shear component of prestress by equivalent loads

David DarwinAssistant Professor of Civil EngineeringUniversity of KansasLawrence, Kansas

Shear Componentof Prestress byEquivalent Loads

The design of prestressed concretebeams for shear using the ACI

Building Code' requires the calcula-tion of the vertical or shear componentof the effective prestress, V.

In most cases, the tendon profile isflat enough so that V P sin a) maybe approximated by P tan a = P dy/dx(Fig. 1), where P equals the effectiveprestress force and a equals the anglemade by the centroid of the prestress-ing steel and the horizontal.

In the past, the common design pro-cedure has been to calculate dy/dx,the slope of the tendon, at points alongthe beam. While this procedure hasproved satisfactory, it is tedious andignores the effects of induced reactionsin continuous beams. An alternativeapproach, presented here, uses the con-cept of "equivalent loading" to obtain

the shear component of the prestress.The equivalent loading concept has

been used successfully for over twodecades for flexural design and is anintegral part of the load balancingtechnique .3

While this technique is used bymany structural engineers, the applica-

Y

x

r Pcosa=P

V.-Psina PP tan a=P dX

Fig. 1. Vertical component ofprestress, Vp, for an inclinedtendon.

64

A design procedure is presented that usesthe concept of equivalent loading to obtainthe shear component of prestress.The prestressing tendons are replaced by aset of equivalent loads.The resulting shear diagram includes theeffects of both the vertical component of theprestress and the induced reactions (if any).Several examples are presented to illustratethe method.The proposed procedure allows the designerto use familiar analysis techniques, reducethe tedium involved in shear design, andobtain a clear picture of the forces in theconcrete.

bility of this approach to shear designis not widely recognized.

Using the procedures presented inthis paper, the designer can quicklycalculate the shear component of theprestress anywhere in the beam simplyby drawing the shear diagram for theequivalent loads. The procedure ap-plies to continuous beams as well assimple spans and automatically in-cludes the effects of induced momentsand reactions due to non-concordanttendons.*

Shear by Equivalent Loads

Using the equivalent load concept, thedesigner looks at the prestressing steelas a cable applying loads to the con-crete. The prestress is represented asa system of equivalent vertical loads

along the beam length (Figs. 2a and2b), and as concentrated moments atthe supports (Fig. 2c).

The equivalent forces on the beamcause a shear in the concrete. Theshear diagram for the equivalent loadsrepresents the shear component of theprestress, V.

In this paper a distinction will bemade between the vertical componentof prestress, P tan a, and the shearcomponent of prestress which also in-cludes the effect of induced reactionsin continuous beams. The distinction isimportant only for continuous beamswith non-concordant tendons.

*A concordant tendon is so located as to pro-duce a line of compressive force in the con-crete at each section that coincides with thecenter of gravity of the steel (cgs). A non-concordant tendon produces a line of compres-sive force that does not coincide with the cgs.

PCI JOURNAL/March-April 1977 65

Prestressed Beams Equivalent Loads

____ -] JP - U niform Load

-- — L 8PhWP = ^2

a) Parabolic Tendon

- P Concentrated Load P

h

L F = 4Ph

b) Bent Tendon L

E31 Support Momentcgc - - - -

ec) End Eccentricity

M = Pe

Fig. 2. Equivalent loads due to prestressing.

Losses due to friction will not be in-cluded in the discussion that follows,but losses should be included in calcu-lations when they are substantial.

Equivalence to P dy/d'xfor Simple Spans

A short example will help to dem-onstrate the equivalence of the pro-posed method with the more commonprocedure of taking Vp approximatelyequal to P tan a = P dy/dx.

Example 1The simply supported beam of length

L, shown in Fig. 3a, has a parabolictendon with sag, h, and end eccen-

tricities, ea and eb at the left and rightsupports, respectively.

For an effective prestress, F, thetendon exerts a uniform upward loadof wp = 8Ph/L2 and concentrated endmoments of Ma = Pe,, and Mb = Pebon the beam as shown in Fig. 3b.

The shear due to the equivalentloads (Fig. 3c) is:

Vp=- 2wpL+wx-}- Mv L MQ (1)

For the purpose of comparison, Eq.(1) is rewritten in terms of the prestressforce and the tendon geometry:

8Ph L 8Ph Peb — Pe„VP-- L= • 2 + L2 x+ L(2a)

66

ea

L,.

a) Prestressed Beam with Parabolic Tendon and Support Eccentricities

P P

hwp = L2 Ma = Pea Mb = Peb

b)Equivalent Loads+Wp L

2

VP

2L

Vp=_pL +WPX+ Mb - Ma

L

Mb- Ma

L

c) Shear Diagrams for Equivalent Loads

Fig. 3. Diagrams for simply supported beam (Example 1).

from which

4h 8hx eb — e^V^ ° P (— L -{- L.^ -{- L 1 (2b)

Eq. (2b) may be compared to thevalue of Vp = P dy/dx.

