CIVE1151 Lect 3 eqivalent load PDF

Preview

Summary

Equivalent Load Concept in Prestressed Concrete • Week 3 Flexure in beams with prestress Chapter 6 text Warner et al. Equivalent load method can be used to evaluate the effects of prestress. This approach provides the basis for a design technique called load balancing, which is useful in the prelimi...

Description

Equivalent Load Concept in Prestressed Concrete • Week 3 Flexure in beams with prestress Chapter 6 text Warner et al.

Equivalent load method can be used to evaluate the effects of prestress. This approach provides the basis for a design technique called load balancing, which is useful in the preliminary design of prestressed concrete structures.

RMIT University©

Lecture Notes prepared by Dr Rebecca Gravina

Forces at Anchorages At its anchorage a prestressing cable exerts on the concrete an equivalent load whose magnitude is that of the prestressing force and whose direction is that of the cable at the anchorage



P

P

Psin Pcos

Equivalent load



P

components

P

For small angles

M=Pe

e

P

e P

equivalent load – eccentric anchorage RMIT University©

components

Lecture Notes prepared by Dr Rebecca Gravina

Equivalent system at centroidral axis

Draped Cables • Pretensioned beams – Tendons are either straight or connected by kinks as below

P

θ

e0

L/3

L/3

Angle at the ends of the cables θ= 3eo/L (assuming for small angles tan θ = θ) Forces applied at each end of beam consist of horizontal component P and vertical component P At a kink, concrete needs to apply a vertical force F = 3Peo/L to balance the cable forces. RMIT University©

Lecture Notes prepared by Dr Rebecca Gravina

Equivalent Loads from a draped cable after prestress 3Peo/L

3Peo/L P

P 3Peo/L

3Peo/L

Moment in mid-span from prestress = -3Peo/L x L/3 = -Peo This moment is same as that we used earlier in section analysis ie Mp=Peo , however, this concept has some advantages in selecting a prestress force

RMIT University©

Lecture Notes prepared by Dr Rebecca Gravina

Forces in curved cables Δθ = curvature x Δx ΔF=2Psin(Δθ/2) = P Δθ ΔF

ΔF/ Δx = P Δθ/Δx = Pk

P

Where k = curvature (change in P

slope per unit length)

Δx

Δθ = kx

ΔF

P

Force triangle

P

RMIT University©

Lecture Notes prepared by Dr Rebecca Gravina

Parabolic cables x h

y L

Equation for the parabola can be established based on a generic expression: y=ax2+bx+c and Boundary conditions, x=0, y=0 and x=L/2, y=h y = 4h[x/L-(x/L)2] Slope= dy/dx=θ=4h/L[1-2(x/L)] Curvature k = d2y/dx2 = -8h/L2 Force per unit length = wp = Pk = -8Ph/L2 wp is a uniformly distributed upward force exerted on the beam due to prestress RMIT University©

Lecture Notes prepared by Dr Rebecca Gravina

Equivalent load –Parabolic cable

P P

P P

wp L wp= 8Ph/L2

RMIT University©

Lecture Notes prepared by Dr Rebecca Gravina

Worked example- Calculation of stresses using the equivalent load method ex 6.2 text A prestressed concrete beam spans 10 m and has a rectangular cross section 800 mm deep and 400 mm wide. It is post-tensioned by a cable having a cross-sectional area of 1000 mm2. The cable eccentricity varies parabolically from zero at each end to a maximum of 250 mm at midspan, where the prestressing force is 1200 kN. f’c = 32 MPa Determine •

The stresses at mid-span due to prestress acting alone

•

The stress at midspan due to prestress, self weight and a 30 kN/m load

RMIT University©

Lecture Notes prepared by Dr Rebecca Gravina

Calculation of stresses using the equivalent load method P

h=250mm

P

wp =24kN/m

P=1200kN

P=1200kN

L P = 1200kN,  = 4h/L = (4 x 0.25)/10 = 0.1 rad Downward force at each anchorage: P = 0.1 x 1200 = 120kN  Udl due to prestress: wp= 8Ph/L2 = (8 x 1200 x 0.25)/ 102 = 24kN/m The stress resultants at mid-span due to equivalent loads are: Bending moment M = -(120 x 5) +(24 x 5 x 2.5) = -300kNm Axial Force N = 1200kN (Compression) The extreme fibre stresses due to prestress are therefore:

 tp 

P Pe 1200x10 3 1200x103 x 250x 200    3.75  7.03  3.28 MPa( tension) yt  Ag I g 320,000 17.07 x10 9

 bp 

P Pe  yt  3.75  7.03  10.78 MPa(compression) Ag I g

RMIT University©

Lecture Notes prepared by Dr Rebecca Gravina

The extreme fibre stresses due to prestress, self weight and live load: To determine conditions under the full load, we add the applied loads, 30 +8 = 38kN/m to the equivalent upward load of 24kN/m exerted by the prestressing cable. That is: Net load can be calculated as: Self weight = 8kN/m Live load = 30kN/m  Prestress equivalent load =wp =24kN/m  Net load = 8+30-24=14kN/m  At mid–span the bending moment is M = wl2/8 = (14 x 102)/8=175kNm

