Flexural Analysis and Design PDF

Flexural Analysis CEPRCD30 (Principles of Reinforced/Prestressed Concrete) Jerome Z. Tadiosa, CE, MSc Assistant Professor 2, Civil Engineering College of Engineering National University – Manila Intended Learning Outcomes Enumerate and desc...

Flexural Analysis CEPRCD30 (Principles of Reinforced/Prestressed Concrete) Jerome Z. Tadiosa, CE, MSc Assistant Professor 2, Civil Engineering College of Engineering National University – Manila Intended Learning Outcomes Enumerate and describe types of RC beam sections. Explain the different failure modes of RC beams in flexure. Determine the nominal and ultimate flexural strength of RC beam sections. Lecture Outline 1. Problems in structural design 2. Flexural analysis a. Concepts and prerequisites b. Singly-reinforced beams c. Doubly-reinforced beams Problems in Structural Design Problems in Structural Design There are two different types of problems in structural design: Analysis/Investigation Problems (1) Given material, configuration and section properties of a member, determine design strength/capacity and/or maximum permissible applied load. (2) Given material, configuration and section properties of a member, and loads applied on it, evaluate or assess whether the member can carry given applied loads. Design Problems Given material and configuration properties of a member, and loads applied on it, determine section properties of the member such that it can carry given applied loads. Flexural Analysis Concepts and Prerequisites Flexural Analysis The strength requirement statement for members subjected to flexure under USD is given by: 𝜙𝑀𝑛 ≥ 𝑀𝑢 Φ: strength reduction factor (varies depending on section type) Mn: nominal moment strength Mu: maximum bending moment (depending on the bending direction) Classification of RC Beam Sections Shape Rectangular Irregular: non-rectangular Reinforcement Make-up Singly-reinforced: reinforced on tension side only Doubly-reinforced: reinforced on both tension and compression sides Bending Direction Positive: tension side at the bottom portion of the section Negative: tension side at the top portion of the section Singly-Reinforced Sections Doubly-Reinforced Sections Positive and Negative Sections Assumptions Sections perpendicular to the axis of bending that are plane before bending remain plane after bending. The strain in the reinforcement is equal to the strain in the concrete at the same elevation or level. The stresses in the concrete and reinforcement can be calculated from the strains by utilizing stress-strain relationships for concrete and steel. The modulus of elasticity of steel is 200000 MPa (29000 ksi) and it assumes an idealized elastoplastic material behavior. Assumptions The tensile strength of concrete is neglected in flexural strength calculations. The section is assumed to have reached its nominal flexural strength when the strain in the extreme concrete compression fiber reaches its maximum useable compression strain (εcu = 0.003). The compressive stress block in concrete may be assumed to be rectangular, trapezoidal, parabolic, or any other shape based on comprehensive flexural tests. Fundamental Equations Notation ECF: extreme compression fiber ETF: extreme tension fiber d’: vertical distance from ECF to centroid of compression steel z: vertical distance from ECF to centroid of concrete compression block a: depth of concrete compression block measured from ECF c: vertical distance from ECF to neutral axis of the beam section d: (effective depth) vertical distance from ECF to centroid of tension steel dt: vertical distance from ECF to centroid of farthest tension steel layer h: (total depth) vertical distance between ECF and ETF Fundamental Equations Notation εcu: concrete compression strain at failure (equal to 0.003) εs’: compression steel strain (measured at d’) εs: tension steel strain (measured at d) εt: steel strain at farthest tension steel layer (measured at dt) εy: yield strain of steel fy: yield strength of reinforcement fs’: compression steel stress fs: tension steel stress fc’: 28-day concrete compressive strength Fundamental Equations Notation Cc: force acting on concrete at compression side Cs: force acting on compression steel T: force acting on tension steel Ac: area of concrete compression block As: area of tension steel As’: area of compression steel Es: modulus of elasticity of steel Fundamental Equations Fundamental Equations Force Equilibrium 0.85𝑓′𝑐 𝐴𝑐 + 𝐴′ 𝑠 (𝑓 ′ 𝑠 − 0.85𝑓 ′ 𝑐 ) = 𝑓𝑠 𝐴𝑠 Strain Compatibility 𝜀𝑐𝑢 𝜀′𝑠 𝜀𝑠 𝜀𝑡 = = = 𝑐 𝑐 − 𝑑′ 𝑑 − 𝑐 𝑑𝑡 − 𝑐 Moment Equilibrium (Nominal Moment Strength) 𝑀𝑛 = 0.85𝑓 ′ 𝑐 𝐴𝑐 𝑑 − 𝑧 + 𝐴′ 𝑠 (𝑓 ′ 𝑠 − 0.85𝑓 ′ 𝑐 )(𝑑 − 𝑑 ′ ) Whitney Stress Block (Rectangular) A uniform compressive stress block with magnitude equal to 0.85fc’, applied throughout the concrete portion of the compression part of the beam section from ECF with depth a. The relationship between a and c is defined by the following equation (Art. 422.2.2.4, Section 422, 2015 NSCP Vol. 1): 𝑎 = 𝛽1 𝑐 0.85 , 𝑓′𝑐 ≤ 28 𝑀𝑃𝑎 0.05(𝑓 ′ 𝑐 − 28 𝑀𝑃𝑎) 𝛽1 = 0.85 − , 28 𝑀𝑃𝑎 < 𝑓′𝑐 < 55 𝑀𝑃𝑎 7 0.65 , 𝑓′𝑐 ≥ 55 𝑀𝑃𝑎 Steel Stress Steel stress is calculated using Hooke’s Law. However, since steel is assumed to be elastoplastic, the steel stress must not exceed fy. Tension Steel Stress 𝑓𝑠 = 𝐸𝑠 𝜀𝑠 ≤ 𝑓𝑦 Compression Steel Stress 𝑓′𝑠 = 𝐸𝑠 𝜀′𝑠 ≤ 𝑓𝑦 𝑓𝑦 Yield Strain of Steel 𝜀𝑦 = 𝐸𝑠 Note: Es = 200000 MPa (29000 ksi) Effect of Section Properties on Strength and Ductility Flexural Analysis Singly-Reinforced RC Beam Sections (SRB) Singly-Reinforced RC Beam Sections This is the commonly used reinforcement make-up for most RC beams. In actual practice, reinforcement is still provided in compression portion of the section but is normally neglected in analysis unless the amount of tension reinforcement is too large. Minimum amount of reinforcement must be provided for RC beam sections for the beam to have its flexural strength more than its cracking strength. Minimum Steel Reinforcement Art. 409.6, Sec. 409, Chapter 4, 2015 NSCP Vol. 1 As,min should be the greater of the following: 0.25 𝑓 ′ 𝑐 𝑏𝑤 𝑑 𝑓𝑦 𝐴𝑠,𝑚𝑖𝑛 = max 1.4𝑏𝑤 𝑑 𝑓𝑦 bw = b for rectangular beams; bw is the web width for T-beams with flange in compression; bw = min {bf, 2bw} for statically-determinate T-beams with flange in tension Steel Ratio It is defined as the ratio of the reinforcement area to the effective area of an RC beam section. Tension Steel Ratio (ρ) 𝐴𝑠 𝜌= 𝑏𝑑 Compression Steel Ratio (ρ’) 𝐴′𝑠 𝜌′ = 𝑏𝑑 Minimum Steel Ratio Using the definition of steel ratio, the minimum steel ratio based on the code may be expressed as follows: 0.25 𝑓 ′ 𝑐 𝑓𝑦 𝜌𝑚𝑖𝑛 = max 1.4 𝑓𝑦 Simplified Fundamental Equations for SRB Force Equilibrium (As’ = 0) 0.85𝑓′𝑐 𝐴𝑐 = 𝑓𝑠 𝐴𝑠 Strain Compatibility 𝜀𝑐𝑢 𝜀𝑠 𝜀𝑡 = = 𝑐 𝑑 − 𝑐 𝑑𝑡 − 𝑐 Moment Equilibrium (Nominal Moment Strength) 𝑀𝑛 = 0.85𝑓 ′ 𝑐 𝐴𝑐 𝑑 − 𝑧 = 𝑓𝑠 𝐴𝑠 𝑑 − 𝑧 Simplified Fundamental Equations for SRB (Rectangular Sections) Force Equilibrium (b = width of beam section) 0.85𝑓′𝑐 𝑎𝑏 = 𝑓𝑠 𝐴𝑠 Strain Compatibility 𝜀𝑐𝑢 𝜀𝑠 𝜀𝑡 = = 𝑐 𝑑 − 𝑐 𝑑𝑡 − 𝑐 Moment Equilibrium (Nominal Moment Strength) ′ 𝑎 𝑎 𝑀𝑛 = 0.85𝑓 𝑐 𝑎𝑏 𝑑 − = 𝑓𝑠 𝐴𝑠 𝑑 − 2 2 Determining Mn of Rectangular SRB Sections 1. Assume tension steel yields (TSY: fs = fy). 2. Check if the given As ≥ As, min. If it is, proceed to Step 3. Else, note that the beam may fail before it reaches its cracking strength. 