at 0.85f c. ΦM n1 =ΦA s1 f y (d h f /2) A s2 =ΦM n2 /Φ f y (d a/2) = (M u ΦM n1 )/Φ f y (d a/2) a = A s2 f y / {0.85f c (b - b o )} A s2 is the steel area which when stressed to f y, is required to balance the longitudinal compressive force in the remaining portion of the section that is stressed uniformly at 0.85f c. Total steel area required (A st ) = A s1 + A s2 Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 60 30
Design of Hollow Rectangular Members Flexural Capacity (Alternate Formulae) ΦM n = M u = ΦA st f y (d x) A st = M u / {Φf y (d x)} x = {b o h f2 /2 + (b b o )a 2 /2}/ {(b b o )a + b o h f } a = {A st f y 0.85f c b o h f }/0.85f c (b b o ) Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 61 Design of Hollow Rectangular Members Ductility Requirements For summation of internal forces, A st f y = 0.85f c ba 0.85f c b o (a h f ) For ε t = 0.005, a = β 1 0.375d, we have A st = A stmax, hollow, therefore, A stmax, hollow = {0.85f c bβ 1 0.375d 0.85f c b o (0.375d h f )}/ f y A stmax, hollow = 0.319(f c /f y )β 1 bd 0.85(f c /f y )b o (0.375d h f ) A stmax, hollow = A smax (singly) 0.85(f c /f y )b o (0.375d h f ) So, for hollow beam to behave in a ductile manner: A st, provided A stmax, hollow Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 62 31
Design of Hollow Rectangular Members Example Design a beam to resist a factored moment equal to 2500 in-kip. The beam has a hollow section with 12 width and overall depth of 24. The hollow part inside the section is 3 wide and 16 deep. Material strengths are f c = 3 ksi and f y = 60 ksi. 16 24 3 12 Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 63 Design of Hollow Rectangular Members Example Solution h = 24 ; d = 21.5 (assumed) b = 12 b o = h f = 4 Check if the beam behaviour is rectangular or hollow rectangular. Let a = h f = 4 A s = M u /Φf y (d a/2) = 2500/{0.90 60 (21.5 4/2)} = 2.37 in 2 a = A s f y /(0.85f c b) = 2.37 60/ (0.85 3 12) = 4.65 > h f Therefore, design as hollow beam. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 64 32
Design of Hollow Rectangular Members Example Solution First calculate A s1, A s1 = 0.85f c b o h f /f y = 0.85 3 3 4/60 = 0.51 in 2 The nominal moment resistance (ФM n1 ), provided by A s1 is, ФM n1 =ФA s1 f y {d h f /2} = 0.9 0.51 60 {21.5 4/2} = 537.03 in-kip The nominal moment resistance (ФM n2 ), provided by remaining steel A s2 is, ФM n2 = M u ФM n1 = 2500 537.03 = 1962.97 in-kip Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 65 Design of Hollow Rectangular Members Example Solution Let a = 4 A s2 =ФM n2 / {Фf y (d a/2)} = 1962.97/ {0.9 60 (21.5 4/2)} = 1.86 in 2 a = A s2 f y / {0.85f c (b b o )} = 1.86 60/ {0.85 3 (12 3)} = 3.65 This value is close to the assumed value of a. Therefore, A st = A s1 + A s2 = 0.51 + 1.86 = 2.37 in 2 Using #8 bar, with bar area A b = 0.79 in 2 # of bars = A st / A b = 2.37/ 0.79 = 3 bars Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 66 33
Design of Hollow Rectangular Members Example Solution Ductility requirements, (A st ) provided <A stmax, hollow A stmax, hollow = A smax (singly) 0.85(f c /f y )b o (0.375d h f ) = 3.48 0.85 (3/60) 3 (0.375 21.5 4) = 2.96 in 2 Therefore, A st = 2.37 in 2 < 2.96 in 2 O.K. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 67 Design of Hollow Rectangular Members Example Solution Design the same beam by alternate approach. Trial 01: Assume a = h f = 4 x = {b o h f2 /2 + (b b o )a 2 /2}/ {(b b o )a + b o h f } = {3 4 2 /2+(12 3) 4 2 /2}/ {(12 3) 4 + 3 4} = 2 A st = M u / {Φf y (d x)} = 2500/ {0.90 60 (21.5 2)} = 2.37 in 2 Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 68 34
Design of Hollow Rectangular Members Example Solution Design the same beam by alternate approach. Trial 02: a = {A st f y 0.85f c b o h f }/0.85f c (b b o ) = {2.37 60 0.85 3 3 4}/ {0.85 3 (12 3)} = 4.87 x = {3 4 2 /2+(12 3) 4.87 2 /2}/ {(12 3) 4.87 + 3 4} = 2.34 A st = 2500/ {0.90 60 (21.5 2.34)} = 2.41 in 2, O.K. This is close to the value calculated previously for hollow-beam. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 69 Reinforced Concrete Members Subjected to Axial Compressive Loads Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 70 35
Contents Axial Capacity Maximum Reinforcement Ratio Example Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 71 Axial Capacity Consider a Rectangular Section Nominal Axial Capacity is given as C s1 + C s2 +C s3 + C c = P n C s1 = A s1 * f s1 C s2 = A s2 * f s2 C s3 = A s3 * f s3 C c = A c * f c A s1 * f s1 + A s2 * f s2 + A s3 * f s3 + A c * f c = P n Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 72 36
Axial Capacity The section will reach its axial capacity when strain in concrete reaches a value of 0.003. The yield strain values of steel for grade 40 and 60 are 0.00138 and 0.00207 respectively. Therefore steel would have already yielded at 0.003 strain. Hence f s1 = f s2 = f s3 = f s4 = f y and f c = 0.85 f c Let A s1 + A s2 + A s3 = A st and A c = A g A st, Then A st f y + 0.85 f c (A g A st ) = P n where A g = gross area of column section, A st = total steel area ΦP n = P u Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 73 Axial Capacity As per ACI code (10.3.6 and 10.3.7 ), the axial capacity for Spiral Columns ΦP n (max) = 0.85Φ [0.85f c (A g A st ) + f y A st ] ; Φ = 0.70 Tied Columns ΦP n (max) = 0.80Φ [0.85f c (A g A st ) + f y A st ] ;Φ=0.65 The ACI factors are lower for columns than for beams, reflecting their greater importance in a structure. The additional reduction factors of 0.80 and 0.85 are used to account for accidental eccentricities not considered in the analysis that may exist in a compression member, and to recognize that concrete strength may be less than f c under sustained high loads. R10.3.6 and R10.3.7 Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 74 37
Maximum Reinforcement Ratio 1 % A st /A g 8 % Practically, however reinforcement more than 6 % is seldom used. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 75 Example Design a 18 18 column for a factored axial compressive load of 300 kips. The material strengths are fc = 3 ksi and f y = 40 ksi. 18 18 Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 76 38
Example Solution Nominal strength (ΦP n ) of axially loaded column is: ΦP n = 0.80Φ{0.85f c (A g A st ) + A st f y } Let A st = 1% of A g ΦP n = 0.80 0.65 {0.85 3 (324 0.01 324) + 0.01 324 40} = 492 kips > (P u = 300 kips), O.K. Therefore, A st = 0.01 324 = 3.24 in 2 Using 3/4 Φ(#6) {# 19, 19 mm}, with bar area A b =0.44 in 2 No. of bars = A s /A b = 3.24/0.44 = 7.36 8 bars Use 8 #6 bars {8 #19 bars, 19 mm} Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 77 Reinforced Concrete Members subjected to Axial Compressive Load with Uniaxial Bending Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 78 39
Contents Behavior of Columns subjected to Uniaxial Bending Axial Capacity Flexural Capacity Design by Trial and Success Method Alternative Approach Interaction Diagram Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 79 Behavior of Columns subjected to Uniaxial Bending Shown in figure, is a vertical rectangular RC member subjected to axial compressive load P u at some eccentricity e x along x-axis of the cross section causing moment M uy. Such a column is called uniaxial column. The bending is called uniaxial bending because the bending exists only about one of the centroidal axis of the cross section. y x Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 80 40
Axial Capacity P u =ΦP n =Φ(C c + C s T) [ F = 0 ] =Φ(0.85f c ab + A s1 f s1 A s2 f s2 ) P u =Φ{0.85f c ab+a s1 f s1 A s2 f s2 }..(1) f s1 = Eε s1 = 0.003E (c d )/c f y f s2 = Eε s2 = 0.003E (d c)/c f y For ε s1 : ε s1 /(c - d ) = ε u /c For ε s2 : ε s2 /(d - c) = ε u /c Note: Negative sign with A s2 shows that steel layer A s2 is under tensile stresses. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 81 Flexural Capacity M u = ΦM n [ M = 0 ] (about geometric center), M u = Φ [C c {(h/2) (a/2)} + A s1 f s1 {(h/2) d } + A s2 f s2 {d (h/2)}] With (d h/2) = {h d h/2} = {(h/2) d } M u = Φ [C c {(h/2) (a/2)} + A s1 f s1 {(h/2) d } + A s2 f s2 {(h/2) d }] (2a) Note: All internal forces are in counter clockwise sense to resist flexural demand caused by P u. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 82 41
Flexural Capacity M u = Φ [C c {(h/2) (a/2)} + A s1 f s1 {(h/2) d } + A s2 f s2 {(h/2) d }] (2a) With, C c = 0.85f c ab ; A s1 = A s2 = A s The equation (2a) becomes (2b) as: M u =Φ[0.425f c ab(h a)+a s {(h/2) d }(f s1 +f s2 )]. (2b) Where, f s1 = Eε s1 = 0.003E (c d )/c f y f s2 = Eε s2 = 0.003E (d c)/c f y & Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 83 Flexural Capacity It is important to note that equation (1) & (2b) are valid for 2 layers of reinforcements only. P u =Φ{0.85f c ab+a s1 f s1 A s2 f s2 } (1) M u =Φ[0.425f c ab(h a)+a s {(h/2) d }(f s1 +f s2 )] (2b) For intermediate layers of reinforcement, the corresponding terms with A s shall be added in the equations. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 84 42
