Structural Analysis of a Prestressed Concrete Beam Under Load

This assignment solution details the analysis of a prestressed concrete member, calculating stresses in concrete, steel, and strands immediately after release. It determines the change in length due to strain, the axial load required to crack the concrete, and the length change associated with that load. The solution also estimates the average crack width and considers the effects of creep, shrinkage, and relaxation on the member over a five-year period without external loads. Finally, it calculates the maximum uniformly distributed load the beam can support based on its flexural capacity, providing a comprehensive structural analysis.

Provided information:
Strain ε =7∗10−3
Reinforcement: four 20M rebars
Four ½ inch diameter 7-wire strands.
Length: L=6200 mm
Use a linear-elastic relationship for concrete in compression.
a) At strand release the young’s modulus of the concrete is 30000 MPa. Calculate the
stresses in the concrete, steel and strands immediately after release.
Concrete stress σ c=Ec∗εc
σ c=30000∗−2.0∗10−3=−60 MPa
The stress resultant of the concrete Nc=σc∗Ac
Ac= A− Ap
Ac= ( 0.3∗0.3∗106 ) − ( 4∗300 )=0.0888 m2
Hence, Nc=−60∗8880=−5328000 MPa
Stress in the strands σ pm=ε p∗Ep
¿ 7∗10−3∗200,000
¿ 1400 MPa
b)
Strain ε = ∆ L
L
7∗10−3 = ∆ L
6200
∆ L=7∗10−3∗6200=43.4 mm
c) What axial load would be required to crack the concrete in tension immediately after
release and by how much would the length of the member change.
Ultimate stress f ps=f py (1− γ
β ρ p
f py
f c ' )

f ps=1600
0.9 (1−
0.28
0.80∗( 4∗0.153
300∗300 )∗1600
35 )=1777.73 MPa
f ps= P
A
Thus, the axial load P=1773.73∗300∗300=1.599∗1082 N
Strain ε = σ
E =1777.73
30,000 =0.0593
Change in length ∆ L=ε∗L
∆ L=0.0593∗6200=367.40 mm
d) ∆ L=4 mm
Strain ε = ∆ L
L = 4
6200 =6.452∗10−4
Stress σ =εE=6.452∗10− 4∗30000=19.355 MPa
Load P=σ∗A=19.355∗300∗300=1.741∗106 N
Average crack width W =3.4 c +0.425 k1 k2
∅
ρeff
¿ 3.4∗( 50− 12.7
2 ) + 0.425∗0.8∗1.0∗12.7
( 300∗10−6 )
0.0888
¿ 242.22 mm
e) No external loads for 5 years
Creep coefficient = 2.1
Shrinkage of concrete = −0.46∗10−3
Relaxation strands = 3%
σ ( t0 ) =εcσ ( t , t0 ) ( 1
E0 ,t 0
+ φ( t , t0 )
E0,28 )−1

¿ 0.03 ( 1
E + 2.1
−0.46∗10−3 )
Thus for concrete,
σ =0.03 ( 1
30000 −4565.22 )=−136.957 MPa
For steel,
σ =0.03 ( 1
200000 −4565.22 )=−136.957 MPa
For strands;
σ =0.03 ( 1
200000 −4565.22 )=−136.957 MPa
ε = σ
E =−136.957
30000 =−0.004565
∆ L=εL=−0.004565∗6200=−28.304 mm
Question 2
Suppose the member is used to support a uniformly distributed load over a span of 6.0m. what is
the maximum value of the load that the beam could support based on its flexural capacity
M max = w l2
8
M n=ρ bd2 f y ( 1−0.59 ρ f y
f c ' )
M n=1777.73∗300∗3002
(1− 0.59∗1777.73∗400
35 )=−1.9177∗1012 Nmm

1.9177∗1012= w ( 62002 )
8
w=399106.174 N
w=399.106 kN