3. Using force equilibrium equation, determine ‘a’ by setting fs = fy. 4. Using relationship between ‘a’ and ‘c’, determine β1 based on given f’c, as well as ‘c’. 5. Using strain compatibility equation, check if tension steel yields by comparing εs and εy (TSY: εs ≥ εy). a. If TSY, proceed to step 6. Determining Mn of Rectangular SRB Sections b. If tension steel does not yield (TSDNY), repeat step 3 but setting fs = Es*εs. Use strain compatibility equation to express εs in terms of ‘c’, and substitute a = β1c. Using the resulting equation, solve for ‘c’ and verify once again if TSDNY. 6. Solve for Mn by substituting appropriate fs value. a. If TSY, use fs = fy. b. If TSDNY, use actual fs value from step 5b. Sample Problem 1 Determine the nominal moment strength of the given section. The material properties are as follows: f’c = 28 MPa, fy = 420 MPa. Use SI units for calculations (1 in = 25 mm). Sample Problem 2 From the previous problem, if the reinforcement make-up of the beam section is changed to 8-#14 bars (with 4 bars for each layer), how much will be the change in its nominal moment strength? Did the tension reinforcement of the new section yield? Use the same material properties as the previous problem. Determining Mn of Irregular SRB Sections 1. Assume tension steel yields (TSY: fs = fy). 2. Check if the given As ≥ As, min. If it is, proceed to Step 3. Else, note that the beam may fail before it reaches its cracking strength. 3. Using force equilibrium equation, determine ‘a’ by setting fs = fy. a. Depending on the shape, cases may be set by defining portions where the expression for finding Ac may change with different values or ranges of ‘a’. In this case, assume a certain case and then verify if it is the right case. b. Also, for more defined shapes like triangles and trapezoids, a more general expression for finding Ac may also be written. 4. Using relationship between ‘a’ and ‘c’, determine β1 based on given f’c, as well as ‘c’. Determining Mn of Irregular SRB Sections 5. Using strain compatibility equation, check if tension steel yields by comparing εs and εy (TSY: εs ≥ εy). a. If TSY, proceed to step 6. b. If tension steel does not yield (TSDNY), repeat step 3 but setting fs = Es*εs. Either 2a or 2b may be repeated depending on the shape of the section. Use strain compatibility equation to express εs in terms of ‘c’, and substitute a = β1c. Using the resulting equation, solve for ‘c’ and verify once again if TSDNY. 6. Solve for Mn by substituting appropriate fs value. a. If TSY, use fs = fy. b. If TSDNY, use actual fs value from step 5b. Sample Problem 3 Determine the nominal moment strength of the given T-beam section. Use the following material properties: f’c = 28 MPa; fy = 420 MPa. The tension reinforcement size is 32 mm. Flanged Sections Flanged sections are special cases of irregular sections that regularly occur in structural concrete construction. These result from monolithic construction, where beams and girders are poured together with slabs during construction. Since they are poured and hardened together, during flexural action, beams and slabs work together to carry bending moments. There are two types of flanged sections for RC beams: T-Beams: slab sections at both sides of the beam section L-Beams: (spandrel beams) slab section at only one side of the beam section Effective Flange Width for Flanged Sections Art. 406.2, Section 406, 2015 NSCP Vol. 1 Analysis Cases for Flanged Sections Case 1: a ≤ hf The depth of compression block is on or above the bottom fiber of the flange portion of the section. In this case, determining Mn of a T-beam section is only the same as that for rectangular sections, but with b = bf. Case 2: a > hf The depth of compression block is below the bottom fiber of the flange portion of the section. In this case, determining Mn of a T-beam section may be performed using either of the following: (1) treating it as an irregular section; or (2) dividing the T-beam into flange and web components. Determining Mn of Flanged SRB Sections (T- Beams) – Method 1 1. Assume tension steel yields (TSY: fs = fy). Determine effective flange width of the section if necessary. 