theoretical design of a nailed or hated joint under

THEORETICAL DESIGN OF A NAILED ORHATED JOINT UNDER LATERAL LOAD

Revised March 1955

INFORMAAND REAHRMED

193C

No. V1951

UNITED STATES DEPARTMENT OF AGRICULTUREFOREST SERVICE

FOREST PRODUCTS LABORATORYMadison 5,WisconsinIn Cooperation with the University of Wisconsin

4

THEORETICAL DESIGN OF A NAILED OR BOLTED JOINT

UNDER LATERAL LOAD-

By

EDWARD W. KUENZI, 2— Engineer

Forest Products Laboratory, 3 Forest ServiceU. S. Department of Agriculture

Summary

This report presents a theoretical method for determining the allowablelateral load of a joint having a single nail or bolt. The design value of thejoint is computed after combining various parameters involving the mate-rial properties of the members being joined and the joining nail or bolt.Included is an appendix giving several examples illustrating application ofthe theoretical method to specific joints.

Introduction

A great deal of experimental work has been done to evaluate the lateralresistance of nailed or bolted joints. Usually the final design value isarrived at for a particular joint by applying a factor of safety to the maxi-mum test load or by determining the test load at some specified amount ofjoint slip or deformation. The former method may lead to structures inwhich joint deformations may be excessive. The latter method may resultin joints in which local stresses in the members or in the nail or bolt will

1—Originally issued in September 1953.

?The author wishes to acknowledge the assistance of W. S. Ericksen informulating the problem and carrying out some , of the mathematicalcalculations.

3—Maintained at Madison, Wis. , in cooperation with the University of

Wis.Report No. 1951 Agriculture-Madison

exceed allowable values because the experimental determination of jointslip may have included much deformation in test members and apparatus;thus small local yielding of the joint would not appreciably affect thetotal deformation as measured, and a higher load, high enough so thatlocal yielding would be great enough to cause nonlinearity, would be theapparent design value.

Certain empirical design criteria, in addition to simplifying assumptions,are in current use for the lateral resistance of nailed or bolted joints. Itis felt, however, that an analysis including the properties of various mate-rials would enable better design of joints. Such an analysis could also beapplied to materials in which no actual joint test data are available for theparticular sizes and materials that it is desired to use.

In 1951 Moller 4— described the failure of a nailed joint by considering thenail under uniformly distributed bearing forces equal to the strength of thewood. His analysis, however, did not consider joint deformation or thepossibility of the nail being permanently distorted at stresses in the woodless than the strength of the wood.

The following analysis considers the nail or bolt as being supported on anelastic foundation. The design value for the joint is finally found by deter-mining the load at which stresses in either the members or the bolts reachproportional limit or yield values, whichever produces the least load.Expressions are given for determining the deformation of the joint at anyload up to this yielding joint load. Extrapolation of the analysis to greaterloads is not expected to produce correct results unless some method bedevised to take into consideration the plastic behavior of the members andthe nails or bolts.

Theoretical Design Criteria

The nail or bolt is assumed to be a beam supported on an elastic founda-tion such that deflection of the nail or bolt is resisted by a pressure propor-

tional to the deflection at any point and such that this pressure can be exertedin both upward and downward directions. The analysis of beams on elasticfoundations of this type was first introduced by E. Winkler in 1867, andsolutions for beams of finite length that are applied here were given byHet6nyi 5 in 1946. The foundation is assumed to be discontinuous(transmitting no shear), but it has been found4 that moments produced in

4—M011er, Torsten. En Ny Metod for Brakning av SpikftSrband. Nr. 117, Trans-actions of Chalmers University of Technology, Gothenburg, Sweden. 1951.

...,Hetenyi, M. Beams on Elastic Foundation, pp. 50-53 and p. 207. Ann

Arbor, Mich. 1946.