Design by Trial and Success Method As discussed in previous lectures, the singly reinforced flexural member can be designed by trial and success method using following formulae: A s = M u / {Φf y (d a/2)} & a = A s f y /0.85f c b In the same way, equations (1) and (2b) may be used for design of RC member subjected to compressive load with uniaxial bending P u =Φ{0.85f c ab+a s1 f s1 A s2 f s2 } (1) M u =Φ[0.425f c ab(h a) + A s {(h/2) - d }(f s1 + f s2 )].(2b) Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 85 Design by Trial and Success Method However unlike equations for beam where f s = f y, here we don t know values of f s1 and f s2. But we do know that steel stress shall be taken equal to or less than yield strength. Therefore f s1 = Eε s1 = 0.003E (c d )/c f y f s2 = Eε s2 = 0.003E (d c)/c f y Equation (1) can be now written in the following form P u =Φ{0.85f c β 1 cb + A s E 0.003(c d )/c A s E 0.003(d c)/c)}---(1) Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 86 43
Design by Trial and Success Method Equation (1) can be transformed into a quadratic equation to obtain the value of c for a particular demand P u and assumed A s : Φ0.85f c β 1 bc 2 + (Φ174A s P u )c Φ87A s (d d ) = 0 However such approach will not be convenient because the check that stresses in reinforcement layers f s1 and f s2 shall not exceed f y can not be applied in the above equation. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 87 Design by Trial and Success Method As an example, with M u = 40 ft-kip, P u = 145 kips, A s = 0.88 in 2, f c = 3 ksi, b = h = 12, d = 9.5 and d = 2.5, c comes out to be 6.08 from quadratic equation. For c = 6.08, now f s1 and f s2 shall be f y f s1 = Eε s1 = 0.003E (c d )/c = 51 ksi ; greater than 40 ksi f s2 = Eε s2 = 0.003E (d c)/c = 49 ksi ; greater than 40 ksi It means that every time when we obtain value of c, we have to check stresses in steel and only that value of c will be used when f s1 and f s2 are f y. Therefore this method of trial and success will not work in members subjected to axial load and flexure together. We now look at another approach. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 88 44
Alternative Approach Instead of calculating c, we assume c and calculate ФP n and ФM n for a given set of data such as follows: ФP n =Φ{0.85f c ab+ A s E 0.003(c d )/c A s E 0.003(d c)/c)} ФM n =Φ[0.425f c β 1 c b (h a) + A s {(h/2) d } (f s1 + f s2 )] For A s = 0.88 in 2, f c = 3 ksi, b = h = 12, d = 9.5 and d = 2.5, all values in the above equations are known except c. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 89 Alternative Approach ФP n andфm n are calculated for various values of c from 0 to h, with the check that during calculations f s1 and f s2 do not exceed f y for both eqns. c (in) 0 c (h = 12) Table 4 ФP n (kips) ФM n (kip-ft) 3.69 0 36.25 5 64.6 41.59 7 133 43.09 9 185.3 36 12 252.64 19.44 Axial capacity 281 0 Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 90 45
Alternative Approach Plot the values and check the capacity of the column for the demand equal to M u = 40 ft-kip and P u = 145 kips Demand point (40,145) Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 91 Interaction Diagram General: For a column of known dimensions and reinforcement, several pairs of P and M from various values of c using equations 1 and 2b can be obtained and plotted as shown. Such a graph is known as capacity curve or interaction diagram. Nominal and Design diagram are given in the figure. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 92 46
Interaction Diagram General: If the factored demand in the form of P u and M u lies inside the design interaction diagram, the given column will be safe against that demand. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 93 Interaction Diagram Important Features of Interaction Diagram Horizontal Cutoff: The horizontal cutoff at upper end of the curve at a value ofαφp nmax represents the maximum design load specified in the ACI 10.3.5 for small eccentricities i.e., large axial loads. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 94 47
Interaction Diagram Important Features of Interaction Diagram: Linear Transition of Φ from 0.65 to 0.90 is applicable forε t f y /E s toε t = 0.005 respectively. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 95 Interaction Diagram Development of Interaction Diagram: Interaction diagram can be developed by calculation of certain points as discussed below: Point 01: Point representing capacity of column when concentrically loaded. This represents the point for which M n = 0. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 96 48
