Question

Transformed-Area Method For Problems 2.8 to 2.14, assume the sections have cracked and use the transformed-area method to compute their flexural stresses for the loads or moments given.

Answer 1

ANSWER ::

  The theory of the transformed-section method is well known and generally accepted as a method of analysis of members loaded in bending. of a composite member, distorted based upon the relative strength of the materials comprising the member, which in turn is used for the analysis of that member.

  
Given Section - the is cracked & tension zone The is Concrete in neglected. steer type #9 Rebat Nominal diameter = 1-128 inch 28.65 mm Nominal ateo - 1 sq. inch 645 mm² qk/f 1) Elastic Analysis :- 130 K A commm qoft mm 6 1 M max = Hookft i BMD - Section AC => Mx = 50x-22 @ A- MADO @CMC = HOOKU

section Bc > Mx Hox-? @c - Mc = 400kft @ B = MB=0 EV=0 VA+VB = 30K + 2x30 - 90K EMA=0 30x10 +8X30x 15 =UBX30 36 (10+ 2x15) = VB X 36 10+30 = VB VB = Hok VA = 50 K lies in section CR moment Maximum Bending Differentiating Mx@Bc 37 * = 40-22=0 (max] x=20 ft Maximum bending moment is under 'C' max MOX90 - 20 = 800 - 400 Hookt

in section 2) Analysis of cracked I) Actual depth of neutral axis i- Ast 6#9 mast Twin Е , Ast = 661 sqin = 658.in Given modular batio n=m=8 16ang, esmed aeg - m Ask - 6x8 = 48 sq.in Formula B x - Em Ast xd-ha) B d width = 16 in effective depth = 28 in gỗ - k < (28- xa) 3. - 24 168 - 6 3a ya +670 -168 = 0

Xa = 10.304 incor) -16.304 in I) Stoesses developed due to Bending Moment - moment onsidering Maximum bending BM = Hoo Kft Bending stress = 5 a = my o M - Bending Moment from y - Distance of fibre Neutral axis W-64 I - Moment of Inertia Neuttal axis Moment of Ineftia of the section in I = 8x la -NA + Bxxax ) + m Ast X 12 = 164/10-304)3 16:4410:301)* (10:30u)* + 48x (98-10.30u)? 1158.66743 +4376-0022 +15031. 1239 = 20865-7935 in a

CG 1/4 Bending Stress - variation pje Kan / Italm Ast -- m Ast - fet Co > Compresswe Stress in concrete (Max) talm concrete @ the Tensile stress in level of Steel fct Tensile stress in concrete (Max Going by equation (1) с = Ca, E. Bн , = Xa са - 3 Ho0 x 10 _ X 10.304 X 12 12 X- 20865.79 - 33333.334 K S 10.304x144X4.79X10 = 2369 Psi = 2370-34 Psi Maximum Compressive stress in Gnosete = 2369 Psi

Going by the linearity of the bending Stress variation diagram, and comparing the similar triangles we obtain - the most talon - fat (d-xo) (Doxa) - talm (d-ta) xa 9369 - talm 10 304 128-10. 304) .. talm = 2 10.304 x 128-10:304) talm = 4068.5 psi Tensile stress in concrete @ Steel level =406815 Psi Tensie stress in steen - taxm we know that m=8 - TensileStress in Steel = ta = 4068.5x8 = 32548 psi - 32574 Psi Gmplessive Stress in Concrete = 9369 PST Tensile stress in steel = 32,574 psi