Latihan 1 - Perilaku Struktur Rangka Baja - Firstka Safira - 25019325 [PDF]

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SI-5211 PERILAKU STRUKTUR RANGKA BAJA TUGAS 1



oleh



25019325 NIM



Firstka Safira NAMA



DOSEN Dyah Kusumastuti, ST, MT, Ph.D.



PROGRAM STUDI TEKNIK SIPIL FAKULTAS TEKNIK SIPIL DAN LINGKUNGAN INSTITUT TEKNOLOGI BANDUNG 2020



Chapter Two – Structural Steel 2.1 A simple structural system consists of the three pin-ended bars shown in Figure 2.46. All bars have a cross-section of 1 in2 (645 mm2). The material is a ductile structural steel which can be idealized as having elasto-perfectly plastic properties, with E of 29000 ksi and y of 36 ksi (200,000 MPa and 250 MPa, respectively). For that structure: a) What is the allowable load, Pall, which can be applied to that structure such that stresses in the bars do not exceed 0.66 Fy (i.e., allowable stress method)? b) What is the elastic limit load, Pu, which can be applied to that structure such that stresses do not exceed Fy anywhere (i.e., limit states design, with elastic analysis)? How would this change if the central member was constructed 1 in (25.4 mm) too short and forced to fit? Is a solution in the elastic range possible? Does it matter if member AC is elongated, or if members AB and AD are shortened, to achieve this fit (support answers by calculations)? c) What is the ultimate load capacity, Pplast, that the structure can carry accounting for plastic behavior (i.e., plastic design philosophy)? d) Draw the force-displacement (P-Δ) graph for the structure up to a maximum displacement of 1 in (25.4 mm) and followed by unloading to zero load. Express the vertical axis of this graph in a ratio of Pstructure over Pym, where Pym is the ultimate capacity of a single member, and the horizontal axis as a ratio of Δ/Δy, where Δy is defined at the occurrence of first yielding in the structure. Calculate and indicate on this plot all important numerical values. e) For the loading history described in (d), also plot the force displacement relation of all three members, superposed on a single graph, the vertical axis of this graph being the ratio of Pmember over Pym, the horizontal axis being the actual structural displacement in inches (mm). Calculate and indicate on this plot all important numerical values; in particular, indicate what are the forces in each bar following the above displacement history. f) If the central member was constructed 1 in (25.4 mm) too short and forced to fit during construction, plot the force displacement relation of all three members corresponding to the following two construction sequences: first, if the central member is elongated by 1 in (25.4 mm), connected to the other two members, and released; second, if the two inclined members are compressed to connect to the central member, and released. Does it matter if member AC is elongated, or if members AB and AD are shortened, to achieve this fit (support answers by calculations)? g) Comment on the differences between elastic and plastic behavior for analysis and design. In particular, comment on whether the principle of superposition is applicable for plastic design (i.e., whether usual computer programs could be used to assess plastic strength), whether lack-of-fit and other displacement-induced forces seem critical to the ultimate strength of the structure, and whether an extremely slender member could buckle under residual stresses upon unloading.



Answer:



Elastic - Perfectly - Plastic 645 mm2



Figure 1.1 All three bars are of same material (E and 𝜎y) and cross-sectional area (A) a) Allowable load, Pall which can be applied to the structure and stresses not exceed 0.66 Fy Equilibrium: Equilibrium conditions express the vanishing of the sum of the forces and moments acting on the system Figure. 2 Free body diagrams



B1 a



P1



C 2 P2



The force and elongation in member 3 will always be equal to that in member 1 as a result of structural and loading symmetry



D 3



P3 =



therefore, 𝑃1 = 𝑃3



P1



y A



∑ 𝐹𝑦 = 0 on point A



F



x



Equilibrium equations of the system, 𝑃1 cos 45𝑜 + 𝑃2 + 𝑃3 cos 45𝑜 − 𝐹 = 0 𝐹 = 2 (𝑃1 cos 45𝑜 ) + 𝑃2 … . (𝐄𝐪. 𝟏) 𝐹 = 1,414(𝑃1 ) + 𝑃2 Geometric compatibility B



C



D



2 3



1



∆ A F Figure 1.3 Deflection



Deflection for member 1 & member 3, ∆1 = ∆3 = ∆ cos 45𝑜 Deflection member 2, ∆2 = ∆