Interaction Diagram Development of Interaction Diagram: Point 02: c = h Point 2 corresponds to crushing of the concrete at the compression face of the section and zero stress at the other face. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 97 Interaction Diagram Development of Interaction Diagram: Point 03: c = (h-d ) At Point 3, the strain in the reinforcing bars farthest from the compression face is equal to zero. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 98 49
Interaction Diagram Development of Interaction Diagram: Point 04: c = 0.68d (Grade 40) c = 0.58d (Grade 60) Point representing capacity of column for balance failure condition (ε c = 0.003 andε t =ε y ). c = d {ε c / (ε c + ε y )} ε c = 0.003 ε y = 0.0013 (Grade 40) ε y = 0.0021 (Grade 60) Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 99 Interaction Diagram Development of Interaction Diagram: Point 05: c = 0.375d Point in tension controlled region for net tensile strain (ε t ) = 0.005, and Φ = 0.90, (ε c = 0.003). c = d {ε c / (ε c + ε t )} ε c = 0.003 ε t = 0.005 Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 100 50
Interaction Diagram Development of Interaction Diagram: Point 06: c = 0.23d Point on capacity curve for which ε t >> 0.005 and ε c = 0.003. c = d {ε c / (ε c + ε t )} ε c = 0.003 ε t >> 0.005 ε t >> 0.005 Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 101 Interaction Diagram Example: Develop interaction diagram for the given column. The material strengths are f c = 3 ksi and f y = 40 ksi with 4 no. 6 bars. 12 Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 102 12 51
Interaction Diagram Solution: Design interaction diagram will be developed by plotting (06) points as discussed earlier. Point 1: Point representing capacity of column when concentrically loaded: Therefore ΦP n =Φ[0.85f c (A g A st ) + f y A st ] = 0.65 [0.85 3 (144 1.76) + 40 1.76] = 281.52 kip ΦM n = 0 Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 103 Interaction Diagram Solution: Point 2: c = h c = 12 (c = h); a =β 1 c = 0.85 12 = 10.2 f s1 = 0.003E (c d )/c = 0.003 29000(12 2.25)/12 = 70.69 ksi > f y, use fy = 40 ksi. f s2 = 0.003E (d c)/c = 0.003 29000(9.75 12)/12 = -16.31 ksi< f y Therefore,ΦP n =Φ{0.85f c ab + A s f s1 A s f s2 } = 0.65{0.85 3 10.2 12 +0.88 40+0.88 16.31} = 235.09 kip ΦM n =Φ[0.425f c ab (h a) + A s {(h/2) d } (f s1 + f s2 )] = 0.65[0.425 3 10.2 12 (12 10.2)+0.88 {(12/2) 2.25}(40-16.31)] = 233.41 in-kip = 19.45 ft-kip Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 104 52
Interaction Diagram Solution: Point 3: c = (h-d ) c =12-2.25=9.75; a = β 1 c = 0.85 9.75 = 8.29 f s1 = 0.003E (c d )/c = 0.003 29000(9.75 2.25)/9.75 = 66.92 ksi > f y, use fy = 40 ksi. f s2 = 0.003E (d c)/c = 0.003 29000(9.75 9.75)/9.75 = 0 ksi< f y Therefore, ΦP n = Φ {0.85f c ab + A s f s1 A s f s2 } = 0.65{0.85 3 8.29 12 +0.88 40} = 187.77 kip ΦM n = Φ [0.425f c ab (h a) + A s {(h/2) d } (f s1 + f s2 )] = 0.65[0.425 3 8.29 12 (12 8.29)+0.88 {(12/2) 2.25}(40)] = 391.67 in-kip = 32.64 ft-kip Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 105 Interaction Diagram Solution: Point 4: Point representing balance failure: The neutral axis for the balanced failure condition is easily calculated from c = d {ε u / (ε u + ε y )} with ε u equal to 0.003 and ε y = 40/29000 = 0.001379, c = 0.68d c b = d {ε u / (ε u + ε y )} = 9.75 0.003/ (0.003 + 0.001379) = 0.68d = 6.68 giving a stress-block depth; a b =β 1 c b = 0.85 6.68 = 5.67 Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 106 53
Interaction Diagram Solution: Point 4: Balance failure: For the balanced failure condition, f s = f y. f s1 = 0.003E (c d )/c = 0.003 29000(6.68 2.25)/6.68= 57.69 ksi > f y, f s2 = 0.003E (d c)/c = 0.003 29000(9.75 6.68)/6.68 = 40 ksi = f y Therefore,ΦP b =Φ{0.85f c ab + A s f s1 A s f s2 } = 0.65{0.85 3 5.67 12 +0.88 40 0.88 40} = 112.77 kip ΦM b =Φ[0.425f c ab (h a) + A s {(h/2) d } (f s1 + f s2 )] = 0.65[0.425 3 5.67 12 (12 5.67)+0.88 {(12/2) 2.25}(40 + 40)] = 528.54 in-kip = 44.05 ft-kip Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 107 Interaction Diagram Solution: Point 5: This point is in tension controlled region for whichε t = 0.005,Φ=0.90: For ε t = 0.005; c = d {ε u / (ε u + ε t )} = 9.75 {0.003/ (0.003 + 0.005)} = 0.375d = 3.66 a =β 1 c = 0.85 3.66 = 3.11 f s1 = 0.003E (c d )/c = 0.003 29000(3.66 2.25)/3.66 = 33.51 ksi < f y f s2 = 0.003E (d c)/c = 0.003 29000(9.75 3.66)/3.66 = 144.76 ksi > f y, use f y = 40 ksi. Therefore,ΦP n =Φ{0.85f c ab + A s f s1 A s f s2 } = 0.90{0.85 3 3.11 12 +0.88 33.51 0.88 40}= 80.50 kip ΦM n =Φ[0.425f c ab (h a) + A s {(h/2) d } (f s1 + f s2 )] = 0.90[0.425 3 3.11 12 (12 3.11)+0.88 {(12/2) 2.25}(33.51+40)] = 599 in-kip = 49.91 ft-kip Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 108 54
