Question

f'm = 1500 psi with inspection and there are 2 #6 Grade 40 rebar 3.5" from the top of the beam. Beam self-weight is 93 lbs/ft.

Calculate the total deflection at B if w- 0.40 kips/ft; L 16 ft; beam section size is 10x40 in, two #6 bars Gr. 40 are placed at 3.5 in. from the top;1500 psi; beam self-weight is 93 Ibs/ft; 8-in. high concrete masonry is used with special inspection provided. 2 kips/ft A B

Answer 1

Ans) According to given loading,

  $\delta$_B = WL⁴ / 8 E I

where, W = load intensity

L = span

  E = modulus of elasticity

I = Effective moment of Inertia

I_g = BD³/12 + [ Bd³/12 + A(D/2 + d/2)²]

= 10 x 40³/12 + [ 10(8)³/12+ 80(24)²] = 99840 in⁴

M_cr = f'_m I_g / (D/2)

= 1500 x 99840 / (40/2)

= 7488000 lb-in or 7488 kip-in

M_a = WL² /2

W = 0.4 kips/ft + 0.093 kips/ft

= 0.493 kips/ft or 0.041 kip/in

=> M_a = 0.041 (16 x 12)² /2

= 7557.1 kip-in

Now, we have to determine I_cr

First calculate modular ratio, n = E_s/E_c = 29000 / 3100 = 9.35

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

bx²/2 + (n-1)A_s ( x - d') = n As ( d - x)

=> 5x² + 8.35( 2 x 0.44) ( x - 6.5) = 9.35 ( 2x0.44) ( 10 - x)

=> 5x² + 7.348 x - 47.768 = 82.28 - 8.288 x

=> 5x² + 15.576 x - 130 = 0

On solving , x = 3.77 in

Hence, I_cr = bx³/12 + xb⁴/4  + (n-1)A_s ( x - d')

= 74465 + 9425 + 2100 = 85991 in⁴

I_eff = (M_cr / Ma)² I_g + [1 - (M_cr / Ma)³ ] I_cr

= 0.99(99840) + 859.91

= 98711.5 in⁴

Puting values,

  $\delta$ = 41 x (16x12)⁴ / 8 (3100000 x 98711.5)

= 0.028 in