The tendon profile in Fig. 3a is de-scribed by the equation:

= LL (x 2̀ — xL) -} - L e ,) x e«, (3)

Differentiating and rearranging terms:

V^ —Pdy—P .4h+8hx+eb—e,,\Ldx ( L L-

(4)

which is identical to the value ob-tained in Eq. (2b) using the proposedprocedure.

The relative advantages of the pro-posed procedure become distinct whenthe usable results for the equivalentloading, Eq. (1) and Fig. 3c, are com-pared with Eq. (4). Fig. 3c is all thatmost designers need. Fig. 3c and Eq.(1) give a clearer picture of the forcesin the concrete than is provided byEq. (4).

Also, importantly, the proposed pro-cedure allows the designer to use fa-miliar analysis techniques to obtain V.


Calculations to obtain the slope of thetendons, which are tedious in com-parison, are no longer required.

As attractive as the approach ap-pears for simple spans, it offers addedadvantages for continuous beams.

Continuous Beams

Two approaches may be taken toobtain the shear in concrete due toprestressing in continuous beams.

Approach 1—The first approach ob-tains the shear by considering the en-tire continuous beam under equivalentloading. It may be used for beams withconcordant or non-concordant tendonsand automatically includes the shearresulting from induced reactions.

Approach 2—The second approachfollows the procedure outlined in Ex-ample 1 and considers the individualspans as simple spans subjected toequivalent loading. This approach, likethe first, may be used for beams withconcordant or non-concordant tendons;but for non-concordant tendons, ityields results that do not include theeffects of induced reactions.

Therefore, for non-concordant ten-dons it applies only when plastichinges form and full moment redis-tribution takes place at ultimate load.

Concordant tendonsThe two approaches are equivalent

for beams with concordant tendons,because in such beams no support re-actions are produced by the prestress.Both approaches produce identical re-sults with those obtained with themore conventional approach, V, = Pdy/dx.

Example 2 illustrates the two ap-proaches for a continuous beam withconcordant tendons.

Example 2A symmetric two span beam with

concordant parabolic tendons is shown

in Fig. 4a. For the left span, the cableis described by:

y = h ('— L -I- Lz ) (5)

The equivalent upward uniform loadis:

wp L~h (6)

1. The first approach considers theentire beam subject to an upward load,wp. The shear diagram for the left spanis easily obtained (Fig. 4b) and gives:

VV = — $ wpL -I- wpx (7)

2. The second approach considersequivalent loads on the individualspans. For the left span, the loads con-sist of an upward uniform load, wp,and a counterclockwise support mo-ment at Point B, Mb = Ph.

The second approach provides re-sults shown in Fig. 4c and Eq. (8).

V11 wpL + wpx • 1 Ph (8)

From Eq. (6):

Ph = wg (9)

Eq. (8) then becomes:

Vp =-8wPL-I-wpx (10)

which is identical to that obtained inEq. (7) using the first approach.

3. The results in Eqs. (7) and (10)are also equivalent to P dy/dx:

Vz,=P dx =Ph(_- +8) (11a)

wLL/ – 3 8x8 L +L2

_ – $ wPL -}- wpx (11b)

68

YI I—eb=h e, =ec=0

/ - cgc

Af - L—hL L

a) Two Span Beam with Concordant Parabolic Tendons

Equivalent Loads 18Ph

Wp L2

5 8 3wp L8

VP

-3wp L -5vv L8 8 Vp = -3 8 + wp x (Left Span)

b) First Approach

^ quivalent Loads M , - PhI

wP - 8 2h

SwPL

VP

-38 P L VP = _2L+WpX+ PL = -38PL +wex

c) Second Approach, Left Span

Fig. 4. Diagrams for continuous beam with concordant tendon (Example 2).


Tendons--_-- Concordant

Non-concordant

A n.

a) Linear Transformation of Tendon at B+PAeb

LVP

b) Change in P dx-PpebL P6eb

M s ^\

c) Induced Secondary Moment, M5+PAeb

L

VP-P6eb

L

d) Shear Caused by Induced Reactions

Fig. 5. Effects of linear transformation of concordant tendon on shear com-ponent of prestress.

