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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
The text passage contains a detailed problem statement and solution for a structural engineering problem involving the calculation of stresses in a prestressed concrete beam. The problem is broken down into multiple parts, each with a specific question and a corresponding answer.
multiple_questions_with_answers
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
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- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
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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
user
The text passage contains a detailed problem statement and solution for a structural engineering problem involving the calculation of stresses in a prestressed concrete beam. The problem is broken down into multiple parts, each with a specific question and a corresponding answer.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS