July 2000 * The Indian Concrete Journal 1

Many structures such as microwave and transmission linetowers, industrial buildings, etc require that the superstructurebe secured safely into the foundations using anchor bolts.However, rational methods are not available in the Indiancode for calculating the tensile and shear load capacity of anchorbolts. Till recently, the methods suggested by the ACI 349-85code were used by designers. Based on extensive experimentalresults, empirical formulae have been suggested by researchersfrom Germany. An innovative method, called the concretecapacity design (CCD), has been found to give reasonableestimates of tensile and shear load capacity of anchor bolts.Moreover, the calculations made using the CCD method aresimple and less complex than the ACI code method. Forfoundations involving high strength concrete, which tend tobe brittle, linear fracture mechanics methods should be applied.In this method, instead of the tensile strength of concrete, thetotal crack formation energy has to be used to find the tensilecapacity of the anchor bolts. The article deals with the recentdevelopments in the design of anchor bolts and presents acomparative study of ACI and CCD methods.

Microwave towers, transmission line towers, towers usedfor oil well derricks and mine shaft equipment, beaconsupports, observation platform towers, etc are examples ofself- supporting towers. Normally, in the case of transmissionline towers, the stub angle is taken inside the pad portion ofthe foundation, and cleat angle and keying rods anchor thisstub angle, Fig 1. But, in the case of microwave towers, thestub angle is connected to the pad portion through base plate

Recent developments in thedesign of anchor bolts

Dr N. Subramanian, Chief Executive, Computer Design Consultants, 191, NorthUsman Road, T. Nagar, Chennai 600 017.

with the help of anchor bolts. Also, in all the industrialbuildings, the columns are connected to the foundationconcrete by means of anchor bolts.

However, methods to design such anchorages are notgiven in the Indian code. Till recently, the only available sourcefor their design was the American codes for nuclear safetyrelated structures1,2. However, a number of researchers allover the world have conducted numerous experiments andbased on them, have suggested formulae for the design ofdifferent kinds of fasteners3,4,5. In India, cast in-situ anchorsare used often in practice, and hence, the design methods ofACI 349 are applicable to them.

It is well known that the concrete cone failure model, assuggested by ACI 349, results in complex calculations,especially when multiple anchorages are used6. Hence, afteran exhaustive number of experiments, Eligehausen and hisassociates have suggested a truncated pyramid failuremodel7,8,9. This model has several advantages over the ACIcode method. They have also suggested modifications to theformula for taking into account cracked concrete10. Theseformulae have recently been incorporated in the Euro code11.A description of these methods is given in this paper alongwith a comparison with the ACI code method. The workedout examples clearly illustrate the ease of applying thismethod for the design of anchorages.

The design philosophy developed so far is based on thetensile strength of concrete. It has been observed that in manykinds of failures where tensile capacity governs, there was adisturbing size-effect that could not be explained12. Accordingto fracture mechanics principles, it has been shown that thefailure load of headed studs anchored in concrete dependson the material parameters, EC and GF and not, as usually

N. SubramanianN. SubramanianN. SubramanianN. SubramanianN. Subramanian

The Indian Concrete Journal * July 20002

assumed, on the tensile strength. This is of great interest,especially in high strength concrete, which tends to be brittle.

Tensile capacity of anchor boltsBased on the results of numerous pullout tests with headed-anchors, an empirical formula for the calculation of themaximum load Nu of fastenings has been derived1,2,3,4,5,6.Generally, this equation is of the form:

Nu = a(fck)b(hef)

c (1)

where, fck = concrete compressive strength, N/mm2

hef = embedment length, mm

and a,b,c = constants.

The influence of embedment length is given by c, foundto be in the range of 1.5 to 1.54, which means that the failureload does not increase in proportion to the surface of thefailure cone, Fig 2. However, in the ACI code, the value of chas been chosen as 2, which anticipates a direct proportionalitybetween the failure load and the size of the failure cone surface.The factor a is used to calibrate the measured failure loadwith the predicted values and to assure the dimensionalcorrectness of eqn. (1). The expression (fck)

b represents the

tensile strength of concrete, derived from the compressivestrength, and a value of 0.5 is assumed for b.