Report No. 1951 -2-

beams supported on an elastic solid may be closely approximated byconsidering the elastic solid to be an elastic foundation with a foundation

beam width modulus equal to its modulus of elasticity x foundation depth'

Additional assumptions in regard to the joint itself are that the load dueto friction between the members is negligible and that tractions are notdeveloped between the members and the nail or bolt. Both of these as-sumptions are somewhat incorrect when joints between wood members arefirst made, but in use the changes in moisture in the wood will cause slightloosening of the joint and thus rectify the assumptions. It is commonpractice in testing joints to allow for shrinkage by spacing the membersslightly and be not joining tightly.

The fit between the bolt and hole is considered to be tight. The hole shouldbe smooth in order that the joint will function most efficiently. It has beendemonstratedb that a rough hole will lower the proportional limit value for ajoint in wood to about one-third of the value obtained for a bolt in a smoothhole.

The differential equation for the deflection curve of a beam supported on anelastic foundation is given byl.

EI d4

Ydx

4 = - ky (1)

where EI = stiffness of the beam; E = modulus of elasticity; I = momentof inertia

y = deflection at point xIc.•= foundation modulus.

The solution of equation I finally results in expressions involving a charac-teristic

k=

4E1

(2)

and expressions for deflections, moments, and shears depend upon thevalue of this characteristic.

6—Goodell, H. R. and Phillips, R. S. Bolt-bearing Strength of Wood and

Modified Wood -- Effects of Different Methods of Drilling Bolt Holes inWood and Plywood. U. S. Forest Products Laboratory Report No. 1523.December 1944. (See Table 1.)

Report No. 1951 -3-

Solutions for two types of joints follow.

I. Nailed or Bolted Joint of Two Members

The joint considered is shown in figure 1. Figure 1, A, shows thechoice of axes and figure 1, B, shows the forces at the joint and theslip between the joints. The thicknesses of the members are desig-nated by a and b. If the nail or bolt does not completely penetrate themember, a or b would be the depth of penetration.

The deflection of the nail or bolt in member 1 is given by

Y1

2PX1 (Sinh Al a Cosh Aix cos ya - x) - sin A l a cos Aix Cosh Xi (a - x)

2M0X12 Sinh A l a Cosh A i x sin X i (a - x) - Sinh Xix cosAlkl

K1 (a - x)]

[+ sin Ala Sinh Al fa - x) cos Xix - Cosh ya - x) sin Xixj

(3)where

41X kl

4E1

= Sinh 2 X / a. - sin2Xia.

The slope of the nail or bolt in member 1 is given by

{ dSinh Ala Cosh X.ix sin Xi (a - x) + Sinh Xix cos Xi (a - x

dy1 2PX. 12

dx

Alkl

+ sin X a [cos Aix Sinh - x) + sin X.ix Cosh Al fa- x)1

4M 0 X 13

P 1 k 1Sinh Al a Cosh X1 x cos Al fa - x) + sin Al a cos Xix Cosh Xi (a - x)

(4)Report No. 1951 -4-

••

By substituting the expression for and further recalling that thed y2

moment M 1 = - El

I , the moment at any point along the nail or bolt

dx2

in member 1 is given by

M = Sinh AlaSinh y sin k(a - x) - sin Ala sin kix Sinh X.1 (a - x)A l X1

MO Sinh X i a Sinh A i x cos X + Cosh k i x sin k i (a -x)]

- sin A l a Icos A i x Sinh A i (a -x) + sin A i x Cosh x

(5)

dMiBy recalling that the shear Q 1 - , the shear at any point along the

dxnail or bolt in member 1 is given by

Q 1 =

Sinh Cosh kix sin Al fa -x) - Sinh kix cos Xi (a - x)1 _J

- sin A l a cos A i x Sinh X i (a - x) - sin k i x Cosh A i (a x)]

2M0 k1Sinh A l a Sinh A i x sin X1 (a - x)+ sin Ala sin Aix Sinh k(a x)

(6)

Similar expressions will be written for the deflections, slopes, moments,and shears in the bolt in member 2.

y 2 -6727-2-- Sinh X. 2b cos X2x Cosh X.2 03- x) - sin Alb Cosh X2x cos X.2 (b - x)2PX2

2M0 X22 Sinh X2b [sin X2x Cosh X2 (b - x) - cos X.2x Sinh X.2 (b - x)A2k21

+ sin X 2b ISinh X2x cos X2 (b - x) - Cosh X. 2x sin X2 (b -x)

Report No. 1951 -5-

••

-1}(7)

where

dy22PX.22

0 2 = Sinh 2 X2b - sin2 X2b X. 2 =

4 k z

4E1

= _

[

Sinh X2 b sin X2 x Cosh X2 (b - x) + cos X2 x Sinh X2 (b - x)dx 0 2 k2

+ sin X2 b [Sinh X

2 x cos X

2- x) + Cosh X2 x sin X2 (b - x)

4M0

X.23

Sinh X2 b cos X2 x Cosh X.-2 (b - x) + sin X.2 b Cosh X2x cos X2(b - )A

2k

(8)

P2M 2 = Sinh X2 b sin X2 x Sinh X2 (b - x) - sin X2 b Sinh X2 x sin X 2 (b - x)

A2k2

inh X2 b cos X2 x Sinh X2 (b - x) + sin K2 x Cosh X2 (b - x

}- sin K2 b [Sinh X2 x cos K2 (b - x)+ Cosh K2 x sin X2 (b - x1

J

2 A2

}[- sin X2 b Cosh X2 x sin X2 (b x) - Sinh X.2 x cos X2 (b - x)

2M0 K2q 2

Sinh X2 b sin K2 xSinh X2 (b - x) + sin X2 b Sinh K2 x sin X2 (3 - x)

MOA

2

(9)

P Sinh X2 b cos X2 x Sinh X2 (b - x) - sin K2 x Cosh X.2 (b -

Report No. 1951 -6-

Sinh k2b Cosh X2b + sin X2b cos X2bK2 k2

2Sinh 2 X2b - sin k2b

At the jancture of members 1 and 2, the slope of the nail or bolt arethe same and

dyi dy2

dxx = a

dx x=0

Substitution of equations 4 and 8 in equation 11 gives

[Sinh 2 k1 a+ sin2 X1 a]4M 0 k13

A l k 1 [Sinh Xi a Cosh Xi a + sin NJ a cos Xia =

2Pk22 4M0 X23 Sinh2 k2b +sin2 k2b + Sinh X2b Cosh X.2b + sin k2b cos X2bA2k2A2k 2

(12)

Sinh2 a + sin Xi ak 1 Sinh 2 k a - sin2 k1 a1

(13)

Sinh2 X2b + sin2 k2lo-2 k2 Sinh2k2b - sin2k2b

3 Sinh Cosh ki a + sin ki a cos X1a

Sinh 2 X1 a - sin2 k, a

LetJ1

2kl

(Curves for use in computing J and K are given in figure 2. )

Report No. 1951 -7-

1:(J 1 - J2)(14)•

and finally

MO 2(KI + K2)

(17

]

Sinh X2b Cosh X2b - sin X2b cos X.2bkz-L2 - 1(2

Sinh2

X.2b - sin2

X2b

Then equation 12 becomes

- 2PJ 1 - 4M0 K1 = - 2PJ 2 + 4M0K2

Also at the juncture of members 1 and 2 the slip, 6, is given by

6 = yl + y2I

(15)

x= a x -=0

and substitution of equations 3 and 7 in equation 15 gives

2PX1=Sinh Xla Cosh X1 - sin Xi a cos Xi a.