Interaction Diagram Solution: Point 6: Point on capacity curve for which εt >> 0.005: Letε t = 2 0.005 = 0.01; c = d {ε u / (ε u +ε t )} = 9.75 {0.003/ (0.003 + 0.01)} = 0.23d = 2.25 a =β1c = 0.85 2.25 = 1.91 fs1 = 0.003E (c d )/c = 0.003 29000(2.25 2.25)/2.25 = 0 < fy fs2 = 0.003E (d c)/c = 0.003 29000(9.75 2.25)/2.25 = 290 ksi > fy, use fy = 40 ksi. Therefore, ΦPn = Φ{0.85fc ab + Asfs1 Asfs2 = 0.90{0.85 3 1.91 12 +0.88 0 0.88 40} = 20.90 kip ΦMn =Φ[0.425fc ab (h a) + As {(h/2) d } (fs1 + fs2)] = 0.90[0.425 3 1.91 12 (12 1.91)+0.88 {(12/2) 2.25}(0 +40) = 384.16 in-kip = 32.01 ft-kip Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 109 Interaction Diagram Solution: M vs P P (kip) 500 450 400 350 300 250 200 150 100 50 Design Interaction Curve Nominal Interaction Curve 0.80φP o 0 0 20 40 60 80 M (kip-ft) Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 110 55
Interaction Diagram Use of Design Aids: The uniaxial columns can be designed using design aids e.g, normalized interaction diagrams such as given in graph A5-A16 (Nilson). These diagrams require the calculation of a dimensionless h constantγ. h =γh+2d d γh d γ = (h-2d )/h b X Once γ is calculated, the interaction diagram corresponding to the value ofγis selected & then column can be designed using steps given on the next slides. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures Y Reference: Design of Concrete Structures 13 th Ed. by Nilson, Darwin and Dolan. 111 Interaction Diagram Use of Design Aids: Graph A.5 to A.16 (Nilson) Calculate γ = (h 2 d ) / h, select the relevant interaction diagram. Given P u, e, A g, f y, and f c Calculate K n = P u /(Φf c A g ) Calculate R n = P u e/(φf c A g h) K n From the values of K n & R n, find ρ from the graph as shown. A st =ρa g R n Reference: Design of Concrete Structures 13 th Ed. by Nilson, Darwin and Dolan. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 112 56
Interaction Diagram Example: Using design aids, design a 12 square column to support factored load of 145 kip and a factored moment of 40 kip-ft. The material strengths are f c = 4 ksi and f y = 60 ksi. 12 Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 113 12 Interaction Diagram Solution: Design Aids (using f c = 4 ksi and f y = 60 ksi) With d = 2.5 in,γ=(12 2 2.5)/12 = 0.60. K n = P u /(Φf c A g ) = 145/(0.65 4 144) = 0.40 R n = P u e/( Φf c A g h) = (40 12)/ (0.65 4 144 12) = 0.11 ρ = 0.007 A st = 0.007 144 = 1.0 in2. < 1 % of A g =1.44 Using #6 bar, No. of bars = A st /A b = 1.44/0.44 4bars Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 114 57
Reinforced Concrete Members subjected to Axial Compressive Load with Biaxial Bending Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 115 Contents Behavior of Columns subjected to Biaxial Bending Difficulties in Constructing Biaxial Interaction Surface Approximate Method for Converting Biaxial case to Uniaxial case Bresler s Approximate Methods for Design of Biaxial Columns Reciprocal Load Method Load Contour Method Circular Columns Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 116 58
Behavior of Columns subjected to Biaxial Bending Column section subjected to compressive load (P u ) at eccentricities e x and e y along x and y axes causing moments M uy and M ux respectively. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 117 Behavior of Columns subjected to Biaxial Bending The biaxial bending resistance of an axially loaded column can be represented as a surface formed by a series of uniaxial interaction curves drawn radially from the P axis. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 118 59
Behavior of Columns subjected to Biaxial Bending (a) uniaxial bending about y axis. (b) uniaxial bending about x axis. (c) biaxial bending about diagonal axis. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 119 Behavior of Columns subjected to Biaxial Bending Force, strain and stress distribution diagrams for biaxial bending Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 120 60
Difficulties in Constructing Biaxial Interaction Surface The triangular or trapezoidal compression zone. Neutral axis, not in general, perpendicular to the resultant eccentricity. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 121 Approximate Method for Converting Biaxial Case to Uniaxial Case For rectangular sections with reinforcement equally distributed on all faces. Biaxial demand can be converted to equivalent uniaxial demand using following equations: (reference PCA) M nxo = M nx + M ny (h/b)(1 β)/β M nyo = M ny + M nx (b/h)(1 β)/β Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 122 61
