pre-stressed concrete = pre-compression concrete · a prestressed concrete beam with an effective...
TRANSCRIPT
• Pre-stressed concrete = Pre-compression concrete
• Pre-compression stresses is applied at the place when tensile stress occur
• Concrete weak in tension but strong in compression
• Steel tendon is first stressed
• Concrete is then poured
• After harden and reaching the required strength, the steel tendon is cut
Pre-tension
• Concrete is first poured in the mould
• After harden, steel tendon is then stressed and anchored at both ends
Post-tension
Phase 1
Phase 2
Phase 3
Phase 1
Phase 2
Phase 3
Between C30 and C60
High workability during wet and high strength when hardened
Concrete strength at transfer, fci 0.6fck(t) N/mm2
Concrete
7 or 19 nos of strand
High strength steel
Steel strength usually 1650 N/mm2
Steel Tendon
Section 5.10.2.2(3): MS EN 1992-1-1: 2004
Stresses Loading StageTransfer Service
Symbol Value or Equation
Symbol Value or Equation
Compressive fct 0.6fck (t) fcs 0.6fck
Tensile ftt fctm fts 0
Some losses are immediate affecting the prestress force as soon as it
transferred. Other losses occur gradually with time. Known as short-term
and long-term losses, they include:
Short-term
(a)Elastic shortening of the concrete
(b)Slip or movement of tendons at anchorage
(c) Relaxation of prestressing steel
(d)Friction at the bend due to curvature of tendons
Long-term
(a)Creep and shrinkage of the concrete under sustained compression
(b)Relaxation of the prestressing steel under sustained tension
Top fibre
Bottom fibre
e
N.A
yb
yt
h
b
Prestressing
tendon
Ungrouted duct
Reinforcement /
bonded tendon
Mmax = wL2/8 = 80 kNm
8 m
w = 10 kN/m
Tensile stress, ft = fc = Mmax/Z = Mmaxy/I
where I = bh3/12 and yt = yb = h/2
Therefore;
ft = fc = (wL2/8)(h/2) (bh3/12)
= 9.65 N/mm2
yt
ft
fcb = 200 mm
h=
500 m
m
yb_
+
P P
Prestress force, P is then applied to eliminate
tensile stress
N.A
ft = 9.64 N/mm2
fc = 9.65 N/mm2
_
+
Stress due to
loading
=
0 N/mm2
14 N/mm2
Total stress
P/A
+
Stress due to
force P
+
9.65 + P/A = 0
Prestress force, P
required to eliminate
tensile stress
= 9.65A
= 965 kN
P P
If the prestress force, P applied is not at the centroid
eN.A
ft = 9.65 N/mm2
fc = 9.65 N/mm2
_
+
Stress due to
loading
=
0 N/mm2
14 N/mm2
Total stress
P/A
+
Stress due to force P
Pe/Zb
++
+
9.65 + P/A + Pe/Z = 0
Prestress force, P required to eliminate tensile stress
= 9.65 / [(1/A) + (e/Z)] = 283 kN
Pe/Zt
(a) At Transfer
P P
Mi/Zb
Mi/Zt
_
+
Stress due to
self weight
e
P/A
+
Stress due to force P
Pe/Zb
++
+
Pe/ZtP/A
=
f1t ftt
Total stress
f2t fct
+
Mi = Moment due to self weight
= Coefficient of short term losses
(a) At Service
P P
Mi/Zb
Mi/Zt
_
+
Stress due to
self weight
e
P/A
+
Stress due to force P
Pe/Zb
++
+
Pe/ZtP/A
ws kN/m (Service Load)
Stress due to
ws kN/m
Ms/Zb
Ms/Zt
_
+
+ =
f1s fcs
Total stress
f2s fts
+
= Coefficient of long term losses
From both Figures:
At Transfer
Mi/Zt + P/A – Pe/Zt –ftt ---------- (1)
–Mi/Zb + P/A + Pe/Zb fct ---------- (2)
At Service
Mi/Zt + Ms/Zt + P/A – Pe/Zt fcs ---------- (3)
–Mi/Zb – Ms/Zb + P/A + Pe/Zb –fts ---------- (4)
Rearranging Eqs. (1) – (4):
P/A(eA/Zt – 1) – Mi/Zt ftt ---------- (5)
P/A(eA/Zb + 1) – Mi/Zb fct ---------- (6)
P/A(1 – eA/Zt) + Mi/Zt + Ms/Zt fcs ---------- (7)
–P/A(1 + eA/Zb) + Mi/Zb + Ms/Zb fts ---------- (8)
Rectangular beam = b h = 300 950 mm. Prestress force, P =
2000 kN acted at e = 200 mm below centroid. The short-term
and long-term losses of the prestress are 10% and 20%,
respectively.