Tensile capacity as per ACI codeFig 2 shows concrete breakout cones for single anchors undertension and shear load, respectively, idealised according toACI 3494. From the figure, it is evident that the concrete conefailure load depends on the tensile capacity of the concrete.ACI committee 349 is concerned with nuclear power-relatedstructures. Because of this, the philosophy has been to designductile fastenings. The cone model as shown in Fig 2 wasdeveloped to obtain a limit to guard against brittle concretefailure.

Under tension loading, the concrete capacity of anarbitrary fastening is calculated assuming a constant tensilestress acting on the projected area of the failure cone, takingthe inclination between the failure surface and concretesurface as 45o, Fig 2(a):

Nu = fct AN (2)

where, fct = tensile strength of concrete and AN = actualprojected area of stress cones radiating toward attachmentfrom bearing edge of anchors. Effective area shall be limitedby overlapping stress cones, intersection of cones withconcrete surfaces, bearing area of anchor heads, and overallthickness of the concrete member.

Substituting for AN and fct, we get for a single anchorunlimited by edge influences or overlapping cones:

Nuo = 0.96 )1(2

ef

hefck

h

dhf + (3)

where, fck = compressive strength measured on 200-mmcubes and hef and dh are defined in Fig 2.

For fastenings with edge effects (c<hef) and/or affectedby other concrete breakout cones (s < 2hef), the average failureload follows from eqn. (3) as under:

Nu =A

ANN

No

uo

where, ANo = projected area of stress cone of a single anchor

Fig 2 Concrete breakout bodies according to ACI 349(a) tensile loading (b) shear loading

45o

dh

he

f

2 h + def h

N

(a)

ANO 45o

c1

Avo

(b)

V

d

Fig 1 Typical foundation for transmission line towers (a)foundation with stub angle (b) foundation with base plate(c) typical stub angle details

Cleat angle

3w

8w

b

w

3w

8w

wStuds

Top of concrete

typicalspacing

(c)

Mat

Pier

Diagonalmember

Stub angle

(a)

Stub angle

Base plate

Anchor bolts

(b)

July 2000 * The Indian Concrete Journal 3

= π ( ) ( ( / )h d hef h ef

21+

where, c = distance from centre of anchor bolt to edgeof concrete

s = distance between anchor bolts (spacing).

Shear load capacity of bolt as per ACI codeThe capacity of an individual anchor failing in shear (Fig 2(b))(provided that the concrete half-cone is fully developed) isgiven by:

Vuo = 0.48 ckf (c1)2 (6)

If the depth of the concrete member is small (h < c1) and/or the spacing is close (s < 2c1) and/or the edge distanceperpendicular to the load direction is small (c2 < c1), theload has to be reduced with the aid of the projectedarea on the side of the concrete member as below:

Vu =A

AVV

vo

uo (7)

AV = actual projected area, and

Avo = projected area of one fastenerunlimited by edge influences, coneoverlapping or member thickness

= ( p/2) (c1)2

The determination of the projected areas AN andAV for tensile and shear loading respectively based onthe ACI code involves complex calculations6. Hence,Eligehausen and his associates have developed amethod called the concrete capacity design (CCD)method based on the results of numerous pullout testsconducted by them on various types of anchors7.

Tensile capacity as per CCD methodIn this method, under tension loading, the concretecapacity of a single fastening is calculated assuming aninclination between the failure surface and surface ofthe concrete member of about 35o, Fig 3. Thiscorresponds to the widespread observation that thehorizontal extent of the failure surface is about threetimes the effective embedment.

As per the above assumption, the concrete cone failureload Nuo of a single anchor in uncracked concrete, unaffectedby edge influences or overlapping cones of neighbouringanchors loaded in tension, is given by

Nuo = k1k2k3 f hck ef( ).15

(8)

where, k1, k2 and k3 are calibration factors. Note that, bybasing the tensile failure load on the effective embedmentdepth hef , the computed load will always be conservative,even for an anchor that might experience pullout8.

Assuming k4 = k1k2k3

Nuo = k4 fck (hef)1.5 (9)

where, k4 = 13.5 for post-installed fasteners

= 15.5 for cast in-situ headed studs and headedanchor bolts

hef = embedment length, Fig 3.

Eligehausen and Balogh10 have further suggested that theabove values of k4 may be multiplied by a factor equal to 0.75to take into account cracked concrete. If headed studs arelocated in the intersection of cracks running in two directions(example in slabs spanning in two directions), the concretecone failure load is about 20 percent lower than the valueaccording to eqn. (9).

Fig 3 Idealised concrete cone for individual fasteningunder tensile loading as per CCD method

N

35o

1.5 hef

h e

f( a )

N

ANo

( b )

Fig 4 Projected areas for different fastenings under tensileloading according to CCD method

ef

1.5hef

1.5hef

3hef

3h

A N

( a )

1

11

1

ef

ef

ef

.5h

.5h

.5h

A N

c

( b )

c1 1

AN

s .5hef1

c1

1s

.5h

ef

1

.5h .5hef ef1 1

.5h

ef

1

1s

1

1

ef ef

ef

= ( c + 1.5 h ) ( 2 x 1.5 h )AN

if : c < _ 1.5 h

1

1

ef

ef

AN

if : s < _ 3.0 h

= ( 2 x 1.5h + s ) ( 2 x 1.5 h ) 1 1

1 2

1 2

ef

ef

ef

ef= ( c + s + 1.5 h ) ( c + s + 1.5 h )

if : c , c 1.5

h

s , s 3.0 h

<

<

_

_

A N

( c )

AN N= A ( single fastening )

= ( 2 x 1.5 h ) ( 2 x 1.5 h )ef efef = 9 h2

o

2 2

.5h

ef

1

( d )

The Indian Concrete Journal * July 20004

Higher coefficients for headed studs and headed anchorbolts in eqn. (9) are valid only if the bearing area of headedstuds and anchors is so large that concrete pressure underthe head is £ 9 fck

(13 f ck for uncracked concrete)8.

Because the concrete pressure under the head of most post-installed undercut anchors exceeds 13 fck at failure, k4 =13.5 is recommended for all post-installed anchors3.

Fracture mechanics theory indicates that, in the case ofconcrete tensile failure with increasing member size, thefailure load increases less than the available failure surface;that means the nominal stress at failure (peak load dividedby failure area) decreases12. This size effect is not unique tofastenings but has been observed for all concrete memberswith a strain gradient. For instance, it is well known that thebending strength of unreinforced concrete beams and shearstrength of beams and slabs without shear reinforcementdecrease with increasing member depth. Size effect has beenverified for fastenings by experimental and theoreticalinvestigations3,5,13. Since strain gradient in concrete forfastenings is very large, size effect is minimum and veryclose to the linear elastic fracture mechanics solutions.Therefore, the nominal stress at failure decreases inproportion to 1 / hef and the failure load increases with(hef)

1.5.

If fastenings are located so close to an edge thatthere is not enough space for a complete concrete coneto develop, the load-bearing capacity of the anchorageis also reduced. (Note that with an edge distance in alldirections c ³ 60d, it may be assumed that no concreteedge failure will occur). This is also true for fastenersspaced so closely that the breakout cones overlap. Oneof the principal advantages of the CCD method is thatcalculation of the changes in capacity due to factors suchas edge distance, spacing, geometric arrangement ofgroupings and similar variations can be readilydetermined through use of relatively simple geometricalrelationships based on rectangular prisms. The concretecapacity can be easily calculated based on the followingequation:

Nu =A

ANN

No

uo Ψ Ψ1 2 (10)

where,

ANo = projected area of one anchor at the concretesurface unlimited by edge influences orneighbouring anchors, idealising the failurecone as a pyramid with a base length scr =3hef, Fig 4(a),

AN = actual projected area at the concrete surface,assuming the failure surface of the individualanchors as a pyramid with a base length scr =3hef. For examples, see Fig 4(b), 4(c), 4(d),

y 1 = factor taking into account the eccentricity ofthe resultant tensile force on tensionedanchor bolts; in the case where eccentricloading exists about two axes, y 1 shall becomputed for each axis individually and theproduct of the factors used as y 1

y 1 = 1/(1 + 2e�N/(3hef)) £ 1 (10a)

e�N = distance between the resultant tensile forceof tensioned anchor bolts of a group andthe centroid of tensioned anchor bolts,

y 2 = tuning factor to consider disturbance of theradial symmetric stress distribution causedby an edge, valid for anchors located awayfrom edges,

y 2 = 1 if c1 ³ 1.5 hef (10b)

y 2 = 0 7 0 315

151

1. ..

