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RIVETED JOINTS
CHAPTER IV
RIVETED JOINTS 54. Introductory.-In the majority of built-up structures the several members are united by riveted joints, and the strength of the joints is just as important as the strength of the members themselves. Unfortunately, the strength of riveted joints cannot be calculated with any degree of certainty, and practical design usually depends on empirical formulre based on experience. Certain theoretical considerations, however, are useful as guides, and provide an interesting application of the principles which we have already studied. R A
7l
:! FIG.
53.
A riveted joint is most commonly made in the manner shown in Fig. 53, where 44 and B are joined together by the rivets R and cover plates C, J). Except in very light work the rivets are generally placed in position red hot and the heads are closed immediately. If the rivets fill their holes ,vhen hot they 'cannot fill them when they have cooled, on account of lateral contraction. Consequently, the members . .4 and B will tend to slide over the cover plates, when a load is applied, until the rivets are all pressed up against their holes. On the other hand, the contraction of the rivets presses the cover plates against the members A and B with considerable force, so that any tendency to slide must be resisted by large friction forces between A, Band C, D. Evidently it is desirable that the load on the joint should be less than that required to overcome friction and produce slip, otherwise there "rill be backlash if the load is reversed, and the joint ,vill become" sloppy." It is also evident that, if the rivets are carrying any load, they nlust be acting as stops to check the relative movements of the plates. But, as long as the friction forces are great enough, there ,vill be no relative movement, and the rivets ,vill not hear any load; as soon as slip occurs, the rIvets are acted on by shearing forces in the planes separating the cover plates from the main plates. At the time of writing it is practically ilnpossible to make any reliable 63
STRENGTH OF MATERIALS
64
calculation of the friction forces which may be reckoned on in any given joint, and in England theoretical calculations are made for the strength of the plates and rivets, friction being ignored. On the Continent, however, the friction theory finds more favour, design resting largely on experimental figures given by Bach. According to Bach, when slipping begins, the friction forces vary from about 14,000 to 30,000 lbs. per square inch of rivet section, for each pair of surfaces in contact, increasing slightly with the length of the rivet, but not quite in proportion to the number of rows of rivets in multiple riveting. In a paper to the Institute of Naval Architects, 1923,* J. Montgomerie gives the following figures for the pull which produces slip, per square inch of rivet section : 0·6" Thickness of plates. Slip pull -:- area of rivet section 7·5 to 8
0·75" 6·2
1·0" 5 tons/in. 2
These figures were obtained for joints made under ordinary shipyard conditions, and show a decrease in the frictional resistance as the length of the rivet increases, which is contrary to Bach's results. The reason for this is probably the increased difficulty of getting good contact between the plates as the thickness increases. When special precautions were taken to ensure good contact the figure for 1" plates was raised to 7 tons/in. 2 In these tests two plates of equal thickness were riveted together,
FIG.
54.
as shown in Fig. 54, and subjected to a load acting in the plane separating the two plates. We shall consider the theoretical calculation of the strength of riveted joints, neglecting friction, which at present is the commonly adopted theory in this country.t Its chief recommendation is that it will produce a joint that will not fail even if slip does occur. 55. Possible Types of Failure of Simple Riveted Joints, Neglecting Friction.-To explain the general principles it will be sufficient to consider the joint shown in Fig. 55. Let d = the diameter of the rivets. x = the pitch of the rivets, i.e. the distance between their centres measured parallel to the line separating the plates. ql = the ultimate shearing stress for the rivets. q2 = the ultimate shearing stress for the plates and cover plates. f = the ultimate tensile stress for the plates and cover plates.
* See
Engineering, June 8, 15, 1923.
t For other papers on the subject see Annales des Ponts and ChaU8sees, articles
by Considere (1886), Dupuy (1895) ;. Zeit. d. Ver. (1897); Iron and Steel Institute, Vol. XCVI (E. B. Wolff); C. Batho in Engineering, Sept. 3t 1920.
RIVETED JOINTS
65
P6= the maximum bearing pressure allo,ved between the plates and the rivets. P = the load per unit length of joint, as shown in Fig. 55. The other dimensions of the joint are as shown in t~e figure.
S~i~. e EJeV~a~ion
...J-
~--~-~--'--~_.-L.-_-"'\~!f-_=--A--Z1.-_~~ . I
~t-T •
Let us consider that portion of the joint which lies between AB and OD: the total force aoting across AO and BD is Px. Then the joint may fail in any of the following ways: (i) The rivets may shear (Fig. 56). Z 't The resistance of each rivet to shear is ,I nd 2q Px . . i' ud Z 2 X -_.. .1, so that the relation : Cfi-q;4 FIG. 56. 2q nd Px < I (1) 2 must be satisfied. (ti) The permissible bearing pressure may be exceeded (Fig. 57). To prevent this we must have Px Px < Pb· d .t l FIG. 57. (2) or < 2Pbd.t' ~ · whichever is the smaller. The bealing pressure between the plates and the rivets arises from the former being pressed up against the latter, and, as we are neglecting friction, the whole load is transmitted in this way, as shown in Fig. 58, which refers to the leftFIG. 58. hand rivets in Fig. 55. (iii) The shaded area FHKG may be pushed out of the plates which are being joined together. This is resisted by the shearing stresses on two rectangular areas perpendicular to the plates, the length of each area being a and the depth t. Hence for safety we must have (3) pz <: 2atq.
-.m:rr/'t f
~L. . -_i_~_b'
_
STRENGTH OF MATERIALS
66
(iv) Similarly, the shaded area LMNQ may be pushed out of the cover plates, and the condition that this should not happen is Px <: (2bt'Q2)
X 2 .
(4)
'rhe factor 2 outside the bracket is due to there being t"ro cover plates. (v) The plates may fail in tension across X the line XY (Fig. 59). The resistance to tension is f X (x - d)t, so that \ve Px ~ ~ __._~ (x-d)tf' require Px <: ft(x - d) (5) y FIG. 59.
1I]:4-1
(vi) The plates may tear in the manner shown in Fig. 60 at Z, but the probability of this is not open to calculation. FIG.
Example.-Two steel plates
60.
i"
thick are to be joined by a single riveted butt joint with cover plates. The rivets are to be til diameter and the tensile and shearing stresses are to be 6 and 4·8 tons/in. 2 respectively. Find the proportions of the joint so that it shall be equally strong in shear and tension, and calculate the bearing pressure between the rivets and plates. The arrangement of the joint \\Till be as sho\vn in Fig. 61. Let x = the "pitch" of the rivets, and let the load on the joint be P tons per inch so that the load on a length x is xP tons. It \\rill be logical to Inake each cover plate half * the thickness of the plates FIG. 61. which are being joined, i.e. k". If the joint fail in tension it will be along the lines . 4 . B. 'rhe area resisting tension is i(x - i) sq. in. \Vith a tensile st ress of 6 tons lin. 2 the total load on ..4 B 6 tons/in.
Hence
\\re
Inust have
2
x
i
(x
Px
-!) in. = 2
=
~x
-
~ ~
~ (x -
i)
\v
ill be
tons.
tons .
(i)
If the joint fail by t he rivets shearing, the area resisting shear, in a length x, is 2 ~< (
n
4
25)
x 64
=
25n . 128 sq. In.
Therefore, \vith a shear stress of 4·8 tons per sq. in. we can have
Px
= 4·8 tons/in. 2
25n 25n x 128 in. 2 = 4·8 x 128 tons
.
(ii)
* In practice the cover plates are usually lnade thicker than this as it is found that the cover plates seem to be \veaker than the plates joined.
RIVETED JOINTS
67
If the joint is to be equally strong in shear and tension we must have, from (i) and (ii), 9 45 25n 4"x - 32 = 4·8 x 128 9
4x
=
2·94
+ 1·4
=
4·34
x = 1·93". Then from (ii), we have Px = 2·95 tons.
The area of each rivet resisting crushing is i" x i H = ii in. 2 Hence the bearing pressure is 2·95 tons/l~ in. I = 12·6 tons/in. 1 Again, since the cover plates are each half the thickness of the plates being joined, equations (3) and (4), p. 65, will give a = band 2a X i" X 4·8 tons/in. 2 = 2·95 tons, whence a = 0·82# = b.