2. Check if the given As ≥ As, min. If it is, proceed to Step 3. Else, note that the beam may fail before it reaches its cracking strength. 3. Using force equilibrium equation, determine ‘a’ by setting fs = fy. a. Assume Case 1 first. Set b = bf. If a ≤ hf , then Case 1; else, Case 2. b. If Case 1, follow Steps 4-6 for rectangular SRB sections with b = bf. c. If Case 2, follow Steps 4-6 for irregular SRB sections. 4. Follow Steps 4-6 depending on whether Step 3b or Step 3c will be done. Determining Mn of Flanged SRB Sections (T- Beams) – Method 2 1. Assume tension steel yields (TSY: fs = fy). Determine effective flange width of the section if necessary. 2. Check if the given As ≥ As, min. If it is, proceed to Step 3. Else, note that the beam may fail before it reaches its cracking strength. 3. Using force equilibrium equation, determine ‘a’ by setting fs = fy. a. Assume Case 1 first. Set b = bf. If a ≤ hf , then Case 1; else, Case 2. b. If Case 1, follow Steps 4-6 for rectangular SRB sections with b = bf. c. If Case 2, follow next steps in this procedure (i-iv). Procedure for Determining Mn of Flanged SRB Sections (T-Beams) – Method 2 i. Considering both components, ℎ𝑓 𝑎 𝑀𝑛 = 𝑀𝑛𝑓 + 𝑀𝑛𝑤 = 𝐴𝑠𝑓 𝑓𝑠 𝑑 − + 𝐴𝑠𝑤 𝑓𝑠 𝑑 − 2 2 ii. Determine Asf (portion of tension steel corresponding to the shaded flange area). 0.85𝑓 ′ (𝑏𝑒 − 𝑏𝑤 )ℎ𝑓 𝑐 𝐴𝑠𝑓 = 𝑓𝑠 iii. Determine Asw (portion of tension steel corresponding to the shaded web area). 𝐴 = 𝐴 − 𝐴 𝐴𝑠𝑤 𝑓𝑦 𝑠𝑤 𝑠 𝑠𝑓 iv. Determine ‘a’. 𝑎= 0.85𝑓′ 𝑏 𝑐 𝑤 Determining Mn of Flanged SRB Sections (T- Beams) – Method 2 4. Determine the following values, in order: Asf, Asw, ‘a’. Using relationship between ‘a’ and ‘c’, determine β1 based on given f’c, as well as ‘c’. 5. Using strain compatibility equation, check if tension steel yields by comparing εs and εy (TSY: εs ≥ εy). a. If TSY, proceed to step 6. b. If tension steel does not yield (TSDNY), repeat step 2 but setting fs = Es*εs. Use strain compatibility equation to express εs in terms of ‘c’, and substitute a = β1c. Using the resulting equation, solve for ‘c’ and verify once again if TSDNY. 6. Solve for Mn by substituting appropriate fs value. a. If TSY, use fs = fy. b. If TSDNY, use actual fs value from step 5b. Sample Problem 5 Determine the maximum allowable uniform service load (dead + live; applied throughout entire span with length of 7.2 m; in kN/m) that one simply-supported T-beam can carry as shown in the section detail below. Assume positive bending with flange under compression. Use the following material properties: f’c = 35 MPa, fy = 420 MPa. Assume that the ratio of live load to dead load is 3.0. Assume ϕ = 0.90. Use SI units for calculations (1 in = 25 mm). Sample Problem 6 Determine the nominal moment strength of the given T-beam section. Use the following material properties: f’c = 28 MPa; fy = 420 MPa. The tension reinforcement size is 32 mm. Failure Modes of RC Beams in Flexure RC beams may fail in flexure as either tension-controlled, balanced, or compression-controlled. Tension-controlled: This failure mode happens when the tension steel yields first before the concrete at ECF cracks. This type of failure is ductile and is therefore considered to be a more favorable failure mode in extreme cases. Sections that fall under this