Non-concordant tendons

Non-concordant tendons are of spe-cial interest because the prestressingforce induces support reactions, shearsand moments in the beam. The lineof compressive force does not coincidewith the center of gravity of steel asit does for concordant tendons.

The induced reactions constitute areal load on the beam, and inducedshears and moments must be includedin design calculations. The induced re-actions are based on an elastic analysis.

If full moment redistribution oc-curs at ultimate load (unusual for mostconstruction), elastic analysis is nolonger applicable. The procedures out-lined above for concordant tendons areeasily adapted to include induced re-

actions or to exclude them in the caseof full moment redistribution.

For elastic behavior, the shearcaused by induced reactions is auto-matically accounted for by using equiv-alent loads. This is demonstrated bythe continuous prestressed concretebeam shown in Fig. 5a.

In the figure, the concordant tendonin Spans AB and BC is linearly trans-formed by reducing the support eccen-tricity at B by an amount Aeb. The re-sult is a non-concordant tendon. Thechange in the value of the verticalcomponent of the prestress, P dy/dx,due to the change in tendon profileequals — PAeb/L and + Pleb/L onSpans AB and BC, respectively (Fig.5b).

rill

The linear transformation has no ef-fect on the location of the line of com-pressive force, but it does cause asecondary moment (Fig. 5c) and a sec-ondary shear (Fig. 5d) due to the in-duced support reactions. The inducedshear equals + PAeb/L and — PAeb/Lon Spans AB and BC, respectively.The induced shear exactly cancels thechange in the vertical component ofthe prestress.

Therefore, the total shear effect ofprestressing remains unchanged and isnot affected by the linear transforma-tion. Since the equivalent loads on thecontinuous beam are not influenced bythe linear transformation of the ten-don, the shear diagram produced forthe equivalent loading will be the samefor concordant and non-concordanttendons and will automatically includeany induced shear.

If full moment redistribution takesplace at ultimate load, the induced re-actions disappear at failure. In such acase the shear component of prestresswill not include induced shear. V. maythen be calculated either by using theconventional approach or by consider-ing each span individually (Approach2), as described in Part 2 of the nextexample.

Although full moment redistribu-tion is rarely used in design, the useof equivalent loads to obtain Vp is justas applicable for plastic design as itis for elastic design.

The amount of moment redistribu-tion actually obtained will vary, de-pending on the type of constructionused. Research by Mattock, Yamazaki,and Kattula4 indicated that beams withunbonded tendons, and an adequateamount of bonded, non-prestressed re-inforcement may exhibit somewhatmore inelastic moment redistributionthan similar beams with bonded ten-dons.

In either case, when the amount ofmoment redistribution is limited to that

allowed by the ACI Code,' the de-signer should use Approach 1 for cal-culating V.

At ultimate load, a properly propor-tioned beam will fail in flexure, notshear. Therefore, the value of prestressused to calculate the equivalent loadsfor shear design should, logically, re-flect the tendon force at the ultimateflexural load. In beams with unbondedtendons, the prestress force will in-crease as the ultimate bending mo-ment is attained at the critical sections.

The increase in prestress is almostuniform throughout the beam due totendon slip and may be estimatedusing ACI Code Eq. (18-4):

fp8 = fse + 10,000 + 100pP

where

fps = stress in prestressing steel atdesign load

fs ^ = effective stress in prestressingsteel

f', = strength of concrete

pp = ratio of prestressed reinforce-ment

This increased stress may be in-cluded in the calculation of the equiv-alent loads and V. In most beams,however, the increase may be con-servatively neglected. For bonded ten-dons, the increase in prestress is quitelocalized and should be neglected forshear design.

Example 3 illustrates the use ofequivalent loads to obtain the shearcomponent of prestress for a beam withnon-concordant tendons. It demon-strates the simplicity with which theproposed procedures may be appliedto shear design.

Example 3A symmetric two-span continuous

T-beam with bonded tendons is shownin Fig. 6a. 5 The beam has non-con-


P = 600K 12"

r-- eb = 8" r-- h = 36" ^

NI 32" / ^.^^Parabolic // 36"

A____________ 80' — 80' C'

a)Two Span Beam with Non-concordant Tendons

Equivalent Loads

ffftf ffffff ftfffffwP = R—' -2.25K/ft

SwpL-112.5K67.5K

VP 8

-38 =- 67• 5 K -112.5'

b) Elastic Behavior

Equivalent Loads My - Pe b - 400 ft - K

It ft H if?w, - 2.25 K/ft

^_ go Mb-5KVP - + L

95K

2_85K

c) Full Moment Redistribution, Left Span

Fig. 6. Diagrams for continuous beam with non-concordant tendon(Example 3).

72

cordant parabolic tendons with sag,h = 3 ft, and prestress force, P = 600kips.

The total upward load is:

8Ph8X600X3wp = = L= 802

= 2.25 kips per ft

1. The shear component of the pre-stress, Vi,, is obtained for elastic be-havior (i.e., less than full moment re-distribution) by calculating the shearin the beam due to the load wp. Thesolution (Fig. 6b) shows that V, varieslinearly from — 3w5L,/8 = — 67.5 kipsat the left support to + 5wPL/8 =112.5 kips at the center support. Ascalculated, Vp includes the effects ofthe induced support reactions.

2. Should the beam be designed forfull moment redistribution at ultimateload, V, may be calculated usingequivalent loads on the individualspans. As shown in Fig. 6c, each spanis considered as a simply supportedspan for purposes of the calculations.Since the tendon is bonded, the in-crease in P, near ultimate can be ne-glected.

Looking at the left span, the equiv-alent loads consist of a uniform load,wp = 2.25 kips per ft, and a supportmoment:

AI I, = Pe,, = 600 X 8/12 = 400 ft-kips

The shear diagrams produced bythese loads are also shown in Fig. 6c.The final results show that Vp varieslinearly from — w,L/2 + M b/L =— 85 kips to + w5L/2 + Mb/L =95 kips. These values are identical withthose obtained from V, = P dy/dx anddo not include induced support reac-tions.

The differences in Solutions 1 and 2are due to the induced support reac-tions. Since most beams are propor-tioned for flexure based on elastic be-

havior, Solution 1 represents the typeof solution that will be commonly usedin practice. Solution 1 should, also, beused if partial moment redistributionis planned, as allowed by the ACICode?

If the beam in the example utilizesunbonded, rather than bonded tendons,the analysis procedures will be identi-caI to those presented above, with theexception that the effective prestress,P, will increase as the ultimate load isattained.

Assuming that fRe = 150,000 psi, f'= 5000 psi and p, = 0.01, and usingEq. (18-4) of the ACI Code:'

p8 =1 C + 10,000 + looPP

=150,000 + 10,000 + 5000100 X 0.01

=165,000 psi

The value of P at ultimate is, there-fore P = (165/150)600 = 660 kips.

This results in a 10 percent increasein the equivalent loads and in the shearcomponent of prestress, V5.

Reversed curvatureThe beams presented in the preced-

ing sections have harped tendons atthe supports for illustrative purposes.In practice, the sharp angle change isreplaced by a portion of the tendon inwhich the direction is graduallychanged over the supports (Figs. 7and 8).

The portion of tendon with the re-versed curvature exerts a downwarddistributed load on the beam. The ef-fect of the reversed curvature must betaken into account in shear design.

Equivalent loads are illustrated forparabolic tendons with reversed curva-ture in Figs. 7 and 8.6 , 7 For an interiorspan, the uniform upward load is:

_ 8Pcwp (1 — 2a)L-1 (12)


I— L; = Span I.

a) Tendon Profile Geometry

1. • L;

b) Equivalent Loads

Fig. 7. Typical interior span.

wherec = drape of tendon profile, high

point to low pointa = ratio of reverse curve length

span length hd

The uniform downward load for thereversed segment is:

w _ 1-2a w (13)

' 2a ^°

An exterior span is likely to have achange in curvature near the midspanwhich results in different upward loadson either side of the low point in thetendon profile. For the exterior spanshown in Fig. 8, the upward uniformload due to the external parabolic seg-ment is:

2Pdwee = b2L2f (14)

whered = drape of tendon profile at end

segment of exterior spanb = ratio of end segment length to

span length in exterior span, L^.

The balance of the uniform upwardload is:

_ 2Pc

WP 1 b)(1--b—a)L2, (15)

The downward load for the reversedsegment is given by Eq. (13).