56. Group-Riveted Joints.*-When two tension members are joined together by cover plates riveted in the manner shown in Fig. 62 J
;
0 00 : o 0 0 0 00 , 0 0 0 ., 0 00 o 0 0 I
r
I I
l
I
I
I
I
FIG.
62.
the joint is said to be group-riveted. The greatest efficiency of joint is obtained when the rivets are arranged as shown in Fig. 63, where it is supposed six rivets are required each side of the join. The loss of cross section in the main members, on the line A, is that due to one rivet hole. If the load is assumed to be equally distributed among the rivets, the
rivet on the line A will take one-sixth of the total load, so that the tension in the main plates, across B, will be a- of the total. But this section is reduced by two rivet holes, so that, relatively, it is' as strong as the section A, and so on: the reduction of the nett cross section of the main plates increases as the load carried by these plates decreases. Thus a·
* See Engineering, Dec. 6, 1918, for an article on the rigidity of riveted joints; the effect of rivet holes is dealt with Sept. 1, 1911, Sept. 19, 1913, Sept. 8, 1922.
68
STRENGTH OF MATERIALS
more efficient joint is obtained than \vhen the rivets are arranged as in Fig. 62. 57. Eccentric Loads.-It is obvious that a uniform distribution of load among the rivets cannot be attained unless the line of action of the resultant force acting on the joint passes through the centroid of the rivet-holes, and, at the same time, the rivets are symmetrically disposed with regard to the resultant force. In very many cases these conditions are not satisfied and it is desirable to form an idea of the load distribution. In Fig. 64 let ABOD be a plate riveted to a member EF, and let P E
a a
D
A
.B
R
,~
G/
0000 0000
,'~ ....
p
pa~
C
D
p~
FIG.
64.
p C
FIG.
65.
be the resultant of the forces acting on AB. Let G be the centroid of the rivet..holes. Let n = the total number of rivets. a = the perpendicular distance of G from the line of action of P. Now we can replace P in Fig. 64 by an equal parallel force P acting through G, combined with a couple Pa, as shown in Fig. 65. We assume that the deformation of the plate is negligible in comparison with that of the rivets, and that the load 011 each rivet is proportional to the displacement of the corresponding hole in ABOD, relative to the member EF, and acts in the same direction. Consider the loads on the rivets due to P and Pa separately:On account of the force P each rivet will carry a load P In acting in a direction parallel to that of P. On account of the couple Pa the displacement of any rivet-hole R will be at right angles to GR (Fig. 65) and proportional to the distance GR. Therefore, according to our assumption, the load on the rivet at R will be proportional to GR and perpendicular to it. Let GR = x" W, = the load on the rivet at R, and let
W, = kx, .
where k is a constant.
(i)
Then we must have
Pa
=
r-=n
1: W,x, .
r-1
. (ii)
RIVETED JOINTS
69
]-'rom (i) and (ii) ,ve get
=
Pa
r-fa
E
7==1
lex; = kr=1E ~ r-n
Pa
= r=n--
:. k
Ex;
r=1
Hence, from (i),
JV~=Pa.~ r
(6)
r=n
Ex;
r=1
This equation gives the load on each rivet due to the couple Pa. The total load on the rivet is the vector resultant of this and the load Pin. Example.-Fig. 66 shows a bracket riveted to a vertical stanchi9n and loaded with a vertical load of 5 tons. Assuming that the total shearing stress in a rivet is proportional to the relative displacement of the bracket and the stanchion in the neighbourhood of the rivet, find the load carried by each of the rivets. (Intercoll. Exam., Cambridge, 1922.)
~
...---+--9"--~~
J.~--I---~
1-.
c
!G
1~1
~
-t2;;-'--i~··
FIG.
66.
The centroid of the rivets is evidently at the point marked G in the figure. For A, x = AG = V9 + 4 = B, x = BG = V13 in. C and D, x = 2 in. E and F, x = V13 in.
Ex; = 13 + 13 + 13 n =6 a = 9 in. P = 5 tons. ... Pa = 45 tons. ins.
+
13
+
4
vIa
+4 =
in.
60 in. 2
Then the loads on rivets A, B, P, E, due to the couple Pa are each
. Vl:~ in. 45 tons/ln. x 60 in. 2
=
2·71 tons,
at right angles to OA, OB, OF, GE respectively.
STRENGTH OF MATERIALS
70
The corresponding loads on the rivets 0 and D a.re each 2 in. 45 tons/in. x 60 in. 2 = 1·50 tons, perpendicular to GO and GD respectively. The load on every rivet, due to the force of 5 t.ons is t ton = 0·833 ton, vertically downwards. Thus the resultant loads on all the rivets are found, as shown in Fig. 67, by drawing parallelograms of forces.
~.11 ~
~
:
B
1
~5",
,,"
,
.........
1:5 C
.<.;>
I
' .. 2:35
~
"/
"" ...
:??
"""'"
!
.P
::·'·"
IO1·5
0·833 FIG.
67.
The resultant load on rivets A and E B F
o
= =
=
=
D
2·35 tons. 3·24 0·667 " 2·333 "
EXAMPLES IV 1. A double-riveted butt-joint connects two i" plates with one cover strap. The diameter of the rivets is 1", and the distance between rivet centres along the pitch line is 5". Assuming that the other unstated dimensions are adequate, calculate the strength of the joint per foot, in tension, allowing 5 tons/in. 2 shear stress in the rivets, and a .tensile stress of 6 tons/in.! in the plates. (Special Exam., Cambridge, 1907.) 2. Two boiler plates i" thick are connected by a double-riveted lap joint, formed by 1" rivets with a pitch of 21". Determine the least tensile stress in the rivets which will enable the joint to remain tight under a tension of 12 tons per foot, along the joint, if the coefficient of friction is 0·2. (Special Exam., Cambridge, 1913.) 3. A riveted joint is formed by two cover plates shaped in the manner represented in Fig. 68. When the joint is subjected to a large pull, it nlay
,"
ZUW I
•
WQ7UI/lUQWII /l~I:~~:I Il!Z!Zl lun+l~ FIG.
68.
be assumed that the resistance is all provided by the shear in the rivets, and that the resistance exerted by a rivet is proportional to the relative slip of the corresponding rivet holes. When the four centre rivets only are in position, it is found that with a pull of 20 tons the slip at the rivet holes is 0·01". With all six rivets in position show that the loads taken by the outer
RIVETED JOINTS
71
the loads taken by each of the four centre rivets in the ratio of 27c:: 25. E for the plates is 14,000 tons/in. 2 (Mech. Sc. Trip., 1922.) by eight rivets in the 4. A fiat steel bar is attached to a gusset pl~te manner represented in Fig. 69. At the section AB the gusset plate exerts on the flat bar a vertical shearing force S and a cOlmter-clockwise couple ],1.
rivets·~xceed
,.---------------------~-----
'--f-(:'--
o
:
t-- -- -
:
:
~x
+
+
~: - - - t . ------- + a a+ a~! ~Z+2-ti + + _i.~
··~-+-------:_------i----,~----.-
+
+
----w.-a-.- a-t! FIG.
'.B
69.
Assuming that the gusset plate, relative to the flat bar, undergoes a minute rotation about a point 0 on the line of the two middle rivets, also that the loads on the rivets are due to and proportional to the relative movement of the plates at the rivet holes, prove that 4M + 3aS x = - a. 4M + 6aS· Prove also that the horizontal and vertical components of the load on the top right hand rivet are 2M + 3aS 4.Lvi + 9aS 24a and 24a respectively. (l\iech. Sc. Trip., 1923.) 5. A steel strip of cross section 2" X t" is bolted to two copper' strips, each of cross section 2" X i", as shown in Fig. 45, there being two bolts on the line of pull. Show that, neglecting friction and the deformation of the bolts, a pull applied to the joint will be shared by the bolts in the ratio 3 to 4. Assume that E for steel is t\vice E for copper. (IntercoU. Exam., Cambridge, 1923.) 6. *Two fiat bars are riveted together in the manner shown in Fig. 70, x being the pitch of the rivets in a direction at right angles to the plane of the figure. Assuming that the rivets themselves do not deform, show
that the load taken by the rivets (I) is free from load.
t f;t" and that the rivets (2) are :Z!
FIG.
70.
* In connection with the distribution of load in riveted joints, see papers by Montgomerie and Batho (see footnote, p. (4); also a paper by the author in Aeronautic8, 1918. .
RIVETED JOINTS 54. Introductory.-In the majority of built-up structures the several members are united by riveted joints, and the strength of the joints is just as important as the strength of the members themselves. Unfortunately, the strength of riveted joints cannot be calculated with any degree of certainty, and practical design usually depends on empirical formulre based on experience. Certain theoretical considerations, however, are useful as guides, and provide an interesting application of the principles which we have already studied. R A
7l
:! FIG.
53.
A riveted joint is most commonly made in the manner shown in Fig. 53, where 44 and B are joined together by the rivets R and cover plates C, J). Except in very light work the rivets are generally placed in position red hot and the heads are closed immediately. If the rivets fill their holes ,vhen hot they 'cannot fill them when they have cooled, on account of lateral contraction. Consequently, the members . .4 and B will tend to slide over the cover plates, when a load is applied, until the rivets are all pressed up against their holes. On the other hand, the contraction of the rivets presses the cover plates against the members A and B with considerable force, so that any tendency to slide must be resisted by large friction forces between A, Band C, D. Evidently it is desirable that the load on the joint should be less than that required to overcome friction and produce slip, otherwise there "rill be backlash if the load is reversed, and the joint ,vill become" sloppy." It is also evident that, if the rivets are carrying any load, they nlust be acting as stops to check the relative movements of the plates. But, as long as the friction forces are great enough, there ,vill be no relative movement, and the rivets ,vill not hear any load; as soon as slip occurs, the rIvets are acted on by shearing forces in the planes separating the cover plates from the main plates. At the time of writing it is practically ilnpossible to make any reliable 63
STRENGTH OF MATERIALS
64
calculation of the friction forces which may be reckoned on in any given joint, and in England theoretical calculations are made for the strength of the plates and rivets, friction being ignored. On the Continent, however, the friction theory finds more favour, design resting largely on experimental figures given by Bach. According to Bach, when slipping begins, the friction forces vary from about 14,000 to 30,000 lbs. per square inch of rivet section, for each pair of surfaces in contact, increasing slightly with the length of the rivet, but not quite in proportion to the number of rows of rivets in multiple riveting. In a paper to the Institute of Naval Architects, 1923,* J. Montgomerie gives the following figures for the pull which produces slip, per square inch of rivet section : 0·6" Thickness of plates. Slip pull -:- area of rivet section 7·5 to 8
0·75" 6·2
1·0" 5 tons/in. 2
These figures were obtained for joints made under ordinary shipyard conditions, and show a decrease in the frictional resistance as the length of the rivet increases, which is contrary to Bach's results. The reason for this is probably the increased difficulty of getting good contact between the plates as the thickness increases. When special precautions were taken to ensure good contact the figure for 1" plates was raised to 7 tons/in. 2 In these tests two plates of equal thickness were riveted together,
FIG.
54.
as shown in Fig. 54, and subjected to a load acting in the plane separating the two plates. We shall consider the theoretical calculation of the strength of riveted joints, neglecting friction, which at present is the commonly adopted theory in this country.t Its chief recommendation is that it will produce a joint that will not fail even if slip does occur. 55. Possible Types of Failure of Simple Riveted Joints, Neglecting Friction.-To explain the general principles it will be sufficient to consider the joint shown in Fig. 55. Let d = the diameter of the rivets. x = the pitch of the rivets, i.e. the distance between their centres measured parallel to the line separating the plates. ql = the ultimate shearing stress for the rivets. q2 = the ultimate shearing stress for the plates and cover plates. f = the ultimate tensile stress for the plates and cover plates.
* See
Engineering, June 8, 15, 1923.
t For other papers on the subject see Annales des Ponts and ChaU8sees, articles
by Considere (1886), Dupuy (1895) ;. Zeit. d. Ver. (1897); Iron and Steel Institute, Vol. XCVI (E. B. Wolff); C. Batho in Engineering, Sept. 3t 1920.
RIVETED JOINTS
65
P6= the maximum bearing pressure allo,ved between the plates and the rivets. P = the load per unit length of joint, as shown in Fig. 55. The other dimensions of the joint are as shown in t~e figure.