are called tension-controlled or underreinforced sections. Balanced: This failure mode happens when the tension steel yields just as the concrete at ECF cracks. This type of failure is brittle and is advised to be avoided in design. Sections that fall under this are called balanced sections. Compression-controlled: This failure mode happens when the tension steel yields after the concrete at ECF cracks. Like balanced sections, this type of failure is brittle and is advised to be avoided in design as well. Sections that fall under this are called compression-controlled or overreinforced sections. Failure Modes of RC Beams in Flexure These are the code-defined section types based on failure modes in flexure (based on the tension steel strain εt): 2015 NSCP Vol. 1 ACI 318-19 (Table 421.2.2) (Table 21.2.2) Compression-Controlled εt ≤ εy εt ≤ εy Transition εy < εt < 0.005 εy < εt < εy + 0.003 Tension-Controlled εt ≥ 0.005 εt ≥ εy + 0.003 Strength Reduction Factors for Flexure Table 421.2.2, Section 421, Chapter 4, 2015 NSCP Vol. 1 For RC beams, use factors under Other column. Strength Reduction Factors for Flexure Table 21.2.2, Section 421, ACI 318-19 For RC beams, use factors under Other column. Determining Ultimate Flexural Strength (φMn) 1. Determine Mn. 2. Using strain compatibility equation, determine εt and classify section depending on the code definitions. a. By default, use code definitions from 2015 NSCP. Only use the code definitions from ACI 318-19 if it is required to do so or if the next edition of NSCP is in full effect. b. If there is only one tension steel layer, εs may be used since dt = d. 3. Determine corresponding value of φ depending on the type of section. 4. Multiply Mn and φ to get φMn. Sample Problem 7 Determine the ultimate moment strength of each beam analyzed in Sample Problems 1 & 2. Use 2015 NSCP Vol. 1 provisions. Flexural Analysis Doubly-Reinforced RC Beam Sections (DRB) Doubly-Reinforced RC Beam Sections Most RC beams are designed to be singly-reinforced. However, there are cases when reinforcements are required on both tension and compression sides due to section restrictions, long spans, larger loads, or combinations thereof. These sections are doubly- reinforced. In cases where RC beams designed as singly-reinforced turned out to have a very high amount of tension steel (i.e. high As), compression steel serves to increase the capacity of the compression portion of the beam, therefore forcing the tension steel to yield, making it more ductile.  Doubly-Reinforced RC Beam Sections There are four reasons for providing compression reinforcement in beams: Fabrication ease: Having compression reinforcement helps in installing stirrups or ties more easily in beams. Reduced sustained-load deflections: Creep in concrete transfers load from the concrete itself to the compression steel, making the concrete stress lower, which leads to lower deflections.  Doubly-Reinforced RC Beam Sections There are four reasons for providing compression reinforcement in beams: Increased ductility: Compression steel forces tension steel to develop more strain, making the beam more ductile. Change of mode of failure from compression to tension: As a consequence of increased ductility, the beam becomes tension- controlled. Balanced Steel Ratio The balanced steel ratio ρbal is given by: 0.85𝛽1 𝑓 ′ 𝑐 𝜀𝑐𝑢 𝐸𝑠 𝜌𝑏𝑎𝑙 = 𝑓𝑦 𝑓𝑦 + 𝜀𝑐𝑢 𝐸𝑠 This is the required minimum steel ratio that guarantees balanced or compression-controlled failure in flexure. Maximum Steel Ratio For a singly-reinforced RC beam to maintain ductility (i.e. to stay tension-controlled), a maximum amount of tension reinforcement is required by the code. However, the maximum steel