, .+ ≤cc h

hef

ef (10c)

Fig 5 Concrete failure cone for individual fastener inthick concrete member under shear loading towardsedge: (a) from test results (b) simplified designmodel according to CCD method

35oo

c 1

1.5 c 1

VV

c 1

1.5

c 1

hh

(b)(a)

@35@

Fig 6 Projected areas for different fastenings under shearloading according to CCD method

1.5

c

c1

A V

1

A = A ( single fastening )v v

= 1.5 c ( 2 x 1.5 c )

= 4.5 c1

11

2

o

3 c

1.5 c1.5 c1

1

1

1.5

c A v

1

2

1

c

c

1.5 c1

A v = 1.5 c ( 1.5 c + c )1

1

1 2

2if : c < 1.5 c _

A v

1.5 c1.5 c11

h

c1

1

1

A v = 2 x 1.5 c x h

if : h < 1.5 c

Av

1.5 c1.5 c 11

h

c1

s

1

1

1

Av = ( 2 x 1.5 c + s ) x h

if : h < 1.5 c

s < 1.5 c

_

_

( a ) ( b )

( d )( c )

July 2000 * The Indian Concrete Journal 5

where, c1 = edge distance to the closest edge,

hef = effective anchorage depth, Fig 3. For faste-nings with three or four edges and cmax £1.5hef

(cmax = largest edge distance), the embedmentdepth to be inserted in eqn. (10) is limited tohef = cmax/1.5. This gives a constant failure loadfor deep embedments8.

Examples for calculation of projected areas are given inFig 4. Note the relatively simple calculation for the CCDmethod compared to that of the ACI 349 method6.

Shear load capacity as per CCD methodThe concrete capacity of an individual anchor in a thickuncracked structural member under shear loading towardthe free edge, Fig 5, is

Vuo = (l/d)0.2 d fck (c1)1.5 (11)

where, l = activated load bearing length of fastener inmm £ 8d

= hef for fasteners with a constant overallstiffness, such as headed studs, undercutanchors and torque-controlled expansionanchors, where there is no distance sleeve

= 2d for torque-controlled expansion anchorswith distance sleeve separated from theexpansion sleeve

d = diameter of anchor bolt in mm

and cl = edge distance in loading direction in mm.

According to eqn. (11) the shear failure load does notincrease with the failure surface area, which is proportionalto (c1)

2. Rather, it is proportional to (c1)1.5. This is again due to

size effect. Furthermore, the failure load is influenced by theanchor stiffness and diameter. The size effect on the shearfailure load has been verified theoretically andexperimentally8.

The shear load capacity of single anchors and anchorgroups loaded toward the edge can be evaluated from eqn.(12) in the same general manner as for tension loading by

Fig 8 Interaction diagram for combined tension and shearloads

1.0

0.8

0.6

0.4

0.2

0.2 0.4 0.6 0.8 1.0

Equation 13

Equation 14

a = 1.5

a = 2.0

u/ V

T / Nu u

0V

Fig 7 Example of multiple fastening with cast-in-situheaded studs close to edge under eccentric shearloading

V

e = ev v

c1

s c1

£ 3 s s

s1

'

Fig 9 Tensile testing of concrete (a) test with load controlwhich gives a brittle failure. (b) test with load -deformation control

Load

Deformation

E Lt

( a )

B

Load

C

DA

GF

Eoarc tan

Aggregate interlock and other frictional effects

Micro cracking starts

crack zone deformation, w 1

( b )

f Atf At

Deformation

Fig 10 Concrete cone failure load of headed studs as afunction of embedment depth13

Nu

a

1200

800

400

00 200 400 600

Ma

x

loa

d,

kN

Nu = 2.1 EG h F ef1.5

Test

embedment depth h , mmef

L B

The Indian Concrete Journal * July 20006

taking into consideration the fact that the size of the failurecone is dependent on the edge distance in the loading direction,while in tension loading, it is dependent on the anchoragedepth7,8

