Question

a hundred random draw with replacement are taken from box

A reinforced concrete beam has square cross section (h x h) h- 280mm. There are 6 steel reinforcing bars (each of diameter d-25mm), 3 evenly spaced towards the top, and 3 evenly spaced towards the bottom. The vertical distance from the center of the bar to top (or bottom) edge of the section is called the cover, and is denoted as e 45mm. It should be taken that psteel-7850 kg/m3, pconcrete 2450 kg/m3, Econcrete 35000 MPa and Esteel-200000 MPa, and that both materials exhibit linear elastic behaviour. The concrete cracks in tension at 2.5 MPa. The beam can be considered as a simply supported beam of length L. a) What is the self weight per unit length of the beam? b) Under self weight only, calculate the length of the beam (L) such that it just cracks under its self weight. c) Now consider the same beam but with a shorter length 0.4L. An additional downwards UDL, w, is then applied along the entire beam. What value of w (in conjunction with self weight) wl induce cracking? Draw the stress and strain distributions for both the steel and the concrete on the critical cross-section when this occurs

Answer 1

a) Total volume of reinforced concrete per m length of beam = 0.28 x 0.28 x 1 m³ = 0.0784 m³

volume of reinforcement per m length = 6 x (3.14/4) x (0.025)² x 1 m^{3 =} 0.00294 m³

So, net volume of concrete in 1 m of Beam = 0.0784 - 0.00294 m³ = 0.0755  m³

Therefore, self weight per unit length of beam = 0.0755 x 2450 + 0.00294 x 7850 kg = 208.1 kg = 2081 N

b) As the concrete cracks in tension,

so, it reaches cracking stress of f_cr = 2.5 MPa = 2.5 x 10⁶ N/m² = 2.5 N/mm²

Cracking moment M_cr = f_cr x (1/6) x bd² = 2.5 x 10⁶ x (1/6) x 0.28 x (0.28 - 0.045)² Nm = 6442.9 Nm

Now, as the beam is simply supported of length L,

So, Cracking moment due to self weight only, M_cr = 2081 x L² / 8

Therefore, 2081 x L² / 8 = 6442.9

L = √(8 x 6442.9 / 2081) = 4.98

Length of the beam, L = 4.98 m = 5 m (approx)

c) Now the length is 0.4L = 0.4 x 5 m = 2 m

additional udl = w N/m

So, total udl acting on beam(in conjunction with self weight) = (w + 2081) N/m

M_cr' = (w + 2081) x 2² / 8 Nm

So, (w + 2081) x 2² / 8 = 6442.9

w + 2081 = 12885.8

w = 12885.8 - 2081 = 10804.8

So, additional udl, w(in conjunction with self weight) which will induce cracking = 10804.8 N/m

Diagrams for stress strain distributions are shown below,

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280 45mm 2 80 ナg.の2 Mm Strass d*agram Section strain dagram