Approximate Method for Converting Biaxial Case to Uniaxial Case 0.55 β 0.7 A value of 0.65 forβis generally a good initial choice in a biaxial bending analysis. For a value of β = 0.65, the equations can be simplified as below: M nxo = M nx + 0.54M ny (h/b) M nyo = M ny + 0.54 M nx (b/h) Pick the larger moment for onward calculations Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 123 Approximate Method for Converting Biaxial Case to Uniaxial Case Design Example Using equations for converting bi-axial column to uni-axial column, design a 12 square column to support factored load of 190 kip and factored moments of 35 kip-ft about x axis and 50 kip ft about y axis. The material strengths are f c = 4 ksi and f y = 60 ksi. Y b=12 X h =12 Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 124 62
Approximate Method for Converting Biaxial Case to Uniaxial Case Design Example Solution: Assuming compression controlled behavior (Φ = 0.65), the required nominal strengths are: M nx = M ux /Φ=35/ 0.65 = 53.84 ft-kip M ny = M uy /Φ = 50/ 0.65 = 76.92 ft-kip M nxo = M nx + 0.54M ny (h/b) Similarly, = 53.84 + 0.54 76.92 1 = 95 ft-kip M nyo = M ny + 0.54 M nx (h/b)=76.92+0.54 53.84 1 = 105.9 ft-kip = 0.65 105.9 = 68.84 ft kip. The biaxial column can now be designed as an equivalent uni-axial column with moment about y-axis. M uy Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 125 Approximate Method for Converting Biaxial Case to Uniaxial Case Design Example Solution: Note: In the original equations developed by PCA, they have used nominal values of moments because the resultant Moment was supposed to be used on the nominal interaction diagram. However if we have factored interaction diagram, the equation can be directly applied on factored moments without any difference in the final output, as follows: M ux = 35, M uy = 50 ; M u = M ux + 0.54M uy (h/b) = 35 + 0.54 50 = 62 ft-kip M u = M uy + 0.54M ux (h/b) = 50 + 0.54 35 = 68.9 ft-kip Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 126 63
Approximate Method for Converting Biaxial Case to Uniaxial Case Design Example Solution: P u = 190 kip and M u = 68.84 fi-kip With 2.25 in. d,γ=(12 2 2.25)/12 = 0.63 0.60. K n = P u /(Φf c A g ) = 190/(0.65 4 144) = 0.51 R n = P u e/(φf c A g h) = 68.84 12/ (0.65 4 144 12) = 0.18 From the graph, with the calculated values of Kn and R n,ρ g = 0.031. Thus, A st = 0.031 144 = 4.46 in2. Using #6 bar, # of bars = A st /A b = 4.46/ 0.44 = 10.33 12 bars Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 127 Approximate Method for Converting Biaxial Case to Uniaxial Case Design Example Solution: Alternatively, we can design the Column from the uniaxial interaction diagram developed for 12 x 12 inch column having 12 no. 6 bars, fc = 4 ksi and fy = 60 ksi. The red dot shows that column is safe for the given values of Pu = 190 kips and Mu = 68.9 Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 128 64
Bresler s Approximate Methods for Design of Biaxial Columns Reciprocal Load Method For P n 0.1f c A g Where P n = P u / Ф Load Contour Method For P n < 0.1f c A g Where P n = P u / Ф Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 129 Reciprocal Load Method Bresler's reciprocal load equation derives from the geometry of the approximating plane. It can be shown that: {(1/P n ) = 1/ (P nxo ) +1/ (P nyo ) (1/P no ) IfФP n P u O.K. Where, = approximate value of nominal load in biaxial bending with eccentricities e x and e y. P n P nyo = nominal load when only eccentricity e x is present (e y = 0), P nxo = nominal load when only eccentricity e y is present (e x = 0), P no = nominal load for concentrically loaded column Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 130 65
Reciprocal Load Method Steps Step 1: M nx = M ux /Ф M ny = M uy /Ф Check if P n 0.1 f c A g Reciprocal Load Method applies Step 2: γ= (h 2d )/h Assuming A s,ρ=a s / bh P no can be determined P no ρ Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 131 Reciprocal Load Method Steps Step 3: e x /h = (M ny /P n )/ h e x /h P nxo can be determined Step 4: e y /b = (M nx /P n )/ b P nyo can be determined P no ρ e y /b Step 5: Using the equation; P nxo {(1/P n ) = 1/ (P nxo ) +1/ (P nyo ) (1/P no ) P nyo IfФP n P u O.K. Note: All values determined from graph shall be multiplied with f c A g Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 132 66