S~i~. e EJeV~a~ion
...J-
~--~-~--'--~_.-L.-_-"'\~!f-_=--A--Z1.-_~~ . I
~t-T •
Let us consider that portion of the joint which lies between AB and OD: the total force aoting across AO and BD is Px. Then the joint may fail in any of the following ways: (i) The rivets may shear (Fig. 56). Z 't The resistance of each rivet to shear is ,I nd 2q Px . . i' ud Z 2 X -_.. .1, so that the relation : Cfi-q;4 FIG. 56. 2q nd Px < I (1) 2 must be satisfied. (ti) The permissible bearing pressure may be exceeded (Fig. 57). To prevent this we must have Px Px < Pb· d .t l FIG. 57. (2) or < 2Pbd.t' ~ · whichever is the smaller. The bealing pressure between the plates and the rivets arises from the former being pressed up against the latter, and, as we are neglecting friction, the whole load is transmitted in this way, as shown in Fig. 58, which refers to the leftFIG. 58. hand rivets in Fig. 55. (iii) The shaded area FHKG may be pushed out of the plates which are being joined together. This is resisted by the shearing stresses on two rectangular areas perpendicular to the plates, the length of each area being a and the depth t. Hence for safety we must have (3) pz <: 2atq.
-.m:rr/'t f
~L. . -_i_~_b'
_
STRENGTH OF MATERIALS
66
(iv) Similarly, the shaded area LMNQ may be pushed out of the cover plates, and the condition that this should not happen is Px <: (2bt'Q2)
X 2 .
(4)
'rhe factor 2 outside the bracket is due to there being t"ro cover plates. (v) The plates may fail in tension across X the line XY (Fig. 59). The resistance to tension is f X (x - d)t, so that \ve Px ~ ~ __._~ (x-d)tf' require Px <: ft(x - d) (5) y FIG. 59.
1I]:4-1
(vi) The plates may tear in the manner shown in Fig. 60 at Z, but the probability of this is not open to calculation. FIG.
Example.-Two steel plates
60.
i"
thick are to be joined by a single riveted butt joint with cover plates. The rivets are to be til diameter and the tensile and shearing stresses are to be 6 and 4·8 tons/in. 2 respectively. Find the proportions of the joint so that it shall be equally strong in shear and tension, and calculate the bearing pressure between the rivets and plates. The arrangement of the joint \\Till be as sho\vn in Fig. 61. Let x = the "pitch" of the rivets, and let the load on the joint be P tons per inch so that the load on a length x is xP tons. It \\rill be logical to Inake each cover plate half * the thickness of the plates FIG. 61. which are being joined, i.e. k". If the joint fail in tension it will be along the lines . 4 . B. 'rhe area resisting tension is i(x - i) sq. in. \Vith a tensile st ress of 6 tons lin. 2 the total load on ..4 B 6 tons/in.
Hence
\\re
Inust have
2
x
i
(x
Px
-!) in. = 2
=
~x
-
~ ~
~ (x -
i)
\v
ill be
tons.
tons .
(i)
If the joint fail by t he rivets shearing, the area resisting shear, in a length x, is 2 ~< (
n
4
25)
x 64
=
25n . 128 sq. In.
Therefore, \vith a shear stress of 4·8 tons per sq. in. we can have
Px
= 4·8 tons/in. 2
25n 25n x 128 in. 2 = 4·8 x 128 tons
.
(ii)
* In practice the cover plates are usually lnade thicker than this as it is found that the cover plates seem to be \veaker than the plates joined.
RIVETED JOINTS
67
If the joint is to be equally strong in shear and tension we must have, from (i) and (ii), 9 45 25n 4"x - 32 = 4·8 x 128 9
4x
=
2·94
+ 1·4
=
4·34
x = 1·93". Then from (ii), we have Px = 2·95 tons.
The area of each rivet resisting crushing is i" x i H = ii in. 2 Hence the bearing pressure is 2·95 tons/l~ in. I = 12·6 tons/in. 1 Again, since the cover plates are each half the thickness of the plates being joined, equations (3) and (4), p. 65, will give a = band 2a X i" X 4·8 tons/in. 2 = 2·95 tons, whence a = 0·82# = b.
