Name: Saifullah Khan Reg # : 11-CT-792 BS.Tech Civil Engineering 6th Semester, UCEST Lahore Leads University Assignment # 1 Question: A prestressed concrete sleeper produced by pre-tensioning method has a rectangular cross-

section of 300mm 250 mm (b h). It is prestressed with 9 numbers of straight 7mm

diameter wires at 0.8 times the ultimate strength of 1570 N/mm2. Estimate the percentage

loss of stress due to elastic shortening of concrete. Consider m= 6.

Solution: 11.8 Area of Concrete = 9.85 11.8

Area of Concrete = 116.23inch

Area of Prestressing steel AP 9.85

Dia of Bar = 0.275inch

Area of 1 bar = (/4) d2

Area of 1 bar = (3.14/4) (0.275)2

Area of 1 bar = 0.596

Area of Prestressing steel AP = 0.596 9

M. I =11.8 (9.85)

12

C. G. C =9.85

12

C. G. C = 4.925

=)

=. ( . ). (. )

.

=..

.

= 4.55inch = . . . .

= 4.925 4.550

Pre-stress force Pi = 0.8 227650 0.536

Pre-stress force Pi = 97616.32/1000

Pre-stress force Pi = 97.616 Kip

fy (top) =

+

.

fy (top) = .

. .+

. .

.. 3.38

fy (top) = -709.35 Psi

fy (bottom) =

+

.

fy (bottom) = .

. .

. .

.. 3.38

fy (bottom) = - 884.52 Psi

Loss of Prestressed at top = mfcAP

Loss of Prestressed at top = 6 516.47 (4 0.0596)

Loss of Prestressed at top = - 738.758 lbs.

Loss of Prestressed at Bottom = mfcAP

Loss of Prestressed at Bottom = 6 884.52 (5 0.0596)

Loss of Prestressed at Bottom = - 1581.52 lbs.

% Loss =

% Loss = ..

.

% Loss = 2.37%

Name: Saifullah Khan Reg # : 11-CT-792 BS.Tech Civil Engineering 6th Semester, UCEST Lahore Leads University Assignment # 2 Question: A post-tensioned beam 4 inch 12 inch (b h) spanning over 10 m is stressed by

successive tensioning and anchoring of3 cables A, B, and C respectively as shown in figure.

Each cable has cross section area of 0.32 inch2 and has initial stress of 1200 MPa. If the

cables are tensioned from one end, estimate the percentage loss in each cable due to friction

at the anchored end. Assume = 0.35, k= 0.000457 / ft.

Given Data b = 4 , d = 12 , L = 32.81 ft.

Area of the Cable = 0.32 in2

Pi = 1200 MPa

Po = 174000 Psi

= 0.35

K = 0.0015/3.281 = 0.000457/ft.

Solution:

Curvature of tendon = =

Po = A Pi

Po = 0.32 174000

Po = 55680 lbs.

For Cable A

A = ( . )

.

A = 0.08 For Cable B

B = ( .)

.

B = 0.04 For Cable C

C = ( )

.

C = 0

For Cable A

xA = A + kx xA = 0.35 x 0.08 0.000457 32.8 xA = 0.043 For Cable B

xB = B + kx xB = 0.35 x 0.04 0.000457 32.8 xB = 0.029 For Cable C

xC = C + kx xC = 0.35 x 0 0.000457 32.8 xC = 0.015

Maximum losses due to friction in Cable A =

= .

= 0.958

Maximum losses due to friction in Cable B =

= .

= 0.971

Maximum losses due to friction in Cable C =

= .

= 0.985

% Losses = (1- ) 100

% Losses for cable A = (1- 0.958) 100 = 4.2%

% Losses for cable B = (1- 0.971) 100 = 2.9%

% Losses for cable C = (1- 0.985) 100 = 1.5%

Assignment # 1Question:Assignment # 2Question: