Wood

P P I • p p i 2 p a s s . c o m

Table 6.4 Time-Effect Factor, λ

load combination λ1.4D 0.61.2D+ 1.6(L+H) + 0.5(Lr or S or R)

0.7 when L is fromstorage

0.8 when L is fromoccupancy

1.25 when L is fromimpact

1.2D+ 1.6(Lr or S or R) + 1.6H+(f1L or 0.5W)

0.8

1.2D+ 1.0W+ f1L+ 1.6H + 0.5(Lr or S or R)

1.0

1.2D+ 1.0E+ f1L+ 0.2S+ 1.6H 1.0

0.9D+ 1.0W+ 1.6H 1.00.9D+ 1.0E+ 1.6H 1.0Note:Where the effect of H resists the primary variable load effect, a loadfactor of 0.9 must be included with H where H is permanent. H mustbe set to zero for all other conditions.Replace f1L with 1.0L for garages, places of public assembly, andareas where L> 100 lbf/ft2. Use f1L = 0.5L for all other live loads.Replace 0.2S with 0.7S for roof configurations that do not shed snow.j

Load Duration Factor, CD (ASD Method)

The load duration factor, CD, is applicable to all refer-ence design values with the exception of compressionperpendicular to the grain and modulus of elasticity.Values of the load duration factor are given in Table 6.5.

Table 6.5 Load Duration Factors

design loadj CDj

dead load 0.90lj

occupancy live load 1.00lj

snow load 1.15lj

construction load 1.25lj

wind or earthquake load 1.60lj

impact load 2.00lj

In a combination of loads, the load duration factor forthe shortest duration load applies for that combination.

Wet Service Factor, CM

When the moisture content of sawn lumber exceeds19%, the adjustment factors given in NDS Supp.Table 4A and Table 4B are applicable to visually gradeddimension lumber; in NDS Supp. Table 4C to machine-graded dimension lumber; in NDS Supp. Table 4D tovisually graded timbers; in NDS Supp. Table 4E todecking; and in NDS Supp. Table 4F to non-NorthAmerican visually graded dimension lumber. When the

moisture content of glued laminated members exceeds16%, the adjustment factors given in NDS Supp.Table 5A through Table 5D are applicable.

The applicability of the wet service factor to the refer-ence design values for sawn lumber and glued laminatedtimber is summarized in Table 6.6.

Table 6.6Wet Service Factor, CM

classification Fb Ft Fv Fc⊥ Fc E Eminsawn lumber <5 ×member

0.85a 1.00 0.97 0.67 0.8b 0.90 0.90

sawn lumber≥5 ×member

1.00 1.00 1.00 0.67 0.91 1.00 1.00

decking 0.85a – – 0.67 – 0.90 0.90

glued laminatedmember

0.80 0.80 0.875 0.53 0.73 0.833 0.833

aWhen FbCP ≤ 1150 lbf/in2, CM= 1.00.bWhen FbCP ≤ 750 lbf/in2, CM= 1.00.

For sawn lumber, CM is applicable when the moisturecontent exceeds 19%. For glued laminated members, CMis applicable when the moisture content exceeds 16%.

Example 6.1

A 2× 10 visually graded, select structural, southernpine, sawn lumber member’s moisture content exceeds19%. The governing load combination is the sum of thedead load, the live load, and the wind load. Determinethe allowable shear capacity (ASD) and the factoredshear capacity (LRFD) of the member.

Solution

The reference design value for shear stress from NDSSupp. Table 4B is

F 175†lbf/inv2=

The applicable wet service factor for shear stress fromTable 6.6 is

C 0.97M =

The temperature factor and the incising factor are

CC

1.0

1.0t

i

==

ASD Method

The applicable load duration factor for a load combina-tion including the wind load from Table 6.5 is

C 1.60D =

6-66-6 S E S T R U C T U R A L E N G I N E E R I N G R E F E R E N C E M A N U A L

Wood

P P I • p p i 2 p a s s . c o m

Figure 6.7 Staggered Fasteners

consider asingle row of4 fastenersconsider a single row of4 fasteners

s

a < s4

Geometry Factor, CΔ

The geometry factor applies to bolts, lag screws, splitrings, and shear plates. The factor is applied, in accord-ance with NDS Sec. 12.5.1 and Sec. 13.3.2, when end oredge distances or spacing are less than the specifiedminimum.

The geometry factor and group action factor are notapplied to nails or screws. NDS Comm. Table C12.1.5.7provides recommended spacing requirements for screws.NDS Comm. Table C12.1.6.6 provides recommendedspacing requirements for nails.

Penetration Depth Factor, Cd

The penetration depth factor applies to lag screws, splitrings, shear plates, screws, and nails. The factor isapplied in accordance with NDS Table 13.2.3 and foot-notes to NDS Table 12J through Table 11R when thepenetration is less than the minimum specified.

End Grain Factor, Ceg

The end grain factor applies to lag screws, screws, andnails. The factor is applied in accordance with NDSSec. 12.5.2 when the fastener is inserted in the end grainof a member.

Ceg is 0.75 for lag screws loaded in withdrawal. Ceg is0.67 for laterally loaded dowel-type fasteners. Spacingand edge distance requirements for laterally loaded fas-teners in the narrow edge of cross-laminated timber aregiven in NDS Table 12.5.1G and NDS Fig. 12I. For lagscrews installed in cross-laminated timber end grain orside grain and loaded laterally, Ceg = 0.67.

Wood screws and nails must not be loaded in with-drawal from the end grain of wood or cross-laminatedtimber.

Metal Side Plate Factor, Cst

The metal side plate factor is applicable to split ringsand shear plates. The factor is applied in accordancewith NDS Sec. 13.2.4 when metal side plates are usedinstead of wood side members.

The effect of metal side plates on the lateral design val-ues of bolts, lag screws, wood screws, and nails areincorporated into the appropriate tables of referencedesign values.

Diaphragm Factor, Cdi

The diaphragm factor applies to nails and spikes. Thefactor is applied in accordance with NDS Sec. 12.5.3when the fasteners are used in diaphragm constructionand Cdi= 1.1.

6-35W O O D D E S I G N 6-35

Table 6.11 Adjustment Factors for Connections*

adjustment factorj boltsjlag

screwsjsplit rings andshear platesj screwsj nailsj

design valuej Zj Wj Zj Pj Qj Wj Zj Wj Zj

CD load duration factor √ √ √ √ √ √ √ √ √

CM wet service factor √ √ √ √ √ √ √ √ √

Ct temperature factor √ √ √ √ √ √ √ √ √

Cg group action factor √ – √ √ √ – – – –

CΔ geometry factor √ – √ √ √ – – – –

Cd penetration depth factor – – √ √ √ – √ – √

Ceg end grain factor – √ √ – – – √ – √

Cstmetal side plate factor – – – √ – – – – –

Cdi diaphragm factor – – – – – – – – √

Ctn toe-nail factor – – – – – – – √ √*Z= lateral design value;W=withdrawal design value; P=parallel to grain design value; Q=perpendicular to grain design value

Forces

P P I • p p i 2 p a s s . c o m

The maximum active pressure at the bottom of the heel is

p K h

(0.27) 120†lbf

ft(15†ft)

486†lbf/ft

A A S

3

2

=

=

=

For a 1 ft length of wall, the total horizontal force on theretaining wall is

Hp h2

486†lbf

ft(15†ft)

(2) 1000†lbfkip

3.65†kips/ft

AA

2

=

=

=

Passive Pressure

For the retaining wall shown in Fig. 1.35(c), which isconstructed on a soil with a density of γS and an angle ofinternal friction of φ, the passive earth pressure that canbe developed in front of the key at a depth z belowground surface is given by Rankine’s theory as

p K zP P S=

In this equation, KP is the Rankine coefficient of passiveearth pressure and is

K1 sin

1 sinP =+

The maximum passive pressure occurs at the bottom ofthe key and is

p K hP P S K=

In this equation, hK is the height of the key. The totalhorizontal force on the key due to passive pressure is

Hp h K h

2 2PP K P S K

2

= =

The line of action of the total horizontal force is located(h + hK)/3 above the bottom of the key (see Fig. 1.35(c)).

Example 1.25

The retaining wall shown in Fig. 1.35(c) is constructedon a soil with a density of 120 lbf/ft3 and an angle ofinternal friction of 35°. The total height of the key is5 ft. Determine the total horizontal force per foot of walldue to passive pressure on the key.

Solution

The Rankine coefficient of passive earth pressure is

K1 sin

1 sin

1 sin 351 sin 353.69

P =+

=+ °

°=

The maximum passive pressure that can be developedat the bottom of the key is

p K h

(3.69) 120†lbf

ft(5†ft)

2214†lbf/ft

P P S K

3

2

=

=

=

For a 1 ft length of wall, the total force due to passivepressure that can be developed on the key is

Hp h

2

2214†lbf

ft(5†ft)

(2) 1000†lbfkip

5.54†kips/ft

PP K

2

=

=

=

Hydrostatic Pressure and Soil Lateral Pressure

A retaining wall is usually provided with a drainage sys-tem behind the wall. Where this is not possible andadjacent soil is below the free-water table, design mustinclude the effects of submerged soil pressure plushydrostatic pressure as shown in Fig. 1.36.

For the retaining wall shown in Fig. 1.36, which has asoil backfill with a dry density of γS, the active earthpressure behind the wall at a depth of hdry below the fillsurface is given by Rankine’s theory as

p K hA A S dry=

1-47V E R T I C A L , I N C I D E N T A L L A T E R A L , A N D O T H E R V A R I A B L E F O R C E S 1-47

Rein

forced

Concrete

P P I • p p i 2 p a s s . c o m

concrete compressive strength of 3000 lbf/in2. Deter-mine whether the corbel is adequate for the applied fac-tored loads indicated.

Nuc = 40 kips

2 in

4 in

20 in

12 in

3–No. 7

3–No. 3(closedstirrups)

anchor bar

Vu = 100 kips

Solution

f b d

V

f b d

Vb d

V

0.2 (0.2)(0.75) 3†kips

in(15†in)(20†in)

135†kips

(480 0.08 ) (0.72)(0.75)(15†in)(20†in)

162†kips

1.6 (1.6)(0.75)(15†in)(20†in)

360†kips

c w

u

c w

u

w

u

2=

=>

+ ==>==>

The corbel conforms to ACI Table 22.9.4.4.

The shear friction reinforcement area is given by ACISec. 22.9.9.2 as

AVf

100†kips

(0.75) 60†kips

in(1.4)

1.59†in

vfu

y2

2

µ= =

=

The tension reinforcement area required is

AN

f40†kips

(0.75) 60†kips

in

0.889†in

nuc

y2

2

= =

=

The factored moment acting on the corbel is

M V a N h d( )

(100†kips)(4†in) (40†kips)(2†in)

480†in-kips

u u v uc= += +=

The area of flexural reinforcement required for φ= 0.75for corbels as given by ACI Sec. R21.2.1 for corbels is

A

bdfM

b d f

f

0.85 1 10.319

0.545†in

f

cu

w c

y

2

2

=

=

The primary reinforcement area required is given byACI Sec. 16.5.5.1 as

A A A

0.545†in 0.889†in

1.434†in

sc f n

2 2

2

= +

= +=

Three no. 7 bars are provided, giving an area of

A 1.80†in

1.434†in [satisfactory]

s2

2

=

>

Also, from ACI Sec. 16.5.5.1, the area of primary rein-forcement must not be less than

AA

A

2

3

(2)(1.59†in )

30.889†in

1.95†in

[unsatisfactory]

vfn

s

22

2

+ = +

=>

The area of closed stirrups required is given by ACISec. 16.5.5.2 as

AA A

2

1.95†in 0.889†in2

0.53†in

hsc n

2 2

2

=

=

=

Three no. 3 closed stirrups are provided, giving an area of

A 0.66†in

0.53†in [satisfactory]

h2

2

=

>

2-282-28 S E S T R U C T U R A L E N G I N E E R I N G R E F E R E N C E M A N U A L

Reinforced

Concrete

P P I • p p i 2 p a s s . c o m

The development length for a grade 60, no. 8 bar, with2.5 in side cover and 2 in end cover and with a standard90° hook, is given by ACI Sec. 25.4.3.1 as

=

=

=>

ld

f

l

(0.7) 1200†lbf

in

(0.7) 1200†lbf

in(1†in)

4500†lbf

in12.5†in

[Anchorage†length†is†inadequate.]

dh

b

c

a

2

2

2

Use an end plate to anchor the bars.

8. CORBELS........................................................................................................................

A corbel is a cantilever bracket supporting a load-bearing member. As shown in Fig. 2.14 and specified inACI Sec. 16.5.1.1, the shear span-to-depth ratio, av/d,and the ratio of horizontal tensile force to vertical force,Nuc/Vu, are limited to a maximum value of unity. Thedepth of the corbel at the outside edge of bearing areamust not be less than d/2.

At the face of the support, the forces acting on the cor-bel are a shear force, Vu, a moment (Vuav+Nuc(h− d)),and a tensile force, Nuc. These require reinforcementareas of Avf, Af, and An, respectively. The shear frictionreinforcement area is derived from ACI Eq. 22.9.4.2 as

µ=A

Vfvfu

y

Also, from ACI Table 22.9.4.4, the factored shear forceon the section is

µ

+

=

=

V f b d

f b d

b d

0.2

(480 0.08 )

1600

coefficient†of†friction†at†face†of†support,

as†given†by†ACI†22.9.4.2

1.4 [for†concrete†placed†monolithically]

u c w

c w

w

The correction factor related to the unit weight of con-crete is defined by ACI Sec. 19.2.4.2 as

==

1.0

0.75

[for†normal†weight†concrete]

[for†all†lightweight†concrete]

In accordance with ACI Sec. 16.5.3.5, the tensile forceNuc may not be less than 0.2Vu, and the correspondingarea of reinforcement required is

=AN

fnuc

y

The required area of primary tension reinforcement isgiven by ACI Sec. 16.5.5.1 as the greater of

= +

=

A A A

Abd

ff

0.04

sc f n

sc c

y

The minimum required area of closed ties distributedover a depth of 2d/3 is given by ACI Sec. 16.5.5.2 as

=AA A

2hsc n

ACI Table 21.2.1 gives the value of the strength reduc-tion factor for corbels as = 0.75.

Example 2.12

The reinforced concrete corbel shown has a width of15 in, is reinforced with grade 60 bars, and has a

2-27R E I N F O R C E D C O N C R E T E D E S I G N 2-27

Figure 2.14 Corbel Details

anchor bar

framing barto anchorstirrups

dh

Nuc ≤ Vu

Asc (primaryreinforcement)

bearingplate

Vu

av

d23

Ah(closedstirrups)

mind2

Rein

forced

Concrete

P P I • p p i 2 p a s s . c o m

concrete compressive strength of 3000 lbf/in2. Deter-mine whether the corbel is adequate for the applied fac-tored loads indicated.

Nuc = 40 kips

2 in

4 in

20 in

12 in

3–No. 7

3–No. 3(closedstirrups)

anchor bar

Vu = 100 kips

Solution

=

=>

+ ==>==>

f b d

V

f b d

Vb d

V

0.2 (0.2)(0.75) 3†kips

in(15†in)(20†in)

135†kips

(480 0.08 ) (0.72)(0.75)(15†in)(20†in)

162†kips

1.6 (1.6)(0.75)(15†in)(20†in)

360†kips

c w

u

c w

u

w

u

2

The corbel conforms to ACI Table 22.9.4.4.

The shear friction reinforcement area is given by ACISec. 22.9.9.2 as

µ= =

=

AVf

100†kips

(0.75) 60†kips

in(1.4)

1.59†in

vfu

y2

2

The tension reinforcement area required is

= =

=

AN

f40†kips

(0.75) 60†kips

in

0.889†in

nuc

y2

2

The factored moment acting on the corbel is

= += +=

M V a N h d( )

(100†kips)(4†in) (40†kips)(2†in)

480†in-kips

u u v uc

The area of flexural reinforcement required for φ= 0.75for corbels as given by ACI Sec. R21.2.1 for corbels is

=

=

A

bdfM

b d f

f

0.85 1 10.319

0.545†in

f

cu

w c

y

2

2

The primary reinforcement area required is given byACI Sec. 16.5.5.1 as

= +

= +=

A A A

0.545†in 0.889†in

1.434†in

sc f n

2 2

2

Three no. 7 bars are provided, giving an area of

=

>

A 1.80†in

1.434†in [satisfactory]

s2

2

Also, from ACI Sec. 16.5.5.1, the area of primary rein-forcement must not be less than

+ = +

=>

AA

A

2

3

(2)(1.59†in )

30.889†in

1.95†in

[unsatisfactory]

vfn

s

22

2

The area of closed stirrups required is given by ACISec. 16.5.5.2 as

=

=

=

AA A

2

1.95†in 0.889†in2

0.53†in

hsc n

2 2

2

Three no. 3 closed stirrups are provided, giving an area of

=

>

A 0.66†in

0.53†in [satisfactory]

h2

2

2-282-28 S E S T R U C T U R A L E N G I N E E R I N G R E F E R E N C E M A N U A L

Steel

P P I • p p i 2 p a s s . c o m

Eleven stress categories are defined in AISC 360Table A-3.1. For stress categories A, B, B′, C, D, E, andE′, the design stress range in the member must notexceed the value given by AISC 360 Eq. A-3-1 as

=FC

n

F

SRf

SR

TH

0.333

For stress category F, the design stress range in themember must not exceed the value given by AISC 360Eq. A-3-2.

=FC

n

F

SRf

SR

TH

0.167

Example 5.34

A tie member in a steel truss consists of a pair of gradeA36 5 in × 5 in × 3

8 in angles fillet welded to a gussetplate. The force in the member, due to dead load only, is90 kips tension. The additional force in the member, dueto live load only, varies from a compression of 7 kips to atension of 50 kips. During the design life of the struc-ture, the live load may be applied 600,000 times. Deter-mine whether fatigue effects are a concern.

Solution

From AISC 360 Table A-3.1, the loading condition ofSec. 4.1 is applicable and the relevant factors are

==

= =××

=

EF

C

n

stress†categoryallowable†stress†range

11 10

6 10

12.21†kips/in

SR

f

SR

0.333 8

5

0.333

2

The area of the tie is

=A 7.22†ins2

The actual stress range is

= =

=<

fT T

A

F

50†kips ( 7†kips)

7.22†in

7.9†kips/in F[exceeds† ]

SRs

TH

SR

max min2

2

This is within the allowable stress range, so fatigueeffects need not be considered.

8. DESIGN OF BOLTED CONNECTIONS........................................................................................................................

Nomenclature

Ab nominal unthreaded body area of bolt in2

C coefficient for eccentrically loaded bolt andweld groups

–

d nominal bolt diameter indm moment arm between resultant tensile and

compressive forces due to an eccentric forcein

dn nominal hole diameter –Du a multiplier that reflects the ratio of the

mean installed bolt tension to the specifiedminimum bolt pretension

–

frv required shear stress kips/in2

fv computed shear stress kips/in2

Fnt nominal tensile stress of bolt kips/in2

Fntnominal tensile stress of a bolt subjected tocombined shear and tension

kips/in2

Fnv nominal shear stress of bolt kips/in2

Fu specified minimum tensile strength kips/in2

hsc modification factor for type of hole –ks slip-critical combined tension and shear

coefficient–

lc clear distance, in the direction of force,between the edge of the hole and the edge ofthe adjacent hole or edge of the material

in

Le edge distance between the bolt center andthe edge of the connected part

in

n number of bolts in a connection –n number of bolts above the neutral axis (in

tension)–

nb number of bolts carrying strength leveltension Tu

–

ns number of slip planes –Pr load on connection kips

Rn nominal strength kipss center-to-center pitch of two consecutive

boltsin

t thickness of connected part in

Ta applied tensile force (ASD) kips

Tb minimum pre-tension force kips

Tu applied tensile force (LRFD) kips

Symbols

μ mean slip coefficient for the applicablesurface

–

φ resistance factor –

5-53S T R U C T U R A L S T E E L D E S I G N 5-53

Steel

P P I • p p i 2 p a s s . c o m

Types of BoltsThere are two categories of bolts: common bolts andhigh-strength bolts. High-strength bolts are additionallygrouped by strength levels into two categories: group Abolts (A325, F1852, A354 grade BC, and A449) andgroup B bolts (A490, F2280, and A354 grade BD).

Common bolts of grade A307 with a nominal tensilestrength of 45 kips/in2 are used in snug-tight (bearing-type) connections only.

High-strength bolts in group A with a nominal tensilestrength of 90 kips/in2, or group B with a nominal ten-sile strength of 113 kips/in2, are used in bearing-type,pretensioned, and slip-critical connections.

Bolts are installed in the following three types ofconnections.

1. Bearing-type or snug-tight connections requirethe bolts to be tightened sufficiently to bring theplies into firm contact. Levels of installed tensionare not specified. Transfer of the load from oneconnected part to another depends on the bearingof the bolts against the side of the holes. Thistype may be used when pretensioned or slip-critical connections are not required.

2. Pretensioned connections require the bolts to bepretensioned to a minimum value of 70% of thebolt’s minimum tensile strength and the fayingsurfaces may be uncoated, coated, or galvanizedregardless of the slip coefficient. Transfer of theload from one connected part to another dependson the bearing of the bolts against the side of theholes. Pretensioned connections are requiredwhen bearing-type connections are used in

• column splices in buildings over 125 ft inheight

• bracing members in buildings over 125 ft inheight

• structures carrying cranes of over 5 toncapacity

• supports of machinery causing impact orstress reversal

3. Slip-critical connections require the bolts to bepretensioned to a minimum value of 70% of thebolt’s tensile strength, and the faying surfacesmust be prepared to produce a specific value ofthe slip coefficient. Transfer of the load from oneconnected part to another depends on the frictioninduced between the parts. Slip-critical connec-tions are required where

• fatigue load occurs

• bolts are used in oversize holes or slotted holesparallel to the direction of load

• slip at the faying surfaces will affect the per-formance of the structure

• bolts are used in conjunction with welds

Bearing-Type Bolts in ShearThe minimum permissible distance and the preferreddistance between the centers of holes is given byAISC 360 Sec. J3.3 as

==

s ds d

2.67

3.0min

pref

The nominal shear strength is based on the nominalunthreaded cross-sectional area of the bolt, Ab, and thenominal shear strength, Fnv. Nominal shear strength offasteners and threaded parts is given inAISC 360 Table J3.2, and for high-strength bolts areduced nominal strength is applicable when threads arenot excluded from the shear planes. No reduction ismade for A307 bolts. For connections longer than 38 in,the nominal strength is reduced. The bolt’s availableshear capacity is obtained from AISC 360 Eq. J3-1.

==

=

=

R F AF A

R F A

F A

0.75

2.00

[LRFD]

[ASD]

n nv b

nv b

n nv b

nv b

Bearing-Type Bolts in Tension and CombinedShear and TensionThe available strength in tension is given by AISC 360Sec. J3.6 as

==

=

=

R F AF A

R F A

F A

0.75

2.00

[LRFD]

[ASD]

n nt b

nt b

n nt b

nt b

Values of the nominal tensile stress Fnt are given inAISC 360 Table J3.2 for all types of bolts. Values of φRnand Rn/Ω are given in AISC Manual Table 7-2.

When a bearing-type bolt is subjected to combinedshear and tension, the available strength in shear isunaffected, and the available strength in tension isreduced in accordance with AISC 360 Sec. J3.7.

5-545-54 S E S T R U C T U R A L E N G I N E E R I N G R E F E R E N C E M A N U A L

Steel

P P I • p p i 2 p a s s . c o m

The value of the reduced available tensile capacity is

=

=

=

=

R F A

F AR F A

F A

0.75

2.00

[LRFD]

[ASD]

n nt b

nt b

n nt b

nt b

Fnt, the modified nominal tensile stress, is calculatedusing AISC 360 Eq. J3-3a for the LRFD method orAISC 360 Eq. J3-3b for the ASD method.

=

=

F FFF

f

FF

Ff

1.3

1.3

[LRFD]

[ASD]

nt ntnt

nvrv

ntnt

nvrv

The required shear stress, frv, is determined using appro-priate load combinations, and must be equal to or lessthan the available shear stress. Values of the nominalshear stress Fnv are given in AISC 360 Table J3.2 for alltypes of bolts.

When either f 30%rv of the available shear stress orf 30%t of the available tensile stress, the effects ofcombined stress do not need to be considered.

Example 5.35

The connection analyzed in Ex. 5.32 consists of 11 gradeA307 7

8 in diameter bolts. Determine the design shearstrength of the bolts in the connection.

Solution

From AISC Manual Table 7-1, the available strength ofthe 11 bolts in single shear is

LRFD Method

=

=

=

R F A n

12.2†kips

bolt(11†bolts)

134†kips

n nv b

ASD Method

=

=

=

R F A n

8.11†kips

bolt(11†bolts)

89†kips

n nv b

Slip-Critical Bolts in ShearSlip-critical bolts are high strength group A or group Bbolt pretensioned to the value specified in AISC 360Table J3.1 of

=T F0.70b u

The pretension produces a clamping force between theparts, and transfers the shear load from one connectedpart to another by friction. At the strength limit state,the connection may slip sufficiently to place the bolts inbearing and AISC 360 Sec. J3.10 requires slip-criticalconnections to also comply with the requirements ofsnug-tight connections.

The frictional resistance developed in a slip-critical con-nection depends on the condition of the faying surfaces.The values of the slip coefficient, μ, for two types of sur-face conditions (class A and class B) are given in AISC360 Sec. J3.8.