P1k

11

22M X.0 1 . 2

+ Sinh2X1 a + sin XI a/I l k1

(16)2PX2

Sinh X2b Cosh X2b - sin X2b cos X2/31P2k2

2M O X22

2Sinh Xzb + sin

2X b.]

k2

]Sinh Xi a Cosh Xi a - sin Xi a cos Xia

Sinh Xi a - sin Xis,2 2

Let

L1xik

1

(Curves for use in computing L are given in figure 2)Report No. 1951 -8-

P2 max. = Pk 2 2L 2 +K1 + K2

J 1 (J - J 2 )(20)

Substituting equations 13, 14, and 17 in 16 finally results in

- J2)2-

= P 2(L 1 + L2) - + K2

The pressure under the nail or bolt at any point is given by

p = ky

Therefore, in member 1 the maximum pressure will occur at x = a andwill be equal to

(18)

p1 max. - Pkl

J 1 (J 1 - J2 )2L 1 K 1 +K2 (19)

and in member 2 the maximum pressure will occur at x = 0 and will beequal to

The maximum moment in the nail or bolt occurs in either member 1 or 2at a point where the shear is zero. By setting the equation for the shearin member 1 (equation 6) equal to zero, after expansion and final reduc-tion the shear will be zero at values of x satisfying the equation

Coth X1 x + cot X1x -

Coth Xi a - cot Xi a - B1

whereXi (J 1 - J2 )

B,- K1 + K2

Similarly, the shear in the nail or bolt in member 2 will vanish at valuesof x satisfying the equation

2 - 13 1 [ Coth + cot Xia.](21)

Report No. 1951 -9-

2 + B 2 [ Coth X2b + cot X 2b ]

Coth X2 (b - x) + cot X 2 (b - x) = Coth X2b - cot X2b + B2

where-X2(j1 J2)

B -2 K1 + K2

A curve of the value of Coth 0 + cote is given in figure 3 for use in solvingfor 0 = X

1 x or 0 = X.2 (b - x) in equations 21 and 22, respectively. The func-

tion Coth 0 + cot 0 is periodic, and only one branch to 0 = it is shown infigure 3. For solutions beyond 0 = it the Coth 0 may be assumed to be onewithout great error and then equations 21 and 22 can be solved directly forcot Xl

x or cot X2 (b - x).

IL Nailed or Bolted Joint of Three Members

The joint considered is shown in figure 4. Figure 4, A, shows the choiceof axes, and figure 4, B, shows the forces at the joint and the slip betweenjoints. The thickness of the inner member is designated by a; each of theouter members, of equal thickness, are of thickness b.

The deflection of the nail or bolt in member 1 is given by

Y1 = - PX1 Cosh X1 x cos Xi (a - x) + cos X1 x Cosh Xi (a. - x)

2M0 X 1

Sinh X1 x cos Xi (a - - Cosh Xi x sin Xi (a -

+ cos X x Sinh X (a - x) - sin X1 x Cosh k 1 (a - x)

1 1

Where 4

X 1 =

Al = Sinh X a + sin X a1 1

(22)

A l k l

(23)

k 1

4E1

Report No. 1951 (10)

The slope of the nail or bolt in member 1 is given by

dyi

dx

p xlz

Al k1Sinh Xi x cos Xi (a - x) + Cosh X1 x sin Xi (a - x)

- cos Xi x Sint]. Xi (a - x) sin Ki x Cosh Xi (a - x

4M0 X 1

Cosh X1 x cos Xl (a -x) - cos 11 x Cosh Xi (a - x) (24)1 k1

The moment at any point along the nail or bolt in member 1 is given by

M 1 =2ASinh Xi x sin Xi (a - x) + sin Xi x Sinh Xi (a - x)

1 1

+ —mo Sinh Xi x cos Xi (a - x) + Cosh Xi x sin Xi (a- x)

+ cos X 1 x Sinh Xi (a - x) + sin Xi x Cosh Xi (a - x)

The shear at any point along the nail or bolt in member 1 is given by

Q1 = -0

Sinh Xi x cos Xi (a - x) - Cosh Xi x sin Xi (a - x)1

+ sin Xi x Cosh Xi (a - x) - cos X i x Sinh Xi (a -x)

2M0X1 Sinh Xl x sin Xl (a - x) sin Xl x Sinh Xl (a -x) (26)Al

The expressions for the deflection, slope, moment, and shear in member 2will be identical with equations 7, 8, 9, and 10, but with P in those equa-tions replaced by P/2.