Relation of displacement for each member, ∆1 = ∆3 = ∆2 cos 45𝑜 = ∆ cos 45𝑜 … . (𝐄𝐪. 𝟐) = 0,707 ∆ Constitutive Law: to express internal forces in terms of member deformations or strains 𝑃1 = 𝑃3 = 𝜎 𝐴 = 𝜀1 𝐸 𝐴 = 𝑃1 = 𝑃3 =



∆1 𝐸𝐴 𝐿1



𝐸𝐴 𝐸𝐴 𝐸𝐴 𝐸𝐴 ∆1 = ∆ cos 45𝑜 = ∆ cos 45𝑜 = ∆ … … (𝐄𝐪. 𝟑) 𝑜 𝑜 𝐿1 𝐿/ cos 45 𝐿/ cos 45 2𝐿 𝑃2 = 𝜎 𝐴 = 𝜀2 𝐸 𝐴 =



∆2 𝐸𝐴 𝐿2



𝐸𝐴 𝐸𝐴 ∆2 = ∆ … . . (𝐄𝐪. 𝟒) 𝐿2 𝐿



𝑃2 =



Substitute (Eq. 3) and (Eq. 4) to (Eq. 2) 𝐹 = 2 (𝑃1 cos 45𝑜 ) + 𝑃2 𝐹 = 2( 𝐹=



𝐸𝐴 𝐸𝐴 ∆ cos 45𝑜 ) + ∆ 2𝐿 𝐿



𝐸𝐴 ∆ (1 + cos 45𝑜 ) = 𝐾∆ … . (𝐄𝐪. 𝟓) 𝐿



Determine displacement ∆ when stress in the bars 𝜎 = 0,66 𝐹𝑦 to calculate Pall, 𝜎 = 𝜀0,66 𝐹𝑦 𝐸 0.66 𝜎𝑦 = 𝐸 ∆=



∆ 𝐿



0.66 𝜎𝑦 𝐿 (0.66) (250)(2500) = = 2.0625 𝑚𝑚 𝐸 (200.000)



Calculate Pall with this following equation, 𝑃𝑎𝑙𝑙 = 𝐹 = 𝑃𝑎𝑙𝑙 =



𝐸𝐴 ∆ (1 + cos 45𝑜 ) 𝐿



(200.000) (645) (2,0625) (1 + cos 45𝑜 ) (2500) 𝑃𝑎𝑙𝑙 = 181.678,839 𝑁



b) Elastic limit load, Pu Determine displacement ∆ when stress in the bars 𝜎 = 𝐹𝑦 𝜎𝑦𝐿 ∆𝑦 = 𝐸 (250)(2500) ∆𝑦 = = 3,125 𝑚𝑚 200.000 𝐸𝐴 ∆𝑦 (1 + cos 45𝑜 ) 𝐿



𝑃𝑢𝑙𝑡 = 𝐹𝑦 = 𝑃𝑢𝑙𝑡 =



(200.000)(645) 3,125 (1 + cos 45𝑜 ) = 275270.968 (2500)



c) Ultimate load capacity Pplast that the structure can carry accounting for plastic behavior. In plastic behavior, member 2 has yielded meanwhile members 1 and 3 have not reached their ultimate capacity/ still in elastic condition. 𝑃1 = 𝑃3 =



𝐸𝐴 ∆ 2𝐿



𝑃2 = 𝑃𝑦 =



;



𝑃1 = 𝑃3 = 𝐴 𝜎𝑦



𝐸𝐴 ∆𝑦 ; 𝑃2 = 𝐴 𝜎𝑦 𝐿



Therefore, 𝑃𝑙𝑎𝑠𝑡 = 2 (𝑃1 cos 45𝑜 ) + 𝑃2 𝑃𝑙𝑎𝑠𝑡 = 2 ( 𝑃 𝑝𝑙𝑎𝑠𝑡 = 2 (



𝐸𝐴 ∆. cos 45𝑜 ) + 𝑃𝑦 2𝐿



𝐸𝐴 2 𝜎𝑦 𝐿 cos 45𝑜 ) + 𝐴 𝜎𝑦 𝐿 𝐸



𝑃 𝑝𝑙𝑎𝑠𝑡 = 389.291,93 𝑁



d) Draw the force displacement (P - ∆) graph of the structure up to maximum displacement of 25,4 mm and followed by unloading to zero load. Calculate and indicate on this plot all important numerical values; in particular. Vertical axis is the ratio of Pstructure over Pym, where Pym is the ultimate capacity of a single member Horizontal axis is the ratio of Δ/Δy, where Δy is defined at the occurrence of first yielding in the structure. 𝑃𝑦𝑚