• Class A surface conditions consist of unpainted cleanmill scale surfaces or blast-cleaned surfaces withclass A coatings. The slip coefficient is

µ = 0.30

• Class B surface conditions consist of unpaintedblast-cleaned surfaces or blast-cleaned surfaces withclass B coatings. The slip coefficient is

µ = 0.50

The nominal slip resistance is identical for the cases ofthreads included or excluded from the shear plane, andis given by AISC 360 Eq. J3-4 as

µ=R D h T nn u f b s

The bolt tension multiplier reflects the ratio of the meaninstalled bolt tension to the specified minimum bolt pre-tension and is given by

=D 1.13u

The modification factor for fillers, hf, is

• 1.00 where bolts are added to distribute loads in thefiller

• 1.00 for one filler between connected parts

• 0.85 for two or more fillers between connected parts

The resistance factor and safety factor adopted dependon the type of hole used in the connection. Connectionsallowing a large amount of slip may cause unacceptableslip and will require a higher safety factor than a

5-55S T R U C T U R A L S T E E L D E S I G N 5-55

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forced

Maso

nry

P P I • p p i 2 p a s s . c o m

Example 7.8

The nominal 16 in square, solid grouted concrete blockmasonry column shown is reinforced with four no. 4grade 60 bars. Determine the required size and spacingof lateral ties.

2–No. 4grade 60 bars

2–No. 4grade 60 bars

16 in nominalCMU column

b = 15.63 in

t = 15.63 in

Solution

As specified by TMS 402 Sec. 5.3.1, lateral ties for theconfinement of longitudinal reinforcement must nothave a diameter of less than 1

4 in. Using no. 3 bars forthe ties, the spacing must not exceed the lesser of thefollowing.

s d

ss d

48 (48)(0.375†in)

18†in16†in

16 (16)(0.5†in)

8†in

[least cross-sectional column dimension]

[governs]

lateral

longitudinal

= ==== =

=

Axial Compression in ColumnsASD Method

The allowable compressive stress in an axially loadedreinforced masonry column is given by

FPAa

a

n=

For columns having an h/r ratio not greater than 99, theallowable axial load is given by TMS 402 Sec. 8.3.4.2.1 as

P f A A Fh

r(0.25 0.65 ) 1

140

[TMS†402†8-21]

a m n st s

2

= +

For columns having an h/r ratio greater than 99, theallowable axial load is given by TMS 402 Sec. 8.3.4.2.1 as

P f A A Fr

h(0.25 0.65 )

70[TMS†402†8-22]a m n st s

2

= +

To allow for accidental eccentricities, TMS 402 Sec.8.3.4.3 requires that a column be designed for a mini-mum eccentricity equal to 0.1 times each side dimension.When actual eccentricity exceeds the minimum eccen-tricity, the actual eccentricity should be used.

For columns with lateral reinforcement, the allowablesteel stress is given by TMS 402 Sec. 8.3.3 as

F

F

20,000†lbf/in

32,000†lbf/in

[grade 40 or grade 50 reinforcement]

[grade 60 reinforcement]

s

s

2

2

=

=

SD Method

The nominal allowable axial compressive strength in anaxially loaded reinforced masonry column, as given byTMS 402 Sec. 9.3.4.11, allows for slenderness effects andaccidental eccentricity of the applied load. The nominalaxial compressive strength must not exceed TMS 402Eq. 9-19 or Eq. 9-20, as appropriate. For members hav-ing an h/r ratio not greater than 99, TMS 402 Eq. 9-19applies and is given by

( )P f A A f Ah

r0.80 0.80 ( ) 1

140n m n st y st

2

= +

For members having an h/r ratio greater than 99,TMS 402 Eq. 9-20 applies and is given by

( )P f A A f Ar

h0.80 0.80 ( )

70n m n st y st

2

= +

Example 7.9

The nominal 16 in square, solid grouted concrete blockmasonry column shown has a specified strength of1500 lbf/in2, a modulus of elasticity of 1,000,000 lbf/in2,and is reinforced with four no. 4 grade 60 bars. The col-umn has a height of 15 ft and is pinned at each end.Neglecting accidental eccentricity, determine the avail-able column strength.

b = 15.63 in16 in CMUcolumn

No. 3 @8 in centers

2.63 in

a =10.37 in

2.63 in

2–No. 4grade 60bars

2–No. 4grade 60 bars

7-207-20 S E S T R U C T U R A L E N G I N E E R I N G R E F E R E N C E M A N U A L

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P P I • p p i 2 p a s s . c o m

Figure 7.6 Column Dimensions

A A

4b < h ≤ 99rmin

8 in ≤ t ≤ 3b

b ≥ 8 in

section A-A

in diameter minimum

2 bars minimum

2 bars minimum

s ≤ 16 × longitudinal bar diameter ≤ 48 × lateral tie diameter ≤ b

14

s2

s2

slab reinforcement

0.0025An ≤ As ≤ 0.04An∑

The design column strength is

P (0.9)(250†kips)

225†kipsn =

=

Combined Compression and FlexureASD Method

The allowable compressive stress in masonry due tocombined axial load and flexure is given by TMS 402Sec. 8.3.4.2.1 as

F f0.45b m=

In addition, the calculated compressive stress due to theaxial load cannot exceed the allowable values given inTMS 402 Sec. 8.2.4.1. For columns having an h/r rationot greater than 99, this is

F fh

r(0.25 ) 1.0

140[TMS†402†8-16]a m

2

=

For columns having an h/r ratio greater than 99, theallowable value is

F fr

h(0.25 )

70[TMS†402†8-17]a m

2

=

When the axial load on the column causes a compressivestress larger than the tensile stress produced by theapplied bending moment, the section is uncracked andstresses may be calculated by using the transformed sec-tion properties.3 To allow for creep in the masonry,6 thetransformed reinforcement area is taken as As(2n − 1),and the resultant stresses at the extreme fibers of thesection, as shown in Fig. 7.7, are given by

f f fPA

MS

m a b

t t

= ±

= ±

Stress in the reinforcement is equal to 2n times thestress in the adjacent masonry.

7-227-22 S E S T R U C T U R A L E N G I N E E R I N G R E F E R E N C E M A N U A L

Figure 7.7 Uncracked Section Properties

As

As(2n − 1) As(2n − 1)

applied forces

column section

transformedreinforcementarea

stress due toaxial load

stress due tobending moment

combined stress

As

P

M

PAt

+MSt

−MSt

− MSt

PAt

+ MSt

PAt

Rein

forced

Maso

nry

P P I • p p i 2 p a s s . c o m

The maximum flexural reinforcement ratio for specialreinforced masonry shear walls with M/(Vdv) ≥ 1.0 andwithP f A0.05 m n> is given by TMS 402 Eq. 8-23 as

nf

f nf

f2

m

yy

m

max =

+

SD Method

When flexural reinforcement is concentrated at the endsof a shear wall and axial loads are comparatively light,the shear wall may be designed in the same manner as abeam in bending. When these conditions are not valid,it is necessary to analyze the wall using basic principles.