As before at the juncture of members 1 and 2, the slope of the nail or boltare the same and

dy1 dy2

dx

dxx= a x =0 (27)

(25)

Report No. 1951 -11-

Substitution of equations 24 and modified'' equation 8 into equation 27gives

px. 21 [Sinh

7Kl a - sin X.l aj

7

-Sinh 2 K 2b+ sin2 X. la]

2+

4M OX.

13 [Gosh X1 a ® cos X a

[Sinh X2b Gosh X2b + sin X2b cos k21D1i

Al k1

PX. 22

A1k1

4M0X23

2k 2 A2k2

(28)

Let

Sinh X a - sin K a1 1Sinh X 1 a+ sin Kia

X2Sinh2X2b + sin2X2b2

2 k2 FSinh 2 X2b - sin2 X2b

k 3 Cosh Xi a - cos X 1 a1

z

k Sinh X a + sin K a1 1

X 23 Sinh X2b Cosh X2b + sin X2b cos K2b

K2 k2 . 2

Sinh2 X2

b sin K2b

(Curves for use in computing J 1 and K1 are given in figure 5 9 for

J 2 and K2 in figure 2. )

Then equation 28 becomes

PJ1 - 4M0

K 1 - P32

+ 4M 0 K2

K1

(29)


and, finally,

P(31 - j2)

M = -

O 4(Ki +K2)

Also at the juncture of members 1 and 2, the slip, 6, is given by

6= - yl + y2 (31)

x= a x=0

and substitution of equations 23 and modified equation 7 in equation 31 gives

(30)

[21 2M

Ok

1 • Cosh X1 a + cos kl a + A k Sinh Al a - sin Ala1 1

PX2- Sinh X2b Cosh X2 - sin X2b cosP2k2 X-2.1"11

2M0 X22

P2k2[Sinh2X2b + sin Xzb] (32)

A l Cosh X a + cos X aL = 1 1 11 k Sinh Xl a + sin X1 a1

X 2 Sinh X b Cosh X - sin X2b b cos X2b2 2 2L 2 =k2 Sinh 2 X2b - sin2 X2b

(Curves for use in computing L 1 are given in figure 5, for L 2 infigure 2. )

(33)

Pk1= Pi k 1

Let


P 2 max. = Pk2 [L2 +(36)]J201 - J2)

2(K1 + K2)

Substituting equations 29, 30, and 33 in 32 finally results in

[6 = P Li + L2 2(K1 + K2) -(J 1 - J2)

The pressure under the nail or bolt at any point is given by

p = ky

Therefore, in member 1, the maximum pressure will occur at x = a orx 0 and will be equal to

P1 max. Pk1 L lJ 1 ( .1 1 - J 2 ) 2(K1 + K2)

(35)

and in member 2 the maximum pressure will occur at x = 0 and will beequal to

(34)

As for the joint of two members, the maximum moment in the nail or boltoccurs in either member 1 or 2 at the point where the shear is zero. Bysetting the expression for the shear in member 1 (equation 26) equal tozero, after expansion and final reduction the shear will be zero at x = a/2or at values of x satisfying the equation

cot X1- —a )2

1 + Coth Al a cot X. + B 1 cot X1 a2 1 2a

1 2

Z a a

1 - Coth X cot X + B 1 Coth X-I 1 -I

where

B- Al (J1- J2 )

1 K1 + K2

( 3 7 )a

Coth X1 (x - 2)


aA curve of cot 0 is given in figure 6 for use in solving for 0 = X1

)(x - —

Coth 0 2

in equation 37. The function cot 0 is periodic, and only one branch toCoth 0

0 = it is shown in figure 6. For solutions beyond 0 = Tr the Coth 0 maybe assumed to be one without great error and then equation 37 can be

solved directly for cot X i (x - _a).

