The point loads are placed at the fixed positions shown in the figure and they are live loads. A k a b a b dn = 1₂ Icr = M B Aso Cross section Asc Ast N.A. Ast Stress profile (concrete) Cross section before cracking Q2) Now, the live load increases gradually and the moment at the critical section just exceeds the cracking moment (Mcr), but the compressive section of the concrete is still under the linear elastic region. Please be reminded that in the flexural design, the crack of the concrete section starts where the tensile stress reaches the tensile strength. It is assumed that the cracks then propagate rapidly up to the entire tension section (up to the neutral axis) and this cracked concrete section cannot resist the tension. It should be noted that in reality, concrete sections between the primary cracks can still resist some tensile stresses as shown in the figure, which should be considered in the displacement design. However, in flexure design, we design the critical section where the largest moment occurs and thus it is reasonable to have such a primary crack in this critical section. 2.1 Find the new neutral axis in the cracked section under the elastic region on the compressive section and also find the cracked second moment of areas (Icr). The transformed area method (dn = Σyi A₁/A₂) is still valid. However, we may need to solve the quadratic equation for calculating the neutral axis distance from the top surface. J 1₂ secondary crack W mm C (centre) f' = 38 Mpa f'ctf = 3.5 MPa fsy = 500 MPa E, = 200 GPa E = 28600 MPa 10⁹ mm¹ a D + b a Neutral axis primary cracks 4₂ A N.A → cracks Ast Cross section just after cracking $ Stress profile (concrete) M → -concrete in tension O Variables 1₁2.5 m 1₂3m a b с ASC Ast 55 mm 600 mm 400 mm 620 mm² 3200 mm² E

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Please do the questions 2.2 and 2.3

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In question 2.1:d_n=215.4 mm

I_cr=2664863900 mm4

Transcribed Image Text:The point loads are placed at the fixed positions shown in the figure and they are live loads. A k a b a b dn = 1₂ Icr= M B 7 Ast Stress profile (concrete) Cross section before cracking Aso Cross section Asc Ast N.A. Q2) Now, the live load increases gradually and the moment at the critical section just exceeds the cracking moment (Mcr), but the compressive section of the concrete is still under the linear elastic region. Please be reminded that in the flexural design, the crack of the concrete section starts where the tensile stress reaches the tensile strength. It is assumed that the cracks then propagate rapidly up to the entire tension section (up to the neutral axis) and this cracked concrete section cannot resist the tension. It should be noted that in reality, concrete sections between the primary cracks can still resist some tensile stresses as shown in the figure, which should be considered in the displacement design. However, in flexure design, we design the critical section where the largest moment occurs and thus it is reasonable to have such a primary crack in this critical section. 2.1 Find the new neutral axis in the cracked section under the elastic region on the compressive section and also find the cracked second moment of areas (Icr). The transformed area method (dn = Σyi A₁/A₂) is still valid. However, we may need to solve the quadratic equation for calculating the neutral axis distance from the top surface. 1₂ secondary crack W mm C (centre) f' = 38 Mpa f'ctf = 3.5 MPa fsy = 500 MPa E, = 200 GPa E = 28600 MPa x 10⁹ mm¹ a D + b a Neutral axis primary cracks 4₂ A N.A → cracks Ast Cross section just after cracking $ Stress profile (concrete) M → -concrete in tension O Variables 1₁2.5 m 1₂3m a b с ASC Ast 55 mm 600 mm 400 mm 620 mm² 3200 mm²

Transcribed Image Text:2.2 The applied loads keep increasing and the concrete section is now under the non-linear regime. The neutral axis moves upward and the concrete compressive section has the curvilinear distribution of concrete stresses, which require the integration of the stress to compute the neutral axis and the ultimate nominal moment. To avoid the complex calculation, the Whitney stress block is introduced during the class. It should be noted that this ultimate strength stage is a condition of the flexural design. Thus, when we design the beam in terms of bending, we can only consider this ultimate condition directly without the previous linear stage. Find the neutral axis distance from the top fibre (dn) at the ultimate strength stage and the ultimate nominal moment (Mu) that this concrete section can carry using the given section dimensions and materials properties. b NA. Ag Stress profile (concrete) Cross section before cracking a dn = M₂= A M* = b Stress profile (concrete) Cross section just after cracking (No answer given) + (No answer given) No Yes N.A → cracks mm kNm kNm 2.3 The beam now has to carry an unfactored dead load (including self-weight) of G = 10 kN/m and an unfactored live load of Q = 20 kN/m. Two concentrated unfactored live loads of P = 15 kN are applied at the positions shown in the figure. Determine if the design moment strength (Mu) exceeds the factored design moment M. p = 0.85 Does the beam satisfy the design requirement? b Note: In flexural design, the ultimate moment in the section occurs when the compressive fibre reached the ultimate concrete strain (&cu = 0.003). The nominal ultimate moment is computed from the property of the concrete section and independent to the applied loads. cracks At Cross section non-linear Stress profile (concrete) U

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