Vu =A

AVv

vo

uoΨ Ψ4 5 (12)

where,

AV = actual projected area at side of concretemember, idealising the shape of the fracturearea of individual anchors as a half-pyramidwith side lengths 1.5c1 and 3c1, Fig 6

AVo = projected area of one fastener unlimitedby corner influences, spacing, or memberthickness, idealising the shape of the fracturearea as a half-pyramid with side length 1.5c1

and 3c1, Figs 5(b) and 6(a)

y 4 = effect of eccentricity of shear load

y 4 =1

1 2 3 1+ e v c' /( )(12a)

e�V = distance between resultant shear force of thegroup of fasteners resisting shear andcentroid of sheared fasteners, Fig 7.

y 5 = tuning factor considering disturbance ofsymmetric stress distribution caused bycorner

= 1 if c2 ³ 1.5c1

= 0 7 0 315

2

1

. ..

+ c

c if c2 £ 1.5c1 (12b)

c1 = edge distance in loading direction, in Fig 6;for fastenings in a narrow, thin member withc2,max < 1.5c1 (c2,max = maximum value of edgedistances perpendicular to the loadingdirection) and h < 1.5c1, the edge distance tobe inserted in eqn. (12a) and (12b) is limitedto c1 = max (c2,max/1.5; h/1.5); this gives aconstant failure load independent of the edgedistance (c1)

8

c2 = edge distance perpendicular to load direction,Fig 6

Examples for calculation of projected areas are shown inFig 6. The relatively simple calculation compared to the morecomplex geometry of the ACI 349 procedures can be easilyidentified.

Resistance to combined tension and shear loadsFor combined tension and shear loads, the followingconditions Fig 8 should be satisfied.

T

N

u

u

≤ 1 (13a)

V

Vu

≤ 1 (13b)

(Tu/Nu) + (V/Vu) £ 1.2 (13c)

where, Tu = tensile load acting on the fixture

V = shear force acting on the fixture.

It has been found that the eqn. (13) yields conservativeresults for steel failure11. More accurate results are obtainedby eqn. (14).

(Tu/Nu)a + (V/Vu)

a £ 1 (14)

The value of a is taken as 2 if Nu and Vu are governed bysteel failure and as 1.5 for all other failure modes.

Comparison of ACI 349 and CCD methodsThe main differences between these two design approachesare summarised in Table 1.

High strength concreteDuring the past 10 to 15 years, high strength concrete (havingstrength greater than 40 N/mm2) has been used increasinglythe world over. In spite of all its enhanced properties, highstrength concrete tends to be brittle than normal concrete. Innormal concrete, the tensile strength of concrete is assumedto give a brittle fracture at a low tensile stress as shown inFig 9(a). However, in failures where the tensile capacitygoverns (as in the case of cone failure of anchorages) there isa size-effect. To explain these failure types, the completeload-deformation curve as shown in Fig 9(b) in tension has tobe taken into account12. Eligehausen and Sawade found thatthe bearing behaviour of anchor bolts with long embedmentdepths can be reasonably predicted by linear fracturemechanics13. Using linear fracture mechanics approach, theyderived the maximum load in tension (for a/1B = 0.45)

Nu = 2.1(EC GF)0.5 (hef)

1.5 (15)

where, GF = total crack formation energy (correspondingto the area below the load-deformation curve as shown in

Table 1: Comparison of the influence of main parameters onmaximum load predicted by ACI 349 and CCD methods

Factor ACI 349-85 CCD method

Anchorage depth, tension (hef)2

(hef)1.5

Edge distance, shear (c1)2

(c1)1.5

Slope of failure cone a = 45o a = 35

o

Required spacing to develop

anchorage capacity 2hef, tension 3hef, tension

2c1, shear 3c1, shear

Required edge distance todevelop full anchor hef, tension 1.5 hef, tension

capacity c1, shear 1.5c1, shear

Small spacing 1 direction nonlinear linearor close to reduction reduction

edge 2 direction

nonlinear reduction

Eccentricity of load ¾ taken into account

July 2000 * The Indian Concrete Journal 7

Fig 9(b). A value of 0.07 N/mm was used.

EC = modulus of elasticity of concrete (a value of 23500 N/mm2

was used).

A comparison of the above equation with the test resultsis shown in Fig 10. It is found that the agreement betweentheory and test result is sufficiently close for practical purposes.

