Question

5. Determine the mid-span short-term deflection of a simply supported beam with the section shown in Figure Q5. Design data: Concrete strength: fcu 30 MPa. Area of tensile steel reinforcement: As 1500 mm Area of compressive steel reinforcement: A,-1500 mm2 Instantaneous static modulus of elasticity of concrete = 25GPa. Span -8.0 m Loading: Dead load 5.0 kN/m (uniformly distributed load); Live load 5.0 kN/m (uniformly distributed load) (Hint: the height of neutral axis of the mid-span section under the service load lies between 200 mm and 250 mm.) (25 marks) d=100 As h-700 d 600 As b-300 Unit: mm Figure Q5 Beam section

Answer 1

Ans) Given,

Depth of beam(h) = 700mm

Width of beam (b) = 300mm

E_c = 25 GPa or 25000 N/mm²  

Step 1) Properties of concrete section:

y = h/2 = 350 mm

I_g = bh³ /12

= 300(700)³ / 12

= 8.575 x 10⁹ mm⁴

Step 2) Properties of cracked section

f_cr = 0.70(30)^0.5

= 3.83 N/mm²

M_r = f_cr I_g /y

= 3.83 x 8.575 x 10⁹ / 350

= 9.40 x 10⁷ N-mm

We know ,E_s = 200000 N/mm²    

m = Es/Ec

= 200000/25000

= 8

Now, let 'x' be distance of neutral axis from bottom of beam ,

As f_y + A_s' f_y - 0.85 f'c b a = 0 (a =0.75 x )

1500(415) + 1500(415) - 0.85 (30) (300) (0.75)x = 0

On solving. x = 217 mm

Now, z = d - x/3

z = 600 - 217/3 = 527.66 mm

I_eff = 300(217)³ /3 + 8(1500) (600 - 217)² = 2.78 x 10⁹ mm⁴  

Step 3 ) Short term deflection :

$\dpi{100} \delta$ = 5W L⁴ / 384 E_c I_eff

   W = 1.5 ( LL + DL)

W = 15 kN/m or 15 N/mm

  $\dpi{100} \delta$ = 15 x (8000)⁴ / (384 x 25000 x 2.78 x 10⁹)

$\dpi{100} \delta$ = 2.30 mm