My 175x106 x 400   4.1MPa I 17.07 x109  t due to applied BM   4.1MPa  b due to applied BM   4.1MPa

stresses due to applied BM 

As before, the uniform compressive stress due to the axial prestress is 3.75MPa so that the total extreme fibre stresses are :  t  3.75  4.1MPa  7.85 MPa(compression)  b  3.75  4.1MPa  0.35MPa(tension)

RMIT University©

Lecture Notes prepared by Dr Rebecca Gravina

Load Balancing Determining the equivalent upward load due to prestress can be useful to determine the required prestress force to balance a given set of applied loads.

24kN/m

120kN 1200kN

120kN 1200kN

24kN/m

Two equal and opposite load systems acting on concrete 1200kN

1200kN

24kN/m

120kN

120kN Cable shown as freebody

Load balancing; internal equivalent loads are equal and opposite to a system of externally applied design loads RMIT University©

Lecture Notes prepared by Dr Rebecca Gravina

Prestressing Force • Up to this point we have simply used prestressing force, P as constant throughout length of beam. • In reality, loss of prestress force in tendon occurs immediately after tendons are stressed and continue to occur throughout the life of prestressed member • In week 5 you will be shown how to calculate loss of prestress at different stages, namely – Immediate loss – occurs immediately after transfer (includes friction loss) – Deferred loss - Time-dependent takes place gradually over time • Load balancing calculations are based on effective prestress, Pe after all losses have occurred. • Calculations of conditions immediately after transfer are based on initial prestress force, Pi

RMIT University©

Lecture Notes prepared by Dr Rebecca Gravina

Pj

Pj

Jacking Pj

RMIT University©

Transfer Immediate losses

Pi

End of design life Deferred losses

Lecture Notes prepared by Dr Rebecca Gravina

Pe

Worked example- Calculation of stresses using the equivalent load method ex 6.3 text

• A post-tensioned one-way slab is 300 mm thick and is required to support a superimposed load of 12kPa over a simply supported span of 10 m. Determine the prestress details to keep the slab level under a superimposed live load of 4 kPa. Also determine the load which will cause decompression of the bottom fibre of the beam. • Assume that the deferred losses of prestress is 18% and the friction loss at mid-span is 5%.

RMIT University©

Lecture Notes prepared by Dr Rebecca Gravina

Prestress force required to balance the self-weight. • Load to be balanced = self-weight. • Consider 1 m width of the slab and treat it as a beam.

300mm

30 + 16mm=46mm

With properties Area = Ag = 300,000mm2/m Second moment of area = Ig = 1000x3003/12 Self-weight = 0.3x1x25=7.5kPa Load to be balanced (ie Upward UDL required from prestress) = 7.5+4.0 = 11.5kPa The cable eccentricity = (300/2)- 46=104mm

wp 

8 Pe L2 wp L2

11.5  102 Pe   1382 kN / m  8e 8  0.104 This is the prestress force requried during service - after all prestress losses have occured Pe RMIT University©

Lecture Notes prepared by Dr Rebecca Gravina

• Since time dependent losses in this case reduce the initial value of the prestressing force by 18% the required initial prestressing force at mid-span is 1382  1685 kN / m 0.82 • The initial prestressing force at the anchorage, allowing for 5 % losses due to friction is: Pi 

Pj 

1685  1775kN / m 0.95

• A four strand cable containing 12.7mm super grade strands (minimum breaking load per 12.7mm strand is 184kN from table 3.3.1 of As3600-2009) stressed to 75 per cent of the minimum breaking stress provides an initial force of 522kN (=0.75*184*4). The required spacing is

1000 x 522  294, ie use 300mm 1775

RMIT University©

Lecture Notes prepared by Dr Rebecca Gravina

A Quick check to ascertain whether the beam can sustain stresses induced at transfer under the selected prestress force • AS3600 requires a rigorous calculation based on limit state design philosophy to check whether the beam is safe under the stresses induced at transfer. However, a simple stress check can usually be performed after selection of the prestress force to ensure that the beam will not fail at transfer. • Clause 8.1.6.2 Prestressed beam at transfer – The maximum compressive stress in the concrete, under the loads at transfer is not to exceed 0.5fcp • Clause 8.6.2 Crack control for flexure in prestressed beams at transfer and under service loads – Flexural cracking controlled if maximum tensile strength in the concrete does not exceed 0.25√f’c (Fully prestressed beam) – if this stress is exceeded need to provide steel reinforcement and limit the maximum flexural tensile stress under short-term service loads to 0.6√f’c (Partially prestressed beam) • This process establishes the upper limit of the jacking force allowed in a section. However, the design check at transfer still needs to be performed as per AS3600 requirement. RMIT University©

Lecture Notes prepared by Dr Rebecca Gravina...