Reciprocal Load Method Design Example Using Reciprocal Load Method, design a 12 square column to support factored load of 190 kip and factored moments of 35 kip-ft each about x and y axis respectively. The material strengths are f c = 4 ksi and f y = 60 ksi. Y b=12 X h =12 Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 133 Reciprocal Load Method Design Example Solution: Design using approximate methods (Reciprocal Load Method): Given demand: M nx = M ux /Ф = 35/0.65 = 53.84 ft-kip M ny = M uy /Ф = 35/0.65 = 53.84 ft-kip; P u = 190 kips Check if P n 0.1 f c A g P n = 190/ 0.65 = 292.31 kip 0.1f c A g = 0.1 4 12 12 = 57.6 kip As P n > 0.1 f c A g, therefore reciprocal load method applies. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 134 67
Reciprocal Load Method Design Example Solution: With d =2.5 in., γ= (12 2 2.5)/12 = 0.60; Graph A.5 of Nilson 13th Ed applies P no Assuming the column to be reinforced with 4 #6 bars, therefore, ρ =0.012 ρ = A s / bh = 4 0.44/ (12 12) = 0.012 P no /f c A g = 1.09 P no = 1.09 f c A g P no = 1.09 4 144= 628 kips Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 135 Reciprocal Load Method Design Example Solution: Consider bending about Y-axis e x /h = 0.18 K n = 0.68 P nyo /f c A g = 0.68 P nyo = 0.68 f c A g P nyo = 0.68 4 144= 391 kips P no P nyo ρ e x /h Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 136 68
Reciprocal Load Method Design Example Solution: Consider bending about X-axis e y /b = 0.18 K n = 0.68 P nxo /f c A g = 0.68 P nxo = 0.68 f c A g P nxo = 0.68 4 144= 391 kips P no P nxo ρ e y /b Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 137 Reciprocal Load Method Design Example Solution: Design using approximate methods (Reciprocal Load Method): Now apply reciprocal load equation, (1/P n ) = 1/ (P nxo ) +1/ (P nyo ) 1/ ( P no ) (1/P n ) = 1/ (391) +1/ (391) 1/ (628) = 0.00372 P n = 284 kip, and the design load is: ΦP n = 0.65 284 = 184 kips 190 kips, O.K. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 138 69
Reciprocal Load Method Design Example Solution: Instead of using Nelson charts, the Interaction diagram developed earlier for 12 x 12 inch column with 4 no 6 bars is used in the next steps of Reciprocal Load Method.. P n = 190/ 0.65 = 292.31 kip M nx = M ny = 53.84 ft-kip Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 139 Reciprocal Load Method Solution: Design using Approximate methods: This interaction curve is for both x and y axes as the column is square Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 140 70
Reciprocal Load Method Design Example Solution: Design using Reciprocal Load Method : From nominal interaction curve, P no = 590 kip, For M nx = 53.84 ft-kip, P nxo = 450 kip For M ny = 53.84 ft-kip, P nyo = 450 kip Now apply reciprocal load equation, (1/P n ) = 1/ (P nxo )+1/ (P nyo ) 1/ ( P no )= 1/ (450) +1/ (450) 1/ (590) = 0.00285 P n = 344.50 kip, and the design load is: ΦP n = 0.65 344.50 = 223.92 kips > 190 kips, O.K. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 141 Reciprocal Load Method Design Example Solution: Designing the same column by converting bi-axial case to uniaxial case. M u = 35 + 0.54 *20 = 53.9kip-ft P u = 190 kip Interaction diagram for 12 x 12 inch column with 4 no 6 bars is given in the figure. The blue dot shows that column is safe under the given demand. P (kip) 650 600 550 500 450 400 350 300 250 200 150 100 50 0 Design Interaction Curve Nominal Interaction Curve 0.80φP o 0 20 40 60 80 100 M (kip-ft) Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 142 71
Load Contour Method The load contour method is based on representing the failure surface of 3D interaction diagram by a family of curves corresponding to constant values of P n. (M nx /M nxo ) α1 + (M ny /M nyo ) α2 1 Where, M nx = P n e y ; M nxo = M n x (when M ny = 0), M ny = P n e x ; M nyo = M ny (when M nx = 0) Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 143 Load Contour Method Whenα1 =α2 =α, the shapes of such interaction contours are as shown for specificαvalues. For values of M nx /M nx and M ny /M ny, α can be determined from the given graph. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 144 72
Load Contour Method Calculations reported by Bresler indicate that α falls in the range from 1.15 to 1.55 for square and rectangular columns. Values near the lower end of that range are the more conservative. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 145 Load Contour Method Steps: Step 1: M nx = M ux /Ф M ny = M uy /Ф Check if P n < 0.1 f c A g Load contour method applies Step 2: γ = (h 2d )/h Assuming A s,ρ=a s / bh ρ Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 146 73
