Answer :
The provided data involves calculations related to the Transformed Sections Method for finding stresses in a structural analysis scenario.The modular ratio (n) is essential to this approach, representing the ratio of the elastic moduli of steel (Es) to concrete (Ec).
In the given data, the Transformed Sections Method is employed to determine stresses in a structural analysis context. This method is commonly used to evaluate the stress distribution in composite materials like concrete and steel. The modular ratio (n) is essential to this approach, representing the ratio of the elastic moduli of steel (Es) to concrete (Ec). The data includes various parameters like concrete and steel properties, applied loads, and dimensions of the structure.
The modular ratio (n) is calculated from the given elastic moduli of steel (Es) and concrete (Ec). The stresses in the transformed sections are determined using the given loadings and dimensions, considering the transformed properties of the composite section. The calculated stresses are compared to the specified code limits to ensure the structural integrity of the system.
The data also provides information about different components of the analysis, such as the moment of inertia (Ig), the transformed moment of inertia (Ig+), and deflection calculations. These elements play a crucial role in accurately evaluating the structural behavior under applied loads.
The calculations follow specific steps involving equations and constants to achieve the desired stress and deflection values. The provided data includes numerical values and formulas to perform these calculations, which enable engineers to assess the safety and performance of the structure.
In conclusion, the Transformed Sections Method serves as a valuable technique in structural engineering to determine stresses and deflections in composite structures. It involves meticulous calculations and considerations of various material properties and dimensions to ensure the structural integrity and safety of the analyzed system.
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Definition S n= E E = = modular ratio fc'(ksi) 3 3.5 4 4.5 5 6 n 9 8.5 8 7.5 7 6.5 Transformed Sections Method & Transformed the fo Concr. Steel Racere ES & Ar = nas I fr = 45 fs=nfi n ex. 6 Purpose: find stresses. 12" М.= 2000 kin Total D+ L 24' оооо 4#10 A, = 4(1.27in?) = 5.08in givesfc'=3ksi Ec=3122ksi 2 n 29000 =9.28 3122 -9.0 alwaysEs=29000ksi Code fy=40ksi 4. 5.4 kен fe = 3 ksi- Коѕикон 1. . . kg 4 ок, - 0.4 10oki gL 0.7 15) 0.43', 51 С. 5 28 29 = 0, + 0 2. 0,3 100 OK Inmin < 13,8 15. - 4" (134- 1971. , 3 12 12 see pg 35 notes -0.3 Ад (s) (ч) - 12 (3) - 1 (11.3-х Ч.5 2. 1562 — 4у. ) О- Ч. 1 ч. – 168,2. 4.57 in 1--14j - ( 5) (-12). 20.5) (4) (467) 4 (1.3 - 4.57 ) - 920. го 1,5 і 25 / 2000) 2 Т24.9 mi e 0.129 ка; Mera hr Ig = 0,129 (1970 2 3.СЯ К = 0, 40% kh hin 3 е Iz = 60 check math -0.1 check math-0.1 е 6.4 rр - 39ket 20, 30% KW ller 0.306 -0.009 34 MO Ica (MOR) Ig + (} Mac ) [er= (01009) ² 1971 1 + (1-(0.0093) (920.4) MO 920.4 in ua 1971.linu C'est E = 57000 Pc = 5700 3000 3122018.6ksi Donau Hie 3.99 x 10 etc = £ 34 (18032 (312 2018.6) 192014) 48 do the same for LL deflection -1 find next-0.5 compare to limit -0.5?