𝑃𝑦𝑚 = 𝐴 𝜎𝑦 = (645)(250) = 161.250 𝑁



𝜎𝑦𝐿 𝐸



∆𝑦 =



(250)(2500) = 3,125 𝑚𝑚 200.000 𝜎𝑦𝐿 𝑃𝑦 ∆𝑦1 (250)(2500) = = 1,7071 ; = 𝐸 = 1 (1, 1.7071) 𝜎𝑦𝐿 𝑃𝑦𝑚 200.000 ∆𝑦0 𝐸 ∆𝑦 =



∆𝑦 𝑃𝑝𝑙𝑎𝑠𝑡 389.291,93 = = 2,414 ; = 𝑃𝑦𝑚 161.250 ∆𝑦1



2 𝜎𝑦𝐿 𝐸 = 2 (2, 2.414) 𝜎𝑦𝐿 𝐸



𝑃𝑝𝑙𝑎𝑠𝑡 389.291,93 ∆𝑚𝑎𝑘𝑠 25,4 = = 2,414 ; = = 8,128 (8.128, 2.414) 𝑃𝑦𝑚 161.250 ∆𝑦 3,125 3 8.128, 2.4142



2, 2.4142



2.5



P/Pym



2 1, 1.7071 1.5 1 0.5 0



0, 0 0



1



2



3



4



5



6



7



8



9



∆/∆y



e) For the loading history described in (d), also plot the force displacement relation of all three members, superposed on a single graph. Vertical axis is the ratio of Pmember over Pym Horizontal axis is the structural displacement in inches 𝑚𝑒𝑚𝑏𝑒𝑟 1 = 𝑚𝑒𝑚𝑏𝑒𝑟 3 𝑃𝑚𝑒𝑚𝑏𝑒𝑟 389.291,93 2𝜎𝑦𝐿 2(250)(2500) = = 1 ; ∆𝑦 = = = 6.25 (6.25, 1) 𝑃𝑦𝑚 389.291,93 𝐸 (200.000)



𝑚𝑒𝑚𝑏𝑒𝑟 2 𝑃𝑚𝑒𝑚𝑏𝑒𝑟 161.250 (250)(2500) = = 1 ; ∆𝑦 = = 3,125 𝑚𝑚 (3,125, 1) 𝑃𝑦𝑚 161.250 200.000



1.2 1



P/Pym



0.8 Member 2



0.6



Member 1 0.4



Member 3



0.2 0 0



5



10



15



20



25



∆/∆y



f)



Force displacement relation of all three members corresponding to the following two construction sequences: first, if the central member is elongated by 1 in (25.4 mm), connected to the other two members, and released; second, if the two inclined members are compressed to connect to the central member, and released.



g) Elastic design is carried out by assuming that at design loads structures behave in a linearly elastic manner. An elastic analysis is performed by applying the design loads and required internal forces in the structural elements (members and connections) are determined and adequate design strength is provided. Since the element forces are determined based on



elastic behavior, the design is governed by elastic stiffness distribution (ratios) among the system elements. In the elastic analysis, superposition can be used for measure displacement and deformation. Structural analysis with elastic analysis example are double integration method, virtual work, conjugate beam, etc. Same with other structural analysis program SAP/ETABS that performed elastic analysis where elasticity modulus is constant. Plastic analysis is to utilize the reserve strength beyond the elastic limit due to the redistribution of internal forces. Therefore, the analysis focuses on the internal forces at the limit level when the yield mechanism forms. Plastic analysis procedures are based on the considerations of equilibrium, yield mechanism, and plastic strength conditions In the plastic analysis, increasing strain in any other member should be maintained because if one of the members reaches yield strength, the elasticity modulus will be changed. Usually, mathematical model should be used when plastic analysis performed in example mathematical model of elastic-perfectly plastic, elastic plastic with strain hardening, curvilinear model (Ramberg-Osfood, Menegotto-Pinto) etc. The defect of the structural member at the construction period probably can change the structural member dimension that occurs the elastic limit load and plastic limit load changed because the defect structural member can develop remaining residual stress. In the slender member design, the plastic condition can be more dangerous than the compact structure design. When the slender element is unloaded in the inelastic condition, it would not be stable because the stiffness is lower than the compact element that can develop the buckling from the residual forces at the unloading process. 2.2 F