Load Contour Method Steps: Step 3: e x /h = (M ny /P n )/ h M nyo can be determined Step 4: e y /b = (M nx /P n )/ b e x /h ρ e y /b M nxo can be determined Step 5: (M nx /M nxo ) α1 + (M ny /M nyo ) α2 1 Note: All values determined from graph should be multiplied with f c A g h M nyo M nxo Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 147 Load Contour Method Design Example Using Load Contour Method, design a 12 square column to support factored load of 30 kip and factored moments of 20 kip-ft each about x axis and 30 kip-ft about y axis. The material strengths are f c = 4 ksi and f y = 60 ksi. Y b=12 X h =12 Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 148 74
Load Contour Method Design Example Solution: Design using Load Contour Method: Given demand: M nx = M ux /Ф = 20/0.65 = 30.76 ft-kip M ny = M uy /Ф = 30/0.65 = 46.15 ft-kip; P n =P u /Ф = 30/ 0.65 = 46.15 kips Check if P n < 0.1 f c A g 0.1f c A g = 0.1 4 12 12 = 57.6 kip As P n < 0.1 f c A g, therefore load contour method applies. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 149 Load Contour Method Design Example Solution: With d =2.5 in.,γ=(12 2 2.5)/12 = 0.60 (graph A.5 of Nilson 13 th Ed applies) ρ Assuming the column to be reinforced with 4 #6 bars, then, ρ = A s / bh = 4 0.44/ (12 12) = 0.012 Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 150 75
Load Contour Method Design Example Solution: Consider bending about Y- axis e x /h = 1 R n = 0.12 M nyo /f c A g h= 0.12 M nyo = 0.12 f c A g h M nyo = 0.12 4 144 12 = 830 in-kip ρ e x /h M nyo Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 151 Load Contour Method Design Example Solution: Consider bending about X- axis e y /b = 0.65 R n = 0.14 M nxo /f c A g h= 0.14 M nxo = 0.14 fc Agh M nxo = 0.14 4 144 12 = 968 in-kip ρ e y /b M nxo Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 152 76
Load Contour Method Design Example Solution: Design using Load Contour Method: Now apply load contour equation, (M nx /M nxo ) α1 + (M ny /M nyo ) α1 = 1 Forα 1.15 (30.76 12/968) 1.15 +(46.15 12/830) 1.15 = 0.95 < 1, OK Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 153 Load Contour Method Design Example Solution: Design by converting bi-axial case to uni-axial case. M u = 30 + 0.54 *20 = 40.8 kipft P u = 30 kip Interaction diagram for 12 x 12 inch column with 4 no 6 bars is given in the figure. The blue dot shows that column is safe under the given demand. P (kip) 650 600 550 500 450 400 350 300 250 200 150 100 50 0 Design Interaction Curve Nominal Interaction Curve 0.80φP o 0 20 40 60 80 100 M (kip-ft) Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 154 77
Circular Columns Behavior Strain distribution at ultimate load. The concrete compression zone subject to the equivalent rectangular stress distribution has the shape of a segment of a circle, shown shaded. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 155 Circular Columns Design Example Design a circular column, using approximate methods, for a factored load of 60 kips and a factored moment of 20 ft-kips about x axis and 30 kip-ft about y axis. The diameter of column is 16. Material strengths are f c = 4000 psi and f y = 60000 psi. 16 diameter Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 156 78
Circular Columns Design Example Solution: Check that which method applies? P n = P u /Ф = 60/0.65 = 90.30 kips M nx = M ux /Ф = 20/0.65 = 30.76 ft-kips M ny = M uy /Ф = 30/0.65 = 46.15 ft-kips Check if P n 0.1 f c A g 0.1f c A g = 0.1 4 π 162/4= 80.42 kip; 92.30 kip > 80.42 kip Therefore, reciprocal load method applies. Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 157 Circular Columns Design Example Solution With d =2.5 in., γ = (16 2 2.5)/16 = 0.70 (graph A.5 of Nilson 13th Ed applies) Take 6 #6 bars, ρ =A s /(A g ) = (6 0.44)/(π 162/4) = 0.013 P no /f c A g = 1.04 P no ρ P no = 1.04 f c A g P no = 1.04 4 201= 836 kips Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 158 79
Circular Columns Design Example Solution Consider bending about Y- axis e x /d = 0.75 K n = 0.15 P nyo /f c A g = 0.15 P no ρ e x /d P nyo = 0.15 fc Ag P nyo = 0.15 4 201= 121 kips P nyo Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 159 Circular Columns Design Example Solution Consider bending about X- axis e y /d = 0.50 K n = 0.25 P nxo /f c A g = 0.25 P no ρ e y /d P nxo = 0.25 fc Ag P nxo = 0.25 4 201= 201 kips P nxo Prof. Dr. Qaisar Ali CE 5115 Advance Design of Reinforced Concrete Structures 160 80