ratio is not explicitly mentioned in the code. NSCP 2010/ACI 318-08M (also with older codes): ρmax = 0.75ρbal NSCP 2015/ACI 318-14M (Art. 409.3.3, Sec. 409, Chapter 4): Minimum tension steel strain (εs, min) for non-prestressed beams with factored axial compression less than 10% of f’c*Ag is 0.004. RC beams are still allowed to be in transition zone if the above condition is satisfied. ACI 318-19(22): εs, min = εy + 0.003 RC beams are now required to be tension-controlled. Maximum Steel Ratio (2015 NSCP Vol. 1) The general equation for ρmax is given by: 0.85𝛽1 𝑓 ′ 𝑐 𝜀𝑐𝑢 𝜌𝑚𝑎𝑥 = ≤ 𝜌𝑚𝑎𝑥,𝑠𝑒𝑖𝑠𝑚𝑖𝑐 𝑓𝑦 𝜀𝑠,𝑚𝑖𝑛 + 𝜀𝑐𝑢 ρmax, seismic is the maximum steel ratio set for beams that are part of an SMRF system. Using the provisions from 2015 NSCP Vol. 1, ρmax becomes: 0.85𝛽1 𝑓 ′ 𝑐 3 𝜌𝑚𝑎𝑥 = ≤ 0.025 𝑓𝑦 7 Maximum Steel Ratio (ACI 318-19) Using the provisions from ACI 318-19, ρmax becomes: 0.85𝛽1 𝑓 ′ 𝑐 3𝐸𝑠 𝜌𝑚𝑎𝑥 = ≤ 𝜌𝑚𝑎𝑥,𝑠𝑒𝑖𝑠𝑚𝑖𝑐 𝑓𝑦 1000𝑓𝑦 + 6𝐸𝑠 For Grade 60 reinforcement, ρmax, seismic = 0.025, while for Grade 75/80 reinforcement, ρmax, seismic = 0.02. This is also the maximum amount of tension reinforcement that guarantees tension-controlled failure in flexure.  0.05000 0.04000 Balanced (G60) rho 0.03000 2010 NSCP (G60) 2015 NSCP (G60) ACI 318-19 (G60) 0.02000 0.01000 21 28 35 42 49 56 f'c (MPa) Simplified Fundamental Equations for DRB (Rectangular) Force Equilibrium 0.85𝑓′𝑐 𝑎𝑏 + 𝐴′ 𝑠 (𝑓 ′ 𝑠 − 0.85𝑓 ′ 𝑐 ) = 𝑓𝑠 𝐴𝑠 Strain Compatibility 𝜀𝑐𝑢 𝜀′𝑠 𝜀𝑠 𝜀𝑡 = = = 𝑐 𝑐 − 𝑑′ 𝑑 − 𝑐 𝑑𝑡 − 𝑐 Moment Equilibrium (Nominal Moment Strength) ′ 𝑎 𝑀𝑛 = 0.85𝑓 𝑐 𝑎𝑏 𝑑 − + 𝐴′ 𝑠 (𝑓 ′ 𝑠 − 0.85𝑓 ′ 𝑐 )(𝑑 − 𝑑 ′ ) 2 Determining φMn of Rectangular DRB Sections – Method 1 1. Assume tension steel yields (TSY: fs = fy) and compression steel does not yield (CSDNY: fs’ = Es*εs’ ≤ fy). 2. Check if the given ρ ≥ ρmax. If it is, proceed to Step 3. Else, analyze the beam as singly-reinforced. Check also if ρ’ ≥ ρmin. 3. Using force equilibrium equation, determine ‘c’. Use strain compatibility equation to express ε’s in terms of ‘c’, and substitute a = β1c. 4. Using relationship between ‘a’ and ‘c’, determine β1 based on given f’c, as well as ‘a’. 5. Using strain compatibility equation, check if compression steel does not yield by comparing ε’s and εy (CSDNY: ε’s < εy). a. If CSDNY, proceed to step 6. Determining φMn of Rectangular DRB Sections – Method 1 b. If CSY, repeat step 3 but setting f’s = fy. Use strain compatibility equation to express εs in terms of ‘c’, and substitute a = β1c. c. Verify as well if TSY (fs = fy). 6. Solve for Mn by substituting appropriate fs value. a. If CSDNY, use actual f’s value. b. If CSY, use f’s = fy. 7. Determine ϕ based on εt from strain compatibility equation. 8. Multiply Mn and ϕ to get ϕMn. Sample Problem 8 Determine the maximum allowable service load (dead + live; in kN/m) that may be applied on a 9-meter simply- supported beam, subjected to triangular load with zero intensity at its left end and maximum intensity at its right end, with beam section shown. Assume that the ratio of live load to dead load is 2.50. Use the following material properties: f’c = 21 MPa; fy = 520 MPa. The size of compression reinforcement is 28 mm while that of tension reinforcement is 36 mm. Use 2015 NSCP Vol. 1 provisions. Determining φMn of Rectangular DRB Sections – Method 2 This method is iterative and is more suitable for being used in spreadsheet programs. The trial value of ‘c’ is recommended to be picked within d/4 and d/3. After determining Mn, follow Steps 7-8 from Method 1 to determine ϕMn. Sample Problem 9 Repeat calculation of ultimate moment strength from Sample Problem 8 using Method 2 (iterative solution). Calculate percent error for f’s and Mn with the answers from Method 1.