2.3 Experimental data has been obtained for the normalized moment-curvature relationship of a steel beam (see below). Use both the Menegotto – Pinto method and the Ramberg - Osgood method to define analytical expressions relating the normalized moments and curvatures for that steel beam. M/My 0 0.5 0.75 1 1.087 1.153



Ф/Фy 0 0.5 0.75 1 1.1 1.2



M/My 1.204 1.245 1.278 1.337 1.375 1.401



Ф/Фy 1.3 1.4 1.5 1.75 2 2.25



M/My 1.42 1.434 1.444 1.469 1.48 1.486



Ф/Фy 2.5 2.75 3 4 5 6



Answer: Based on Table 2.1 below, moment - curvature curve for the steel beam is shown in Figure 2.1: M/My 0 0.5 0.75 1 1.087 1.153



Ф/Фy 0 0.5 0.75 1 1.1 1.2



1.204



1.3



1.245



1.4



1.278



1.5



0.6



1.337



1.75



0.4



1.375



2



0.2



1.401



2.25



1.42



2.5



1.434



2.75



1.444



3



1.469



4



1.48



5



1.486



6



Table 2.1 Normalized moment – curvature relationship of steel beam



Normalized Moment - Curvature (Experimental Data) 1.6 1.4 1.2



M/My



1 0.8



0 0



1



2



3



4



5



Ф/Фy Figure 2.1 Normalized moment – curvature relationship of a steel beam



6



a. Menegotto – Pinto method The general form of the Manegotto-Pinto functions is as follows:



…. (Eq. 1) Where: 0 = yield strength 0 = yield strain b = tangent slope’s ratio, ratio of the final to initial tangent stiffnesses d = value that is graphically defined in Figure 2.2 n = curvature parameter (transition between elastic and post-yield slopes)



Initial slope Final slope



Figure 2.2 Menegotto – Pinto expression for the stress-strain relation of reinforcing steel



In the normalized space of stress and strain, the initial stiffness has a slope 1, the slope of the final tangent stiffness is b, and d varies from zero to (1 – b) as /0 progressively increases from zero to maximum value at the last of data point.



Step-by-step. 1. Draw the initial and final tangent slopes to the data curve. The intersection of these two curves gives 0 ~ (M/My)𝑜 and 0 ~ (Ф/Ф𝑦 )𝑜



Initial tangent slope First point Last point Final tangent slope First point Last point



M/My



Ф/Фy



0 1.6



0 1.6



1.45 1.486



Linear equation for initial slope: y = x …. (1) Linear equation for final slope: y - y1 = m (x – x1) … (2)



0 6



Table 2.2 Initial slope and final slope data point



Intersection: 𝒚 − 𝒚𝟏 = 𝒎 (𝒙 − 𝒙𝟏 )



𝒙=𝒚



𝑦 − 1.45 = 0.006 (𝑥 − 0)



𝑥 = 0.006𝑥 + 1.45 = (1 − 0.006)𝑥 = 1.45



𝒚 = 𝟎. 𝟎𝟎𝟔𝒙 + 𝟏. 𝟒𝟓



𝒙 = 𝟏. 𝟒𝟓𝟗 = 𝒚



Normalized Moment - Curvature 1.6



Intersection



1.4



y = 0.006x + 1.45 R² = 1



1.2



y = 0.1761x + 0.7984 R² = 0.4897



M/My



1 0.8 0.6



y=x R² = 1



0.4 0.2 0 0



1 Moment - Curvature



2



3 Ф/Фy initial slope



4



5



6



final slope



Figure 2.3 Initial slope and final slope of normalized moment – curvature curve using Menegotto – Pinto Method



Ф/Фy



0 0.5 0.75 1 1.087 1.153 1.204 1.245 1.278 1.337 1.375 1.401 1.42 1.434 1.444 1.469 1.48 1.486



0 0.5 0.75 1 1.1 1.2 1.3 1.4 1.5 1.75 2 2.25 2.5 2.75 3 4 5 6



(M/My)/ (M/My)o 0 0.3427 0.5141 0.6854 0.745 0.7903 0.8252 0.8533 0.8759 0.9164 0.9424 0.9602 0.9733 0.9829 0.9897 1.0069 1.0144 1.0185



(Ф/Фy)/ (Ф/Фy)o 0 0.3427 0.514051 0.685401 0.753941 0.822481 0.891021 0.959561 1.028101 1.199452 1.370802 1.542152 1.713502 1.884853 2.056203 2.741604 3.427005 4.112406