(a) If fck = 40 N/mm2 draw the stress distribution diagram at mid-
span during transfer and service. Also check that these
stresses are within the allowable limits.
P Pe
ws = 20 kN/m
15 m
Cross sectional area, Ac = b h
= 300 950
= 2.85 105 mm2
Moment of inertia, Ixx = bh3/12
= 300 9503/12
= 2.14 1010 mm4
Modulus of section, Zt = Zb = Ixx/y
= 2.14 1010 / 475
= 4.51 107 mm3
Stress limit for fck = 40 N/mm2 and fci = 28 N/mm2
At Transfer
fct = 0.6fck(t) = 0.6(28) = 16.8 N/mm2
ftt = fctm = 0.30 (28)2/3 = 2.77 N/mm2
At Service
fcs = 0.6fck = 0.6(40) = 24.0 N/mm2
fts = 0.0 N/mm2
Stress Distribution
At Transfer
Self weight, wi = (Ac)(25) = (2.85 105)(25) 10-6 = 7.12 kN/m
Mi at mid-span = wiL2/8 = (7.12)(152)/8 = 200.2 kNm
Short-term losses, = (1 – 0.10) = 0.90
Stress at top fibre
f1t = Mi/Zt + P/A – Pe/Zt
= 5.20 N/mm2
Stress at bottom fibre
f2t = – Mi/Zb + P/A + Pe/Zb
= 17.84 N/mm2
PASS
FAILf2t (17.84) fct (16.8)
f1t (5.20) ftt ( 2.77)
+
_
Stress Distribution
At Service
Service load, ws = 20 kN/m
Ms at mid-span = wsL2/8 = (20)(152)/8 = 562.5 kNm
Long-term losses, = (1 – 0.20) = 0.80
Stress at top fibre
f1s = Mi/Zt + Ms/Zt + P/A – Pe/Zt
= 8.33 N/mm2
Stress at bottom fibre
f2s = –Mi/Zb – Ms/Zb + P/A + Pe/Zb
= 3.23 N/mm2
PASS
PASSf1s (8.33) fcs (24.0)
f2k (3.23) fts (0)
+
_
Eq. (5) + Eq. (7)
Mi/Zt – Mi/Zt + Ms/Zt (fcs + ftt)
Zt ( – )Mi + Ms (fcs + ftt) ---------- (9)
Eq. (6) + Eq. (8)
Mi/Zb – Mi/Zb + Ms/Zb (fts + fct)
Zb ( – )Mi + Ms (fts + fct) ---------- (10)
A prestressed concrete beam with an effective length of 20 m is
simply supported at both ends. During service, a characteristics
load 20 kN/m is applied to the beam apart from its self weight. The
concrete strength is 50 N/mm2 and the transfer is done when the
concrete achieve the strength of 30 N/mm2. The prestressing force
applied is 2000 kN at the eccentricity of 500 mm at mid-span. The
short-term and long-term losses is 10% and 20%, respectively.
Design the suitable beam cross section if a rectangular section is
used.
Stress limit;
At Transfer
fct = 0.6fck(t) = 0.6(30) = 18.0 N/mm2
ftt = fctm = 0.30 (30)2/3 = 2.90 N/mm2
At Service
fcs = 0.6fck = 0.6(50) = 30.0 N/mm2
fts = 0.0 N/mm2
Ms = (20) 202/8 = 1000 kNm (not including the self weight)
From Eq. (9):
(fcs + ftt) = 0.9(30) + 0.8(2.90) = 29.32 N/mm2
Zt ( – )Mi + Ms (fcs + ftt)
Zt (0.1Mi + 900) 106 / 29.32
From Eq. (10):
(fts + fct) = 0.9(0) + 0.8(18.0) = 14.4 N/mm2
Zb ( – )Mi + Ms (fts + fct)
Zb (0.1Mi + 900) 106 / 14.4
Since Zb Zt, only Zb is used & checked to find
the suitable cross section of the beam
For rectangular cross section
Zt = Zb = Ixx/y = bh3/12 (h/2) = bh2/6
Try 300 mm 1000 mm
Self weight, wi = 0.3 1.0 25 = 7.5 kN/m
Moment due to self weight, Mi = 375 kNm
Zb (0.1 375 + 900) 106 / 14.4
65.1 106 mm3
Z = 300 10002/6 = 50 106 mm3 Zb Increase size
Try 300 mm 1400 mm
Self weight, wi = 0.3 1.4 25 = 10.5 kN/m
Moment due to self weight, Mi = 525 kNm
Zb (0.1 525 + 900) 106 / 14.4
66.1 106 mm3
Z = 300 14002/6 = 98.0 106 mm3 Zb OK
TA
RM
AC
TO
PF
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TA
RM
AC
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