ConclusionMethods for the design of anchorages are not available in theIndian code. The methods suggested in the ACI 349-85 involvecomplex calculations. Based on the extensive experimentalstudy, equations for the ultimate tensile capacity and ultimateshear capacity of headed anchor bolts have been derived byEligehausen and his associates. These equations are easy toapply for multiple anchorages, since they involve square-based, truncated pyramids. The differences between thisapproach and the ACI approach have been brought out. Forfoundations using high strength concrete, instead of thetensile strength of concrete, the total crack formation energyhas to be taken to compute the ultimate tensile load.

AcknowledgementsThe author is highly indebted to Prof. Eligehausen, professorand head for fastening techniques at the Institute for BuildingMaterials, University of Stuttgurt, Germany for makingavailable numerous publications, based on which this articleis written.

References1. ______ Code requirements for nuclear safety related concrete structures (ACI

349-85). ACI Committee 349, American Concrete Institute, Detroit, Michigan48219, USA, 1985.

2. ______ Design Guide to ACI 349-85, ACI Committee 349, American ConcreteInstitute, Detroit, Michigan 48219, USA, 1985.

3. ELIGEHAUSEN, R., MALLEE, R., and REHM, G., Befesttigungstecknik, Beton-Kalender 1997, Ernst & Sohn, Berlin, 145 pp.

4. HAWKINS, N. Strength in shear and tension of cast-in place anchor bolts inAnchorage to Concrete, SP-103, American Concrete Institute, Detroit, 1985,pp. 233-255.

5. BODE, H., and ROIK, K. Headed studs � embedded in concrete and loaded intension in Anchorage to Concrete, SP-103, American Concrete Institute, Detroit,1993, pp. 61-88.

6. SUBRAMANIAN, N., and VASANTHI, V., Design of anchor bolts in concrete. TheBridge and Structural Engineer. September 1991, Vol. XXI, No.3, pp. 47-73.

7. FUCHS, W., ELIGEHAUSEN, R., and BREEN, J.E., Concrete capacity design (CCD)approach for fastenings to concrete, ACI Structural Journal. January-February1995, Vol 92, No 1, pp. 73-94.

8. FUCHS, W., and ELIGEHAUSEN, R., Das CC-Verfahren fur die Berechnung derBetonausbruchlast von Verankerungen, Beton und Stahlbetonbau, H. 1/1995,pp. 6-9, H. 2/1995, pp. 38-44, H. 3/1995, pp. 73-76.

9. ELIGEHAUSEN, R. et al., Tragverhalten von Kopfbolzenverankerungen beizentrischer Zugbeanspruchung, Bauingenieur, 1992, pp. 183-196.

10. ELIGEHAUSEN, R. and BALOGH, T., Behaviour of fasteners loaded in tension incracked reinforced concrete, ACI Structural Journal. May-June 1995, Vol 92,No 3, pp. 365-379.

11. Comite Euro-International du Beton, Design of Fastenings in Concrete, DesignGuide, Thomas Telford, UK, 1997.

12. BEZANT, Z.P., Size effect in blunt fracture: concrete, rock, metal, Journal of the

Engineering Mechanics Division. ASCE, April 1984, Vol 110, No 4, pp. 518-535.

13. ELIGEHAUSEN, R., and SAWADE, G., Fracture mechanics based description of thepull-out behaviour of headed studs embedded in concrete in �RILEM report onFracture Mechanics of Concrete Structures From theory to Applications, Elfgren,L, Ed, Chapman & Hall, London, 1989, pp. 281-299.

Illustrative examplesExample 1Calculate the tensile and shear capacities of a single headed anchorbolt as per ACI and CCD method. Assuming that concrete halfcone is fully developed in case of shear failure and there is no edgeinfluences or overlapping cones.

dh = 45.2 mm (for d = 27mm)

c1 = 300 mm (for headed anchors c1 = hef)

hef = 300 mm

fck = 20 N/mm2

Tensile capacity as per ACI code

Nu = 0 96 2 0 300 145 2

300

2. (

.)× + = 444,609 N

Tensile capacity as per CCD method

Nuo = 15.5 20 (300)1.5 = 360,187 N

Shear capacity as per ACI method

Vu = 0.48 20 (300)2 = 193,196 N

Shear capacity as per CCD method

Vu = (300/27)0.2 27 20 (300)1.5 = 195,445 N

It is clearly seen that the values predicted by the CCD methodare quite reasonable as compared to the ACI code method.