56. Group-Riveted Joints.*-When two tension members are joined together by cover plates riveted in the manner shown in Fig. 62 J
;
0 00 : o 0 0 0 00 , 0 0 0 ., 0 00 o 0 0 I
r
I I
l
I
I
I
I
FIG.
62.
the joint is said to be group-riveted. The greatest efficiency of joint is obtained when the rivets are arranged as shown in Fig. 63, where it is supposed six rivets are required each side of the join. The loss of cross section in the main members, on the line A, is that due to one rivet hole. If the load is assumed to be equally distributed among the rivets, the
rivet on the line A will take one-sixth of the total load, so that the tension in the main plates, across B, will be a- of the total. But this section is reduced by two rivet holes, so that, relatively, it is' as strong as the section A, and so on: the reduction of the nett cross section of the main plates increases as the load carried by these plates decreases. Thus a·
* See Engineering, Dec. 6, 1918, for an article on the rigidity of riveted joints; the effect of rivet holes is dealt with Sept. 1, 1911, Sept. 19, 1913, Sept. 8, 1922.
68
STRENGTH OF MATERIALS
more efficient joint is obtained than \vhen the rivets are arranged as in Fig. 62. 57. Eccentric Loads.-It is obvious that a uniform distribution of load among the rivets cannot be attained unless the line of action of the resultant force acting on the joint passes through the centroid of the rivet-holes, and, at the same time, the rivets are symmetrically disposed with regard to the resultant force. In very many cases these conditions are not satisfied and it is desirable to form an idea of the load distribution. In Fig. 64 let ABOD be a plate riveted to a member EF, and let P E
a a
D
A
.B
R
,~
G/
0000 0000
,'~ ....
p
pa~
C
D
p~
FIG.
64.
p C
FIG.
65.
be the resultant of the forces acting on AB. Let G be the centroid of the rivet..holes. Let n = the total number of rivets. a = the perpendicular distance of G from the line of action of P. Now we can replace P in Fig. 64 by an equal parallel force P acting through G, combined with a couple Pa, as shown in Fig. 65. We assume that the deformation of the plate is negligible in comparison with that of the rivets, and that the load 011 each rivet is proportional to the displacement of the corresponding hole in ABOD, relative to the member EF, and acts in the same direction. Consider the loads on the rivets due to P and Pa separately:On account of the force P each rivet will carry a load P In acting in a direction parallel to that of P. On account of the couple Pa the displacement of any rivet-hole R will be at right angles to GR (Fig. 65) and proportional to the distance GR. Therefore, according to our assumption, the load on the rivet at R will be proportional to GR and perpendicular to it. Let GR = x" W, = the load on the rivet at R, and let
W, = kx, .
where k is a constant.
(i)
Then we must have
Pa
=
r-=n
1: W,x, .
r-1
. (ii)
RIVETED JOINTS
69
]-'rom (i) and (ii) ,ve get
=
Pa
r-fa
E
7==1
lex; = kr=1E ~ r-n
Pa
= r=n--
:. k
Ex;
r=1
Hence, from (i),
JV~=Pa.~ r
(6)
r=n
Ex;
r=1
This equation gives the load on each rivet due to the couple Pa. The total load on the rivet is the vector resultant of this and the load Pin. Example.-Fig. 66 shows a bracket riveted to a vertical stanchi9n and loaded with a vertical load of 5 tons. Assuming that the total shearing stress in a rivet is proportional to the relative displacement of the bracket and the stanchion in the neighbourhood of the rivet, find the load carried by each of the rivets. (Intercoll. Exam., Cambridge, 1922.)
~
...---+--9"--~~
J.~--I---~
1-.
c
!G
1~1
~
-t2;;-'--i~··
FIG.
66.
The centroid of the rivets is evidently at the point marked G in the figure. For A, x = AG = V9 + 4 = B, x = BG = V13 in. C and D, x = 2 in. E and F, x = V13 in.
Ex; = 13 + 13 + 13 n =6 a = 9 in. P = 5 tons. ... Pa = 45 tons. ins.
+
13
+
4
vIa
+4 =
in.
60 in. 2
Then the loads on rivets A, B, P, E, due to the couple Pa are each
. Vl:~ in. 45 tons/ln. x 60 in. 2
=
2·71 tons,
at right angles to OA, OB, OF, GE respectively.
