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  Problem 19.
  
  21
  24
  
  0.9 in2
  
  14
  
  The beam shown has 30-ft simple span; fc = 5000 psi, fpu = 250,000 psi, and the initial prestress is 160,000 psi and
  
  fpy = 0.85 fpu
  
  Given:
  
  b = 14 in L = 30 ft or
  h = 24 in L = 360 in
  d = 21 in fc = 5000 psi
  As = 0.9 in2 fpi = 160,000 psi
  
  a) Calculate the concrete stress in the top and bottom of the beam at midspan immediately after the tendons are cut.
  (Ans. ftop= +0.183 ksi, fbottom= -1.041 ksi)
  
  Solution:
  
  (@ midspan)
  
  3
  Pi = fpi As I=
  12
  (14")(24")3
  Pi = (160,000 psi)( 0.9 in2) I=
  12
  Pi = 144,000 lb or 144 kips I = 16,128 in4
  
  c = 12 in Ac = (14)(24)
  e = 9 in Ac = 336 in2
  
  1
  W b = (14)(24)(150 3 ) ( 12 )3
  
  W b = 29.167
  
  2
  Md = 8
  
  (29.167 )(360 )2
  
  Md = 8
  
  Md = 472,505.4 lb-in
  
  ftop = - +
  -
  
  144,000 lb 144,000 lb 9 (12 ) 472,505.4 lb in (12 )
  
  ftop = - +
  336 in 2 16,128 in 4 16,128 in 4
  
  fbottom = - -
  +
  
  fbottom = -1,041.29 or -1.04129 --------------------------- answer
  
  b) Recalculate the stresses at midspan after assumed prestress losses in the tendons of 20%. (Ans. ftop = +0.076 ksi,
  fbottom = - 0.762 ksi)
  
  Peff = Pi - 0.20(Pi)
  
  Peff = 144,000 lb - 0.20(144,000 lb)
  
  Peff = 115,200 lb
  
  ftop = - +
  -
  
  115,200 lb 115,200 lb 9 (12 ) 472,505.4 lb in (12 )
  
  ftop = - +
  336 in 2 16,128 in 4 16,128 in 4
  
  fbottom = - -
  +
  
  fbottom = -780.139 or -0.780139 ------------------------------ answer
  
  c) What maximum service liveload can this beam support in addition to its own weight if allowable stresses of 0.45fc in
  compression and 6 fc in tension are permitted? (ans. 1.182k/ft)
  
  Peff = 115,200 lb
  
  Allowable:
  
  ftop = 0.45 fc --------- compression MTotal = MD + ML
  
  2 ( ) 2
  fbottom = 6 fc --------- tension Mt = 8
  + 8
  
  ftop = - +
  
  115,200 lb 115,200 lb 9 (12 ) (12 )
  
  -0.45 fc = - +
  336 in 2 16,128 in 4 16,128 in 4
  
  (12 )
  - 0.45 (5000 psi ) = - 342.857 + 771.429 16,128 in 4
  
  (12 )
  - 2,250 psi = - 342.857 + 771.429 16,128 in 4
  
  (12 )
  - 2,678.572 psi = 16,128 in 4
  
  Mt = + 3,600,000.768 lb-in
  
  2 ( ) 2
  Mt= 8
  + 8
  
  29.167 (360 )2 (360 )2
  
  3,600,000.768 lb-in = 8
  + 8
  
  29.167 (360 )2 (360 )2
  
  3,600,000.768 lb-in = 8
  + 8
  
  (360 )2
  3,600,000.768 lb-in = 472,505.4 lb-in +
  8
  
  (360 )2
  3,127,495.369 lb-in = 8
  
  = 193.055 or
  
  1 12
  = 193.055 (1000 )( 1
  ) = 2.317
  
  fbottom = -
  +
  
  115,200 lb 115,200 lb 9 (12 ) (12 )
  
  6 fc = - - +
  336 in 2 16,128 in 4 16,128 in 4
  
  (12 )
  6 5000 psi = - 342.857 - 771.429 +
  16,128 in 4
  
  (12 )
  424.264 psi = - 342.857 - 771.429 + 16,128 in 4
  
  (12 )
  1538.55 psi = 16,128 in 4
  
  Mt = 2,067,811.2 lb-in
  
  2 ( ) 2
  Mt= 8
  + 8
  
  29.167 (360 )2 (360 )2
  
  2,067,811.2 lb-in = 8
  + 8
  
  29.167 (360 )2 (360 )2
  
  2,067,811.2 lb-in = 8
  + 8
  
  (360 )2
  2,067,811.2 lb-in = 472,505.4 lb-in +
  8
  
  (360 )2
  1,595,305.8 lb-in = 8
  
  = 98.476 or
  
  1 12
  = 98.476 (1000 )( 1
  ) = 1. 182
  
  ( the least governs)!!
  
  Therefore, = = 1. 182 ----------------- answer