The position of zero shear is given in member 2 by the value of x satis-fying the equation

2 + B 2 [ Coth X 2b + cot X2b]Coth X 2 (b - x) + cot X (b - x) =

where

X2 (J

1 - J

2)

B 2 =K

1 + K

2

A curve of the value of Coth 0 + cot 0 is given in figure 3 for use in solvingequation 38 for 0 = X2 (b - x). It should be recalled, as for the two-member

joint, the Coth 0 + cot 0 is a periodic function.

Application of Theoretical Design Criteria

For determining the design value of a joint, it is necessary to examine allpossible points where high stress can begin to cause failure. Thus, acomplete analysis must be carried out for any particular joint. Perhapsafter calculating a series of joints involving certain changes of membersor nails or bolts it may be possible to present curves of design valuesdependent on the various parameters involved.

For members having definite compressive failing stresses, such as woodloaded in a direction parallel to the grain, the load at which the compressiveproportional limit stress is reached at the point of maximum nail or boltdeflection will often be the design value of the joint if the nail or bolt isrelatively stiff. The foundation modulus, k, for wood loaded parallel tothe grain direction may be taken as the compressive modulus of

widthbeam w elasticity x because deformations occur mainly infoundation depth

directions parallel to the grain of the wood. The load at which yielding

Coth X2b - cot X 2b + B2

(38)


of the nail or bolt occurs, due to maximum moment, should be calcu-lated to be sure it is not less than the load necessary to exceed thecompressive proportional limit stress of the member.

For members having no definite compressive failing stress, such aswood loaded in a direction perpendicular to the grain, the design valueof the joint will be the load necessary to cause yielding of the nail orbolt. It should be remembered that some permanent joint slip due toplastic deformation of the members may occur at this design load. Itmight be expected that the nail or bolt loaded in a direction perpendicu-lar to the grain would be supported by a foundation modulus in excess ofthe compressive modulus of elasticity perpendicular to the grain becauseof some supporting action in a direction parallel to the grain. Previoustests,' however, have shown such supporting action to be insignificant,so that the foundation modulus may be taken to be the compressive modu-

lus of elasticity perpendicular to the grain x beam widthfoundation depth

7Trayer, George W. The Bearing Strength of Wood Under Bolts. U. S.D. A. Technical Bulletin No. 332. October 1932.


Appendix

Example 1

Determine the design value of a two-member joint of nominal 1- and2-inch Douglas-fir with an eightpenny common nail. The joint is loadedlaterally and in directions parallel to the grain in each member.

Members -- a = 1. 625 inch, b = 0. 781 inch, E = 2, 112, 000 pounds persquare inch, SPL = 6, 450 pounds per square inch.

Nail -- d = 0. 131 inch, E = 30, 000, 000 pounds per square inch, EI =434 lb. -in. 2, SEL = 60, 000 pounds per square inch,

MEL

= 13. 25 in. -lb.

0. 131

k = 1 x 2, 112, 000 = 277, 000 pounds per dquare inch

(assuming effective foundation depth of 1 inch).

X 1 = X 2 = 3. 55 then X la = 5. 77 and X

2b = 2. 77

Sinh X a = 160. 3, Cosh X 1a = 160. 3, Coth Xla = 1. 000, sin X a =-0. 4901 1 1

cos X a = 0. 873,

1cot X

1a = -1. 783

Sinh X2b = 7. 948, Cosh X

2b = 8. 011, Coth X

2b = 1. 008, sin X b = 0. 361

2

cos X2b = -0. 932, cot X 2b = -2. 583

2 2 3 3X X X X

1 2 1 2J, = . J= , K

1 = , K=

1 k 2 1

2 k21

k 2 k k

X. 1 X2L 1 =

L2 k1 k '=

1 2


Then

and since

then

Finally,

then

X.1 = X

2 and k = k

1 2

J1 = J

2 and M

o = 0 and B

1 =13

2 = 0

= n - 21D k1 max. '2 max.