To prevent brittle failure of a lightly reinforced shearwall, TMS 402 Sec. 9.3.4.2.2.2 requires the nominalflexural strength of the shear wall to be greater than 1.3times the nominal cracking moment strength. Therequired nominal moment is

M M1.3 cr

In order to provide adequate ductile response in a shearwall, TMS 402 Sec. 9.3.3.5.1 through TMS 402 Sec.9.3.3.5.4 limit the maximum reinforcement ratio inaccordance with a prescribed strain distribution. Themasonry compressive strain is defined as 0.0025 for con-crete masonry. The tensile strain, in the extreme tensionreinforcement, depends on the seismic design category,the shear wall type, the response modification factor, R,and the value of Mu/(Vdv). The tensile strain is limitedto the following values.

• 1.5s y= for R ≥ 1.5 andMu/(Vudv) ≤ 1.0

• s is not limited for R ≤ 1.5 andMu/(Vudv) ≤ 1.0

• 3.0s y= for an intermediate reinforced masonrywall withMu/(Vudv) ≥ 1.0

• 4.0s y= for a special reinforced masonry wall withMu/(Vudv) ≥ 1.0

The response modification factor, R, is defined inASCE/SEI75 Table 12.2-1. For masonry shear wallsused in bearing wall structural systems, the values of Rare given in Table 7.1.

The limit on maximum tensile reinforcement ratio forshear walls is waived if special boundary elements areprovided to the shear wall in compliance with TMS 402Sec. 9.3.6.5. Special boundary elements need not be pro-vided in shear walls meeting the following conditions.

The factored axial load does not exceed the valueP A f0.10u g m for geometrically symmetrical wall sec-

tions, or P A f0.05u g m for geometrically unsymmet-rical wall sections. In addition, one of the followingconditions must apply.

MV d

V A f

1.0

3M

V dwhen† 3.0

u

u v

u n mu

u v

Table 7.1 Response Modification Factor

masonry shear wall type R

special reinforced 5.0

intermediate reinforced 3.5special reinforced 5.0

ordinary reinforced 2.0

detailed plain 2.0

ordinary plain 1.5

Example 7.12

The nominal 8 in solid grouted concrete block masonryshear wall shown in the illustration has a specifiedstrength of 1500 lbf/in2 and a modulus of elasticity of1,000,000 lbf/in2. An in-plane wind load of 32 kips actsat the top of the wall, as shown, and this is the govern-ing shear load. The wall is located in a structureassigned to seismic design category C and is laid in run-ning bond. Determine the reinforcement required in thewall. Axial load may be neglected.

W = 32 kips

h = 15 ft

dv = 15 ft

Solution

ASD Method

The wall is located in seismic design category C. Anordinary reinforced masonry wall may be used sinceh 160†ft< .

7-307-30 S E S T R U C T U R A L E N G I N E E R I N G R E F E R E N C E M A N U A L

Reinforced

Masonry

P P I • p p i 2 p a s s . c o m

The allowable stresses, in accordance with TMS 402Sec. 8.3.3 and Sec. 8.3.4.2.2, are

F f

F

0.45

(0.45) 1500†lbf

in

675†lbf/in

32,000†lbf/in

b m

s

2

2

2

=

=

=

=

Using IBC Eq. 16-15, the factored load is

V W0.6(0.6)(32†kips)

19.2†kips

===

The bending moment, produced by the wind load, atthe base of the wall is

M Vh(19.2†kips)(15†ft)

288†ft-kips

===

Use Mu/(Vdv) = 1.0 per TMS 402 Comm. Sec. 8.3.5.1.2.Assuming that two no. 6 reinforcing bars are located4 in from each end of the wall, the relevant parametersof the wall are

( )b

d

A

nEE

Abd

n

7.63†in

(15†ft) 12†inft

4†in 176†in

(2)(0.44†in ) 0.88†in

29,000,000†lbf

in

1,000,000†lbf

in29

0.88†in(7.63†in)(176†in)

0.000655(0.000655)(29)

0.0190

s

s

m

s

2 2

2

2

2

=

= =

= =

=

=

=

=

=

===

From App. 2.B, the wall stresses caused by the windload, in accordance with TMS 402 Sec. 7.3.2, are

( )

k n n n

jk

fM

jkbd

2 ( )

(2)(0.0190) (0.0190) 0.0190

0.177

13

10.177

30.9412

(2)(288†ft-kips) 12†inft

1000†lbfkip

(0.941)(0.177)(7.63†in)(176†in)

176†lbf/in

675†lbf/in [satisfactory]

b

2

2

2

2

2

2

= +

= +

=

=

=

=

=

=

=

<

( )

fM

jdA

(288†ft-kips) 12†inft

1000†lbfkip

(0.941)(176†in)(0.88†in )

23,713†lbf/in

32,000†lbf/in [satisfactory]

ss

2

2

2

=

=

=

<

The flexural reinforcement provided is adequate.

The shear stress in the masonry wall is given byTMS 402 Eq. 8-24 as

( )f

VA

(19.2†kips) 1000†lbfkip

(7.63†in)(15†ft) 12†inft

14†lbf/in

vnv

2

= =

=

The allowable stress is obtained by applying TMS 402Sec. 8.3.5.1.3.

7-31R E I N F O R C E D M A S O N R Y D E S I G N 7-31

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Maso

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P P I • p p i 2 p a s s . c o m

The allowable shear stress in an ordinary reinforcedshear wall without shear reinforcement is given byTMS 402 Eq. 8-29 as

( )

FMVd

fPA

f

4.0 1.75 0.25

12

4.0 (1.75)(1.0) 1500†lbf

in0†

lbf

in

43.6†lbf/in

14†lbf/in [satisfactory]

vm mn

v

1

2

2 2

2

2

= +

= +

=

> =

Masonry takes all the shear force, and nominal rein-forcement is required as described in TMS 402Sec. 7.3.8.1.

SD Method

The structure is assigned to seismic design category Cand an ordinary reinforced masonry wall may be usedsince h < 100 ft.

Using IBC Eq. 16-6, the factored load is

V W1.0

(1.0)(32†kips)

32†kips

u ===

The bending moment produced by the wind load at thebase of the wall is

M V h(32†kips)(15†ft)

480†kips-ft

u u===

Assuming two no. 6 reinforcing bars are located 4 infrom each end of the wall, the relevant parameters of thewall are

( )b

d

A

7.63†in

(15†ft) 12†inft

4†in 176†in

(2)(0.44†in ) 0.88†ins2 2

=

= =

= =

The stress block depth is

aA f

bf0.80

(0.88†in ) 60†kips

in1000†

lbfkip

(0.80)(7.63†in) 1500†lbf

in5.77†in

s y

m

22

2

=

=

=

The nominal strength is

( )M A f da2

(0.88†in ) 60†kips

in176†in

5.77†in2

12†inft

762†ft-kips

n s y

22

=

=

=

The design strength is

M

M

(0.9)(762†ft-kips)

686†ft-kips[satisfactory]

n

u

==>

The flexural reinforcement provided is adequate.