STRENGTH OF MATERIALS
70
The corresponding loads on the rivets 0 and D a.re each 2 in. 45 tons/in. x 60 in. 2 = 1·50 tons, perpendicular to GO and GD respectively. The load on every rivet, due to the force of 5 t.ons is t ton = 0·833 ton, vertically downwards. Thus the resultant loads on all the rivets are found, as shown in Fig. 67, by drawing parallelograms of forces.
~.11 ~
~
:
B
1
~5",
,,"
,
.........
1:5 C
.<.;>
I
' .. 2:35
~
"/
"" ...
:??
"""'"
!
.P
::·'·"
IO1·5
0·833 FIG.
67.
The resultant load on rivets A and E B F
o
= =
=
=
D
2·35 tons. 3·24 0·667 " 2·333 "
EXAMPLES IV 1. A double-riveted butt-joint connects two i" plates with one cover strap. The diameter of the rivets is 1", and the distance between rivet centres along the pitch line is 5". Assuming that the other unstated dimensions are adequate, calculate the strength of the joint per foot, in tension, allowing 5 tons/in. 2 shear stress in the rivets, and a .tensile stress of 6 tons/in.! in the plates. (Special Exam., Cambridge, 1907.) 2. Two boiler plates i" thick are connected by a double-riveted lap joint, formed by 1" rivets with a pitch of 21". Determine the least tensile stress in the rivets which will enable the joint to remain tight under a tension of 12 tons per foot, along the joint, if the coefficient of friction is 0·2. (Special Exam., Cambridge, 1913.) 3. A riveted joint is formed by two cover plates shaped in the manner represented in Fig. 68. When the joint is subjected to a large pull, it nlay
,"
ZUW I
•
WQ7UI/lUQWII /l~I:~~:I Il!Z!Zl lun+l~ FIG.
68.
be assumed that the resistance is all provided by the shear in the rivets, and that the resistance exerted by a rivet is proportional to the relative slip of the corresponding rivet holes. When the four centre rivets only are in position, it is found that with a pull of 20 tons the slip at the rivet holes is 0·01". With all six rivets in position show that the loads taken by the outer
RIVETED JOINTS
71
the loads taken by each of the four centre rivets in the ratio of 27c:: 25. E for the plates is 14,000 tons/in. 2 (Mech. Sc. Trip., 1922.) by eight rivets in the 4. A fiat steel bar is attached to a gusset pl~te manner represented in Fig. 69. At the section AB the gusset plate exerts on the flat bar a vertical shearing force S and a cOlmter-clockwise couple ],1.
rivets·~xceed
,.---------------------~-----
'--f-(:'--
o
:
t-- -- -
:
:
~x
+
+
~: - - - t . ------- + a a+ a~! ~Z+2-ti + + _i.~
··~-+-------:_------i----,~----.-
+
+
----w.-a-.- a-t! FIG.
'.B
69.
Assuming that the gusset plate, relative to the flat bar, undergoes a minute rotation about a point 0 on the line of the two middle rivets, also that the loads on the rivets are due to and proportional to the relative movement of the plates at the rivet holes, prove that 4M + 3aS x = - a. 4M + 6aS· Prove also that the horizontal and vertical components of the load on the top right hand rivet are 2M + 3aS 4.Lvi + 9aS 24a and 24a respectively. (l\iech. Sc. Trip., 1923.) 5. A steel strip of cross section 2" X t" is bolted to two copper' strips, each of cross section 2" X i", as shown in Fig. 45, there being two bolts on the line of pull. Show that, neglecting friction and the deformation of the bolts, a pull applied to the joint will be shared by the bolts in the ratio 3 to 4. Assume that E for steel is t\vice E for copper. (IntercoU. Exam., Cambridge, 1923.) 6. *Two fiat bars are riveted together in the manner shown in Fig. 70, x being the pitch of the rivets in a direction at right angles to the plane of the figure. Assuming that the rivets themselves do not deform, show
that the load taken by the rivets (I) is free from load.
t f;t" and that the rivets (2) are :Z!
FIG.
70.
* In connection with the distribution of load in riveted joints, see papers by Montgomerie and Batho (see footnote, p. (4); also a paper by the author in Aeronautic8, 1918. .