Example 2Calculate the tensile and shear capacities of a headed anchor boltarrangement as shown in Fig 11.

d = 27 mm

dh = 45.2 mm

c1 = 500 mm

hef = 300 mm

fck = 25 N/mm2

Tensile load capacity

Fig 11 A typical headed anchor bolt arrangment

The Indian Concrete Journal * July 20008

c1 = 500 > 1.5 hef

s1 = 600 £ 3 hef

therefore, Referring Fig 4a and 4c

AN = (2 ´ 1.5 ´ 300 + 600) (2 ´ 1.5 ´ 300)= 1,350,000 mm2

ANo = 9h2ef = 9 ´ 3002 = 810,000 mm2

Nuo = 15.5 25 (300) 1.5 = 402,701 N

Nu =1350 000

810 0001 1 402 701

,

,,× × × = 671,168 N

Shear load capacity (see Fig 6)

Vuo = (300/27)0.227 25 (300)1.5 = 218,516 N

Av = (2 ´ 1.5 ´ 300 + 600) 300 = 450,000 mm2

Avo = 4.5 ´ 3002 = 405,000 mm2

c2 > 1.5 ´ 300 = 450 therefore, y 5 = 1

Vu = 450 000

405 0001 1 218 516

,

,,× × × = 242,795 N

Example 3Design the post-installed anchor system as shown in Fig 12 for thegiven data. Assume that there is no eccentricity.

Tensile force = 300,000 N

Shear force= 60,000 N

fy = 240 N/mm2

fut = 420 N/mm2

fck = 20 N/mm2

Assuming hef = 250-mm and 20-mm diameter bolts (corearea = 244 mm2)

dh = 32.95 mm

c1 = 300 mm < 1.5hef

c2 = 300 mm < 1.5hef

s1, s2 = 600 mm < 3hef

Nuo = 13.5 20 (250) 1.5 = 238,648 N

A N = (300 + 600 + 1.5 ´ 250) (300 + 600 + 1.5 ´ 250)

= 1,625,625 mm2

ANo = 9 ´ 2502 = 562,500 mm2

y 2 = 0.7 + 0.3(300/(1.5 ´ 250)) = 0.94

N u = 1 625 625

562 5001 0 94 238 648

, ,

,. ,× × × = 648312 N

Assuming the concrete has cracked

Nu = 0.7 ´ 648312 = 453, 818 N > 300,000 N

Tensile capacity of bolts = 4 ´ 244 ´ 420

= 409,920 N > 300,000 N

Vuo = (250/20)0.2 20 20 (300)1.5 = 172,224 N

y 4 = 1

y 5 = 0.7 + 0.3(300/(1.5 ´ 300)) = 0.9

Avo = 4.5 ´ 3002 = 405,000 mm2

h < 1.5 c1

s > 1.5 c1

therefore, Av = 2 ´ 1.5 ´ 300 ´ 250 ´ 2 = 450,000 mm2

Vu = 450 000

405 0001 0 9 172 224

,

,. ,× × × = 172,224 N

Assuming that there is no reinforcement and the concrete willhave cracks

Vu = 0.7 ´ 172,224 = 120,556 N

Shear capacity of bolts = 0.6 ´ 244 ´ 240

= 35,136 N > 20,000 N

Check for combined tension and shear8

T

N

u

u=

300 000

453 818

,

, = 0.661 < 1

V

Vu=

60 000

120 556

,

,= 0.498 < 1

(Tu/Nu) + (V/Vu) = 0.661 + 0.498 = 1.159 < 1.2

(Tu/Nu)1.5 + (V/Vu)

1.5 = (0.661) 1.5 + (0.498) 1.5

= 0.537 + 0.351

= 0.888 < 1

Hence, 4 nos. 20-mm diameter anchor bolts with thearrangement as shown in Fig 12 is safe.

These examples clearly show the ease of calculation by usingthe CCD method.

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Fig 12 A typical post-installed anchor system