TMS 402 Comm. Sec. 9.3.4.1.2 permits the adoption ofMu/Vudv= 1.0. The nominal shear strength of a shearwall without shear reinforcement is given by TMS 402Eq. 9-24. Since Pu= 0 lbf,

( )( )

VM

V dA f P

V

V

4.0 1.75 0.25

4.0 (1.75)(1.0) (7.63†in)(15†ft)

1500†lbf

in12†

inft

0†lbf

1000†lbfkip

120†kips

(0.8)(120†kips)

96†kips32†kips [satisfactory]

nmu

u vnv m u

nm

u

2

= +

=× +

===> =

The masonry takes all the shear force, and nominal rein-forcement is required as detailed in TMS 402Sec. 7.3.8.1.

8. DESIGN OF SLENDERWALLS........................................................................................................................

Nomenclature

a depth of equivalent rectangular stress block in

Ag gross cross-sectional area of member in2

Amax maximum area of reinforcement that willsatisfy TMS 402 Sec. 9.3.3.5.1

in2

An net cross-sectional area of member in2

As cross-sectional area of reinforcing steel in2

Ase equivalent area of reinforcing steel in2

b width of section in

7-327-32 S E S T R U C T U R A L E N G I N E E R I N G R E F E R E N C E M A N U A L

Reinforced

Masonry

P P I • p p i 2 p a s s . c o m

The applied strength level moment at midheight of thewall is given by TMS 402 Eq. 9-30 as

( )

Mw h P e

P

MM

8 2

†

0.03†kip

ft(20†ft)

8

(0.36†kip)(7.5†in)

(2) 12†inft

†(1.04†kips)(0.20†in)

12†inft

† 1.63†ft-kips†

†

[satisfactory]

[TMS†402 Eq. 9-30 applies]

uu uf u

u u

n

cr

1

2

1

2

= + +

= +

+

=<<

The deflection corresponding to the factored moment isdetermined in accordance with TMS 402 Sec. 9.3.5.4.2.The moment of inertia of the net wall section is

( )I

bt12

†(1†ft)(7.63†in) 12†

inft

12† 444†in

n

3

3

4

=

=

=

The modulus of elasticity of reinforcement is given byTMS 402 Sec. 4.2.2.1 as

E 29,000†kips/ins2=

The modulus of elasticity of concrete masonry is givenby TMS 402 Sec. 4.2.2.2.1 as

E f900

† (900)1500†

lbf

in

1000†lbfkip

† 1350†kips/in

m m

2

2

=

=

=

The modular ratio is

nEE

29,000†kips

in

1350†kips

in21.5

s

m

2

2

= =

=

The distance from the extreme compression fiber to theneutral axis is given by TMS 402 Eq. 9-35 as

cA f P

f b0.64

(0.15†in ) 60†kips

in1.04†kips 1000†

lbfkip

(0.64) 1500†lbf

in(12†in)

0.87†in

s y u

m

22

2

=+

=

+

=

Allowing for the axial load, the moment of inertia of thecracked transformed section about the neutral axis isgiven by TMS 402 Eq. 9-34 as

( )

Ibc

n APf

d c3

( )

†(1†ft)(0.87†in) 12†

inft

3

† (21.5) 0.15†in1.04†kips

60†kips

in

†7.63†in

20.87†in

† 33.8†in

cr su

y

32

3

2

2

2

4

= + +

=

+ +

×

=

Since M Mu cr1 < , the midheight deflection correspond-ing to the factored moment is derived from TMS 402Eq. 9-29 as

( ) ( )

M hE I

5

48

†

(5)(1.69†ft-kips) 12†inft

(20†ft) 12†inft

(48) 1350†kips

in(444†in )

† 0.203†in

†This is approximately equal to

the assumed deflection; further

iterations are unnecessary.

ucr

m n

2

2

24

=

=

=

So,

MM

1.63†ft-kips

[satisfactory]u

n

=<

The flexural capacity is adequate.

7-39R E I N F O R C E D M A S O N R Y D E S I G N 7-39

Reinforced

Masonry

P P I • p p i 2 p a s s . c o m

Example 7.19

As given in Ex. 7.17, a glulam crosstie between dia-phragm chords is supported at one end in a steel beambucket. The bucket is attached to a solid grouted con-crete block masonry wall with a masonry compressivestrength of 1500 lbf/in2. The four headed anchor boltsare †in3

4 diameter ASTM A 307 type C, with a mini-mum specified yield strength of 36 kips/in2 and an effec-tive cross-sectional area of 0.334 in2. The effectiveembedment length of the anchor bolts is 6 in, and thethreaded portion of the bolts is not located within theshear plane. The ASD governing load combination givesa tension force on the beam bucket of T = 13 kips and ashear force ofV 1.5†kips= . The SD governing load com-bination gives a tension force of T 21.7†kipsu = and ashear force ofV 2.25†kipsu = . Determine if the bolts areadequate for the combined tension and shear forces.

Solution

The applied tension force on one bolt is

bT413†kips

43.25†kips

a =

=

=

The applied shear force on one bolt is

bV41.5†kips

40.38†kip

v =

=

=

For combined tension and shear, TMS 402 Eq. 8-10must be satisfied.

bB

bB

1

3.25†kips

5.25†kips

0.38†kip

1.66†kips0.85

† 1 [satisfactory]

a

a

v

v+

+ =

<

SD Method

The factored tension force on one bolt is

bT421.7†kips

45.43†kips

afu=

=

=

The factored shear force on one bolt is

bV42.25†kips

40.56†kips

vfu=

=

=

For combined tension and shear, anchor bolts must com-ply with TMS 402 Eq. 9-10, which is

b

B

b

B1

5.43†kips

8.41†kips

0.56†kip

2.48†kips0.87

1 [satisfactory]

af

an

vf

vn+

+ =

<

10. DESIGN OF PRESTRESSEDMASONRY........................................................................................................................

