m3l5 - Module 3 Limit State of Collapse Flexure (Theories...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Module 3 Limit State of Collapse - Flexure (Theories and Examples) Version 2 CE IIT, Kharagpur
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Lesson 5 Determination of Neutral Axis Depth and Computation of Moment of Resistance Version 2 CE IIT, Kharagpur
Background image of page 2
Instructional Objectives: At the end of this lesson, the student should be able to: determine the depth of the neutral axis for a given cross-section with known value of A st and grades of steel and concrete, write and derive the expressions of x u,max , p t,lim , M u,lim /bd 2 and state the influences of grades of steel and concrete on them separately, derive the corresponding expression of M u when x u < x u,max , x u = x u,max and x u > x u,max , finally take a decision if x u > x u,max . 3.5.1 Introduction After learning the basic assumptions, the three equations of equilibrium and the computations of the total compressive and tensile forces in Lesson 4, it is now required to determine the depth of neutral axis (NA) and then to estimate the moment of resistance of the beams. These two are determined using the two equations of equilibrium (Eqs. 3.1 and 3.3). It has been explained that the depth of neutral axis has important role to estimate the moment of resistance. Accordingly, three different cases are illustrated in this lesson. 3.5.2 Computation of the Depth of Neutral Axis x u From Eqs. 3.1, 9 and 14, we have 0.87 f y A st = 0.36 b x u f ck (3.15) or ck st y u f b . A f . x 36 0 87 0 = (3.16) We can also write: ck st y u f d b . A f . d x 36 0 87 0 = (3.17) Version 2 CE IIT, Kharagpur
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Substituting the expression of d x u from Eq. 3.17 into Eq. 3.13 of Lesson 4, we have lever arm = d b f . A f . . d ck st y 36 0 87 0 42 0 1 = d b f f A . d ck y st 015 1 1 (3.18) Ignoring multiplying factor 1.015 in Eq. 3.18, we have lever arm = d b f f A d ck y st 1 (3.19) 3.5.3 Limiting Value of x u (= x u , max ) Should there be a limiting or maximum value of x u ? Equation 3.17 reveals that d x u increases with the increase of percentage of steel reinforcement d b A st for fixed values of f y and f ck . Thus, the depth of the neutral axis x u will tend to reach the depth of the tensile steel. But, that should not be allowed. However, let us first find out that value of x u which will satisfy assumptions (ii) and (vi) of sec. 3.4.2 and designate that by x u , max for the present, till we confirm that x u should have a limiting value. Version 2 C IIT, Kharagpur
Background image of page 4
Figure 3.5.1 presents the strain diagrams for the three cases: (i) when x u = x u , max ; (ii) when x u is less than x u , max and (iii) when x u is greater than x u , max . The following discussion for the three cases has the reference to Fig. 3.5.1. (i)
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/17/2012 for the course CENG 3012 taught by Professor Prof.j.n.bandopadhyay during the Summer '01 term at Indian Institute of Technology, Kharagpur.

Page1 / 14

m3l5 - Module 3 Limit State of Collapse Flexure (Theories...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online