Pmax.P =2 x 3. 55

In member 1 the shear is zero if

6450 x . 131

7. 10- 119 pounds.

2Coth X 1 x + cot X x - - 0. 719

1 1 + 1. 783

and from figure 3, 0 = 1. 88, X 1 = 3. 55, then x = 0. 530, which is obviously

not the position of maximum moment since the maximum moment willoccur nearer to the joint. Since X.

1x must be greater than 4, then

Coth X 1 x = 1, and finally

cot X 1 x = -0. 279

which has solutions of

X x = 4. 9861

x = 1. 405

x = 8. 127 x = 2. 29, which is physicallyX1

impossible; therefore the maximum moment occurs at x •= 1. 405.

X 1 x = 4. 986, then Sinh X x = 73. 17 sin X x = -0. 9631 1

X.1 (a - x) = 0. 780, then Sinh X (a - x) = 0. 861 sin X 1

(a - = 0. 7031

and finally

73. 16 x 0. 703 P

M 1 max. 160.2 x 3. 55 11. 1


and

P = 11. 1 x 13. 25 = 147 pounds

In member 2 the shear is zero if

Coth x2 (b - x) + cot x2 (b - x) - 2 = O. 5571. 008 + 2. 583

and from figure 3

0 = 2. 02, then k2 = 3. 55, b - x = 0. 569, x = 0. 212,

Another solution is obtained for 0 = 5. 127 or x = b - 1. 445, which isphysically impossible.

Then the maximum moment occursat x = 0. 212 and

X x = 0. 754 Sinh2x = 0. 828 sin

2x = 0. 6852

k2 (b - x) = 2. 02 Sinh X2 (b - x) = 3. 703 sin k (b - x) = 0. 9002

and finally

and

M2 max.3. 703 x O. 685 P

= _

7. 948 x 3. 55 11. 1

P = 11. 1 x 13. 25 = 147 pounds.

Therefore the design value for the joint is determined by the compressionproduced in the wood and the design load is

P = 119 pounds.

At this load a joint slip of

119 x 4 x 3. 55 6 - 277, 000 - 0. 0061 inchL3

will occur.8—Experiments often give several times this slip, but much of the deforma-

tion of the members and apparatus is usually included in such test data.Report No. 1951 -19-

Sinh la = 4. 596 Cosh kla = 4. 704 Cothl a = 1. 240

2

sin k1 a = 0. 789 cos k1Xa = -0. 614 cot = 0. 491Ala

2

Figure 7 shows the shear, moment, and deflection diagrams for thisexample.

Example 2

Determine the design value of a three-member joint of nominal 2-inchDouglas-fir members on each side of a nominal 4-inch Douglas-firmember. The joint is made with a 1/2-inch steel machine bolt. The2-inch side members are loaded parallel to the grain direction and thecenter member is loaded perpendicular to the grain direction.

Member 1 -- a = 3. 625 inch, E l = 105, 000 pounds per square inch

Member 2 -- b = 1. 625 inch, E 2 = 2, 112, 000 pounds per squareinch, SPL = 6, 450 pounds per square inch