Nomenclature

An net cross-sectional area of masonry in2

Aps area of prestressing steel in2

As area of nonprestressed reinforcement in2

b width of section inCm force in masonry stress block lbf

d distance from extreme compression fiberto centroid of prestressing tendon orreinforcement

in

D dead load or related internal moments orforces

lbf

e eccentricity of axial load in

Em modulus of elasticity of masonry lbf/in2

Emi modulus of elasticity of masonry attransfer

lbf/in2

Eps modulus of elasticity of prestressing steel lbf/in2

fa calculated compressive stress in masonrydue to axial load

lbf/in2

fai calculated compressive stress in masonryat transfer due to axial load

lbf/in2

fb calculated compressive stress in masonrydue to flexure

lbf/in2

fbi calculated compressive stress in masonryat transfer due to flexure

lbf/in2

fmspecified compressive strength of masonry lbf/in2

fmispecified compressive strength of masonryat time of prestress transfer

lbf/in2

fpj stress in prestressing tendon due tojacking force

lbf/in2

fps stress in prestressing tendon at nominalstrength

lbf/in2

fpsi initial stress in prestressing tendon lbf/in2

7-49R E I N F O R C E D M A S O N R Y D E S I G N 7-49

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Maso

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P P I • p p i 2 p a s s . c o m

For a centrally located, laterally restrained or unre-strained, unbonded prestressing tendon, the stress inthe tendon at nominal load has been determined empiri-cally, and it is given by TMS 402 Eq. 10-3 as

f fE d

l

A fP

f bd

ff

0.03 1 1.56

†

†

ps seps

p

ps psu

m

se

py

= ++

The force in the prestressing tendon at nominal load is

P f Aps ps ps=

A rectangular compression stress block is assumed inthe face shell on the compression face of the wall with amagnitude of f0.80 m and a depth of a. The compressionforce on the stress block acts at mid-depth of the stressblock and is given by

C abf0.80m m=

Stress in the face shell on the tension face of the wall isignored. Allowing for the factored axial load, Pu, on thewall, the depth of the stress block is obtained by equat-ing forces on the section to give

af A

P

f b0.80

ps psu

m

=+

When the wall also contains concentrically placed,bonded, nonprestressed reinforcement in grouted cores,the depth of the stress block is given by TMS 402Eq. 10-1 as

af A f A

P

f b0.80

ps ps y su

m

=+ +

The nominal flexural capacity of the section is obtainedby taking moments about the line of action of the com-pression force to give TMS 402 Eq. 10-2 which is

( )M f A f AP

da2n ps ps y s

u= + +

The design flexural capacity is given by TMS 402Sec. 10.4.3.3 as

M M0.8n n=

To ensure a ductile failure in concrete block walls sub-ject to out-of-plane loading, TMS 402 Sec. 10.4.3.6 spec-ifies a maximum depth for the stress block of

a d0.36=

In addition, the depth of the stress block may not exceedthe thickness of the face shell.

Example 7.22

The nominal 8 in ungrouted concrete block masonry walldescribed in Ex. 7.20 has face-shell mortar bedding oftype S portland cement/lime mortar, a specified strengthof f 2500†lbf/inm

2= at 28 days, and a specified strength

of fmi = 2000 lbf/in2 at transfer. The wall is post-tensionedwith †in7

16 diameter steel rods at 16 in centers placed cen-trally in the wall, and the rods are laterally restrained.The wall has an effective height of h = 20 ft and is simplysupported at the top and bottom. The roof live load is200 lbf/ft, and the roof dead load is 400 lbf/ft applied tothe wall without eccentricity. The masonry wall has aweight of γw = 37 lbf/in2. The properties of the steel rodare Eps = 29,000 kips/in2, Aps = 0.142 in2, fpy = 100 kips/in2, fpu = 122 kips/in2. Wind load governs and has a valueof w = 25 lbf/ft2. Assume the loss of prestress at transfer is5% and the final loss of prestress is 30%. Determine if thewall design flexural capacity is adequate. Ignore P-deltaeffects.

Solution

The following calculations are based on a 1 ft length ofwall and on information obtained from Ex. 7.20 andEx. 7.21.

The weight of the wall above midheight plus the roofdead load is

P 770†lbf=

In determining the moment demand on the wall, thegoverning load combination is IBC Eq. 16-6, which is

Q D W0.9 1.0u = +

The factored axial load on the wall is

P P0.9 (0.9)(770†lbf) 693†lbfu = = =

7-587-58 S E S T R U C T U R A L E N G I N E E R I N G R E F E R E N C E M A N U A L

Forces

P P I • p p i 2 p a s s . c o m

The maximum active pressure at the bottom of the heel is

p K h

(0.27) 120†lbf

ft(15†ft)

486†lbf/ft

A A S

3

2

=

=

=

For a 1 ft length of wall, the total horizontal force on theretaining wall is

Hp h2

486†lbf

ft(15†ft)

(2) 1000†lbfkip

3.65†kips/ft

AA

2

=

=

=

Passive Pressure

For the retaining wall shown in Fig. 1.35(c), which isconstructed on a soil with a density of γS and an angle ofinternal friction of φ, the passive earth pressure that canbe developed in front of the key at a depth z belowground surface is given by Rankine’s theory as

p K zP P S=

In this equation, KP is the Rankine coefficient of passiveearth pressure and is

K1 sin

1 sinP =+

The maximum passive pressure occurs at the bottom ofthe key and is

p K hP P S K=

In this equation, hK is the height of the key. The totalhorizontal force on the key due to passive pressure is

Hp h K h

2 2PP K P S K

2

= =

The line of action of the total horizontal force is located(h + hK)/3 above the bottom of the key (see Fig. 1.35(c)).

Example 1.25

The retaining wall shown in Fig. 1.35(c) is constructedon a soil with a density of 120 lbf/ft3 and an angle ofinternal friction of 35°. The total height of the key is5 ft. Determine the total horizontal force per foot of walldue to passive pressure on the key.

Solution

The Rankine coefficient of passive earth pressure is

K1 sin

1 sin

1 sin 351 sin 353.69

P =+

=+ °

°=

The maximum passive pressure that can be developedat the bottom of the key is

p K h

(3.69) 120†lbf

ft(5†ft)

2214†lbf/ft

P P S K

3

2

=

=

=

For a 1 ft length of wall, the total force due to passivepressure that can be developed on the key is

Hp h

2

2214†lbf

ft(5†ft)

(2) 1000†lbfkip

5.54†kips/ft

PP K

2

=

=

=

Hydrostatic Pressure and Soil Lateral Pressure

A retaining wall is usually provided with a drainage sys-tem behind the wall. Where this is not possible andadjacent soil is below the free-water table, design mustinclude the effects of submerged soil pressure plushydrostatic pressure as shown in Fig. 1.36.

For the retaining wall shown in Fig. 1.36, which has asoil backfill with a dry density of γS, the active earthpressure behind the wall at a depth of hdry below the fillsurface is given by Rankine’s theory as

p K hA A S dry=

1-47V E R T I C A L , I N C I D E N T A L L A T E R A L , A N D O T H E R V A R I A B L E F O R C E S 1-47