(M/My)/(M/My)o vs (Ф/Фy)/(Ф/Фy)o 1.2 1



(M/My)/(M/My)o



M/My



0.8 0.6 0.4 0.2 0 0



0.5



1



1.5



2



2.5



3



3.5



4



(Ф/Фy)/(Ф/Фy)o Experimental Data



initial slope



final slope



Figure 2.4 Initial slope and final slope of (M/My)/(M/My)o vs (Ф/Фy)/(Ф/Фy)o curve



Table 2.3 Moment – curvature relation with (M/My)o and (Ф/Фy)o



2. Calculate the ratio of tangent slopes (b), where b is the result of slope from the last pair of data points in Table 2.1 𝑏=



𝑏=



M/My2 − M/My1 Ф/Фy2 − Ф/Фy1



1,486 − 1,48 = 0,006 6− 5



3. Calculate d* from (Eq. 1), with /0 = 1 (as in Figure 2.1) M/M𝑦 Ф/Ф𝑦 = 𝑏 ( ) + 𝑑∗ … … (𝐄𝐪. 𝟐) (M/My)𝑜 (Ф/Ф𝑦 )𝑜 In order to determine 𝑀/𝑀𝑦 value when /0 ~



Ф/Ф𝑦 (Ф/Ф𝑦 )𝑜



= 1 , data from Table 2.3



should be used as the surrounding points for the interpolation.



Interpolation to find 𝑀/𝑀𝑦 when



Ф/Ф𝑦 (Ф/Ф𝑦 )𝑜



X = X1 + ( X = 1.245 + (



= 1,



Y − Y1 ) (X2 − X1) Y2 − Y1



1 − 0.959561 ) (1.278 − 1.245) 1.028101 − 0.959561 X = 1.264 = 𝑀/𝑀𝑦



Using Eq. 2 to determine d*, M/M𝑦 Ф/Ф𝑦 = 𝑏 ( ) + 𝑑∗ (M/My)𝑜 (Ф/Ф𝑦 )𝑜 M/M𝑦 = 𝑏 (1) + 𝑑∗ (M/My)𝑜 𝑑∗ =



1.264 − 0.006 1.45



𝑑∗ = 0.860 4. Now calculate the value of n (curvature parameter) by using the value of d* that was found previously,



…. (Eq. 3) 𝑛=



log 2 = 4.786 log(1 − 0.006) − log 0.860



5. Menegotto-Pinto equation as follow, M/M𝑦 Ф/Ф𝑦 = 𝑏 ( ) + 𝑑∗ (M/My)𝑜 (Ф/Ф𝑦 )𝑜



(M/My)𝑜 𝑏 ( ) Ф/Ф𝑦 + (Ф/Ф𝑦 )𝑜



M/M𝑦 = (



M/M𝑦 = 0.006 Ф/Ф𝑦 +



(1 − 𝑏) (



(M/My)𝑜 ) Ф/Ф𝑦 (Ф/Ф𝑦 )𝑜 1



Ф/Ф𝑦 𝑛 𝑛 ) ) (1 + ( (Ф/Ф𝑦 )𝑜 (0.994) Ф/Ф𝑦 1/4.786



Ф/Ф𝑦 4.786 (1 + ( 1.459 ) )



)



Comparison table and curve of moment – curvature relation between Experimental Data Vs Menegotto-Pinto Method is shown below. M/My Ф/Фy Experimental Menegotto Data - Pinto 0 0 0 0.5 0.5 0.4993849 0.75 0.75 0.7437097 1 1 0.9689558 1.1 1.087 1.0486655 1.2 1.153 1.1202776 1.3 1.204 1.182891 1.4 1.245 1.2362294 1.5 1.278 1.2806343 1.75 1.337 1.3585313 2 1.375 1.4029813 2.25 1.401 1.4282854 2.5 1.42 1.4432278 2.75 1.434 1.4525682 3 1.444 1.4588073 4 1.469 1.4718301 5 1.48 1.479413 6 1.486 1.4858976



Normalized Moment - Curvature 1.6 1.4 1.2



M/My



1 0.8 0.6 0.4 0.2 0 0



1



2



3



Experimental Data



6



Menegotto-Pinto Method



Figure 2.5 Comparison curve of moment – curvature relationship between Experimental Data & Menegotto - Pinto Method



b. Ramberg - Osgood method The general form of the Ramberg-Osgood functions is expressed as:



…. (Eq. 1)



E = initial elastic modulus =



5



Ф/Фy



Table 2.4 Comparison table of moment – curvature relationship between Experimental Data & Menegotto - Pinto Method



Where: 0 = yield strength 0 = yield strain



4



 



n = strain hardening exponent a = yield offset



Figure 2.6 Ramberg – Osgood material model for structural steel



Step-by-step 1. Determine the initial elastic modulus, E M/M𝑦 1.087 = = 0.988 Ф/Ф𝑦 1.1 2. Calculate 0 and 0 -



-



a) Use the 0.2% offset method to determine 0 ~ (M/My)𝑜 = 1.278 b) Find 0 by this following equation, 0 (M/My)𝑜 1.278 0 = ~ (Ф/Ф𝑦 )𝑜 = = = 1.293 𝐸 𝐸 0.988



3. Select point A and calculate its slope from the origin (0,0) 𝑚𝐴 =



(M/My)A 1,278 = = 0.852 (Ф/Ф𝑦 )𝐴 1,5



4. Select point B (last point for which an increase in stress measured) 𝑚𝐵 =



(M/My)B 1,486 = = 0,248 (Ф/Ф𝑦 )𝐵 6



Normalized Moment - Curvature 1.6 1.4



M/My



1.2 1 0.8 0.6 0.4 0.2 0 0



1



2



3



4



5



6



Ф/Фy Experimental Data



Point B



Point A



Figure 2.7 Slope mA and slope mB of moment – curvature curve using Ramberg – Osgood method



5. Calculate n with this following equation, (1 − 𝑚𝐴)𝑚𝐵 (1 − 𝑚𝐵)𝑚𝐴 … . (𝐸𝑞. 2) (M/My)𝑜 𝑙𝑜𝑔 (M/My)𝐵



𝑙𝑜𝑔 𝑛 =1+



𝑙𝑜𝑔 𝑛=1+



(1 − 0.852)0,248 (1 − 0,248)0.852 = 19.976 1.278 𝑙𝑜𝑔 1,486



6. Calculate a with this following equation, 1 − 0.852 1 − 𝑚𝐴 𝑚𝐴 0.852 𝑎= = 0.00131 𝑛−1 = 1.278 19.976−1 (M/My)𝑜 ( ) ( ) 0.988 𝐸 7. Resulting Ramberg – Osgood equation as follow, Ф/Ф𝑦 (Ф/Ф𝑦 )𝑜



=(



M/My a M/My 𝑛 ) +( )( ) (M/My)𝑜 𝐸𝐼 (Ф/Ф𝑦 )𝑜



M/My M/My 𝑛 Ф/Ф𝑦 = ( ) (Ф/Ф𝑦 )𝑜 + 𝑎 ( ) (M/My)𝑜 𝐸 M/My M/My 19.976 Ф/Ф𝑦 = ( ) 1.293 + 0.00131 ( ) 1.278 0.988



Comparison table and curve of moment – curvature relation between Experimental Data Vs Ramberg - Osgood Method is shown below.



0 0.5 0.75 1 1.087 1.153 1.204 1.245 1.278 1.337 1.375 1.401 1.42 1.434 1.444 1.469 1.48 1.486



Ф/Фy Experimental Ramberg Data Osgood 0 0 0.5 0.5058685 0.75 0.7588081 1 1.0134044 1.1 1.1085835 1.2 1.1951818 1.3 1.2861459 1.4 1.3923856 1.5 1.5169066 1.75 1.9042713 2 2.356614 2.25 2.8210805 2.5 3.2737064 2.75 3.6855988 3 4.0285097 4 5.1039536 5 5.6964552 6 6.0559869



Normalized Moment - Curvature 1.6 1.4 1.2



M/My



M/My



1 0.8 0.6 0.4 0.2 0 0



1



2



3



4



5



Ф/Фy Experimental Data



Ramberg - Osgood



Figure 2.8 Comparison curve of moment – curvature relationship between Experimental Data & Ramberg Osgood Method



Table 2.5 Comparison table of moment – curvature relationship between Experimental Data & Ramberg Osgood Method



REFERENCE 1. Bruneau, Michel et all., 2001, Ductile Design of Steel Structures, 2nd Ed., McGraw Hill, New York. 2. Wong, M. Bill., 2009, Plastic Analysis and Design of Steel Structures, Elsevier Ltd, Oxford.



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