These equivalent Ioads are used tocalculate the shear component of the

74

e

NeutralAxis

i--- bL e —'

1

a) Tendon Profile Geometry

I--- aLe--

Le

H1U1W,

M

ft ft t ff1 } t ft ft tttlWpe Wp

bLe

-- aLe-

L

b) Equivalent Loads

Fig. 8. Typical exterior span.

prestress, as is demonstrated in Ex- drape, c = 40 in., upward load is:ample 4. 2

Example 4The symmetric two-span beam in

Example 3 is redesigned with a re-verse curve in the tendon over thecenter support as shown in Fig. 9a.For the external segment, with tendondrape, d = 32 in., and prestress force,P = 600 kips, the upward uniform loadis:

2Pd _ 2 X 600 x 32/12w°` _ b=L2^ (0.50)2 x 802

= 2.0 kips per ft

For . the inner segment, with tendon

_ Pcwp (1 - b) (1 - b - a)L20

_ 2 x 600 x 40/12(0.50) (0.40) x 802

= 3.125 kips per ftThe downward load for the re-

versed segment is:

_1-2aW. 2a wP

_1-2x0.10 X2x0.10 3.125

= 12.5 kips per ftAs before, Vp can be obtained by

PCI JOURNAL/March -April 1977 75

P-600`

12"

a) Two Span Beam with Reversed Curvature

w,-12.5K/ft

Equivalent Loads 1ft ltftt tI li t tftfWP. - 2.O K/ft wP - 3.125 K/ft

.115.7K

V 15.7K 15.7KP L----64.3K

b) Elastic Behavior, Left Span

wr-125K/ft

Equivalent Loads ) Mb - 400 ft - K

I t it t twp,-2.0K/ft wp -3.125Klft

100KOK

VPOK

-8oK

c) Full Moment Redistribution, Left Span

Fig. 9. Diagrams for continuous beam with reversed curvature (Example 4).

76

constructing the shear diagram for thebeam subjected to the equivalent loads.

The calculation of Vp for beams withreversed curvature can be greatly sim-plified by using the moment coeffi-cients provided in References 6 and 7.The solution, shown for the left span(Fig. 9b), includes the shear due to theinduced reactions.

The effect of the change in the ten-don profile on V, may be seen by com-paring Figs. 6b and 9b. The shear dia-gram for full moment distribution (Fig.9c) can be obtained by loading eachspan individually, using Approach 2.

Conclusions

1. In a prestressed concrete beam,Vp, the shear component of the pre-stress due to both the vertical compo-nent of prestress and the induced re-actions (if any), may be obtained byreplacing the tendon . with a set ofequivalent loads on the beam and cal-culating the shears so produced.

2. While the examples presented inthis paper are limited to parabolic ten-dons and two-span continuous beams,the methods illustrated may be appliedto any tendon profile.

3. The use of equivalent loads tocalculate the shear component of theprestress, Vi,, has several advantagesover the conventional method:

a. The proposed procedures allowthe designer to use familiar anal-ysis techniques to obtain V, andthereby reduce the tedium in-volved in shear design.

b. The use of equivalent loads givesa clearer picture of the forces inthe concrete.

c. The proposed method (Approach1) automatically includes the ef-

fects of induced reactions in con-tinuous beams with non-concor-dant tendons. The method isalso, easily adapted (Approach 2)for full moment redistribution.

References

1. ACI Committee 318, "Building CodeRequirements for Reinforced Concrete(ACI 318-71)," American Concrete In-stitute, Detroit, Michigan, 1971, 78 pp.

2. Moorman, R. B. B., "Equivalent LoadMethod for Analyzing Prestressed Con-crete Structures," ACI Journal, V. 23,No. 5, January 1952, pp. 405-416.

3. Lin, T. Y., "Load Balancing Methodfor Design and Analysis of PrestressedConcrete Structures," ACI Journal, V.60, No. 6, June 1963, pp. 719-742.

4. Mattock, A. H., Yamazaki, J., and Kat-tula, B. T., "Comparative Study of Pre-stressed Concrete Beams, With andWithout Bond," ACI Journal, V. 68,No. 2, February 1971, pp. 116-125.

5. Lin, T. Y., and Thornton, K., "Second-ary Moment and Moment Redistribu-tion in Continuous Prestressed Con-crete Beams," PCI JOURNAL, V. 17,No. 1, January-February 1972, pp. 8-20.

6. Post-Tensioning Manual, Post-Tension-ing Institute, Glenview, Illinois, 1976,288 pp.

7. Turula, P., and Freyermuth, C. L.,"Moment Influence Coefficients forContinuous Post-Tensioned Structures,"PCI JOURNAL, V. 17, No. 1, Janu-ary-February 1972, pp. 35-57.

Discussion of this paper is invited.Please forward your comments toPCI Headquarters by Sept. 1, 1977.


shear component of prestress by equivalent loads

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