Bolt -- E = 30, 000, 000 pounds per square inch, EI = 92, 100lb. -in. 2,

= 30, 000 pounds per square inch, MEL = 368 in. -lb.SEL

Then0. 50

k 1 = 105, 000 x 1 = 52, 500 pounds per square inch

52, 500

X1 CI 4 x 92, 100= O. 615 X 1 a = 2.23

0. 378 x 3. 807 J = = 5. 10 x 10 -6

1 52, 500 x 5. 385

O. 232 x 5. 318 K4. 36 x 10-6

1= 52, 500 x 5. 385


=N2 4 x 92, 100 - 1. 30

1 056 000 4

Xb = 2. 11

L2 =1, 056, 000 x 16. 15

= 1. 36 x 10-61. 30 x 17. 79

0. 615 x 4. 090 L 1 - = 8. 91 x 10 -652, 500 x 5. 385

0. 50k2 2 , 112 , 000 x = 1, 056, 000 pounds per square inch1

Sinh X 2b = 4. 064

Cosh X2b = 4. 185 Coth X 2b = 1. 031

sin X b = 0. 857 cos X 2b = -0. 515 cot X2b = -0. 6012

1. 69 x 17. 61 -6J = 1. 75 x 102 1,056,000 x 16. 15

2.20 x 16.91 -62

K = - 2. 19 x 101, 056, 000 x 16. 15

Finally,

5. 10 - 1. 75M -0 4(4. 36 + 2. 18)

P = -0. 128 p

O. 615 x 3. 35B

1 - - O. 315

6. 54

1. 30 x 3. 35 B

2 - - 0. 6666. 54

In member 2 the maximum pressure under the bolt will be

-6 1. 75 x 3. 35 x 10 5= 1, 056, 000 1. 36 x 10 +Z max.2 x 6. 54

= 1. 91 P2 max.


SixthM 1 max. Al k 1 Si aSinh k 1 2- cos X a2 A 1 21

2 sin

which gives

6, 450P = ' - 1, 690 pounds.2 x 1. 91

Find maximum moment in member 1. The positions of zero shear willoccur for

cot k 1 (x - -a-2 ) 1 +0.61 +0.155 1. 765

1 - 0. 610 + 0.3 90 0. 780Coth X 1 (x -

2

_ 2.26

The curve of figure 6 shows a solution exists for 0> ir. ThereforeCoth A ti 1 and finally 0 = 3. 56, from which x = 7. 60, which is physicallyimpossible since the member is only 3. 625 inches thick.

ZThe maximum moment in member 1 must then occur at x = — = 1. 8122and is equal to

k l a k 1a 2 x 0.128P

kla+ Cosh sin a

A l aSinh =

M

1. 361

max.

k 1 2

asin X = 0. 898 cos1 2

2 x 0.128 (0.599 + 1.518)

X 1 2

= 0.Cosh

2

A l a - 1.6892

and

2

1.361 x O. 898

5. 385 x O. 615 5. 385

= 0.268 Pml max.

which gives

368P - 1,370 pounds.0.268

440


Find maximum moment in member 2. The positions of zero shearwill occur for

2.+ O. 666 (1. 031 - O. 601) _ 0. 995Coth X . (b - x) + cot X2 (b - x) =1. 031 + 0. 601 + 0. 666

and from figure 3

1. 660 = 1. 66 then b - x = = 1. 275 x = 0. 350 inch.1. 30

Another solution of this equation exists at 0 = 4. 729 when Coth 9 = 1. 00.1Then x = b - 3. 63, which is physically impossible because the memberis only 1. 625 inches wide. Therefore, the -maximum .moment occurs inmember 2 at x = 0. 350 inch.

X2x = 0. 455, X (b - x) = 1. 662

Sinh X2x = 0. 471 Cosh X2x = 1. 105 sin X x = 0. 439 cos X

2x = 0. 898

2

Sinh X2 (b - x) = 2. 534 Cosh X (b - x) = 2. 725 sin X (b - x) = 0. 996

2

cos X2 (b - x) = -0. 091

and

M2 max. 4. 064 x O. 439 x 2. 534 - 0. 857 x O. 47131. 55 x 1. 30

x 0. 99E]

0. 128P4.4. 064 (0. 898 x 2. 53 + 0. 439 x 2. 725) - 0. 857 (-0. 471

31. 55

x D. 091 + 1. 105 x 0. 996)i

M=2 max.

P = 6: 50 x 368 = 2, 390 .pounds.or- 6. 50


Therefore the design load for the joint is controlled by the maximummoment in member 1 and the design load is 1, 070 pounds.

The joint slip at this design load is

5 = 1,3 70 8. 91 + I. 36[

2

x 106

= 0.0129 inch.3

.35

2x 6.55


theoretical design of a nailed or hated joint under

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