When a ring is subjected to two similar point loads, but these loads act in opposite outward directions, its deformation can be analysed by exactly the same method as above. One key difference is that the position of the two side plastic hinges, corresponding to the maximum bending moment, moves as deflection increases (hinge B in Fig. 4.3(a)). This moving hinge is the
P P
D D
o
=
+ -Ê Ë ˆ
¯ È
ÎÍ ˘
˚˙ 1 1 2
2 1
d d 2
P M
R M
o D
p p
=4 = 8
P M
R
= p
Ê - Ë
ˆ
¯ 2 2 cos p4
q
P Mp
2Mp
Mp P
2
P 2
P 2 P 2
A B
q P
P
d2
(a) (b)
4.1 Collapse mechanism of a ring under two inwards acting point loads (a). Forces acting on a quadrant (b). Four plastic hinges are necessary.
same as that encountered in Section 2.3. The undeformed segment is always tangential to the deformed straightened portion at the current plastic hinge B. Furthermore, the total length of the ring remains the same during defor- mation. These considerations, together with the equivalent structure tech- nique described above, lead to (Yu, 1979)
[4.6]
or
[4.7]
and
[4.8]
Equations [4.7] and [4.8] give the load–displacement curve shown in Fig.
4.2 by the long dashed line.
The above analysis does not consider the effect of axial force on yield- ing, which can be important when the ring is thick and when d/Dis close to 1. The effect of axial force on yielding for a rectangular cross-section (b¥ h) was discussed in Section 2.2.2. One approximate way of incorporating
d =2R(cosq q+ -1) P
Po = -
1 1 sinq
PR Mp
2 (1-sinq)=2
6 5 4 3 2 1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
Eq. [4.5]
d/D Eqs [4.7] and [4.8]
P = 2Np
P/Po
4.2 Non-dimensional load–displacement curves for a ring of rectangular cross-section under two point loads. D/h=5. --- Eq. [4.5] (compressive loads); – – – Eqs [4.7] and [4.8] (without axial force effect); —— with axial force effect. The force is limited to P=2Np(i.e. P/Po=D/h=5).
this axial force effecthere is to modify Eq. [4.6] by considering Eq. [2.22].
The axial force at each hinge is P/2 and the bending capacity now reduces from Mpto
Thus, Eq. [4.6] becomes
[4.9]
As before,Np =Ybhis the axial yielding force. Combining Eqs [4.9] and [4.8] leads to a load–displacement curve with a maximum load equal to 2Np when q=p/2.
Nevertheless, strictly speaking, this analysis is only approximate because the associated flow (normality) rule has not been taken into account, although the axial force was considered in the yielding criterion. The total length of the ring remains the same in this case. One method of overcom- ing this shortcoming is presented here (Lu, 1993a).
The normality rule(associated flow rule) described in Section 2.2.2 (Eq [2.23]) can be understood here, alternatively, from the actual stress distrib- ution over the cross-section, instead of the resultant axial force Nand resul-
PR M P
p N
2 1 2 1 p2 2
( - )= -Ê
ËÁ ˆ
¯˜ sinq
M M P
p N
p
= -Ê
ËÁ ˆ
¯˜
1 2 2
w a
o
o2 o1
P
Y
2
e2 e1
P 2 d
2
o¢ o'
o
2
o'1
q bh/ 2 q
P P A R
R B
d 2R +
qR
(a) (b) (c)
4.3 Collapse mechanism of a ring under two outwards acting point loads (a). Stress state of a plastic hinge with axial force (b). Details of forces acting on a segment (c). During collapse the two middle hinges split into four hinges such as at point B. (Lu, 1993a)
tant bending moment M. It is clear that corresponding to such a stress state, deformation of the cross-section is a rotation about point O¢ rather than the midpoint O, see Fig. 4.3(b). O¢is also called the ‘pivot point’. Thus when the cross-section rotates about point O¢by angle q, fibres at the mid-point O stretch by qbh/2. Here bis a fraction and bh/2 is the distance between O and O¢. It is easy to see that the resultant axial force is N= bhband the bending moment is M=(1 -b2)Yh2b/4. The axial force at the upper hinge is zero and at the lower hinge it is P/2. For a ring under tension (Fig. 4.3(a)), the total external force is
[4.10]
The position of the line of action of forces in the equivalent structure (Fig. 4.3(c)) is given by
[4.11]
and
[4.12]
Again from equilibrium considerations, the two opposite forces must act along the same line. From the geometry at the onset of collapse, displace- ment dis zero and we have
[4.13]
Substituting Eqs [4.11] and [4.12] into [4.13], we obtain the expression for the positive value of b:
[4.14]
For example consider the two cases D/h =10 and D/h=5. Equation [4.14]
leads to b=0.0995 and 0.1962, respectively.The initial collapse load becomes [4.15]
This formula is identical to that given by de Runtz and Hodge (1963) for the initial collapse load of a ring compressed by two rigid flat plates when the effect of axial force on yielding is considered.
As mentioned previously, during the collapse process, the middle hinge will split into two, one moving upwards and the other downwards. This is required by the condition that a plastic hinge always forms at the position of maximum bending moment. The deformation in the upper hinge remains a pure rotation about point O1¢, which is the mid-point of the cross-section.
P Y hb D h
h
D Yhb
o = = + Ê
Ë ˆ
¯ - Ê
ËÁ
ˆ
¯˜
2 2 1 2 1
2
b b= - + Ê
Ë ˆ
¯ + D
h D
h
2
2 e1+e2=R=D 2 e M
P H
2 2 4
= = Ê1-
Ë ˆ b b¯ e M
P
p H
1= 2=4
b P=2Y hbb
From the stress state for the post-collapse stage as shown in Fig. 4.3(c), Eq. [4.13] becomes
[4.16]
where w is the horizontal distance between O1¢ and O2. Substituting in Eqs [4.11] and [4.12]
[4.17]
Let the total rotation of the rigid segment at this instant be q. We have [4.18]
and
[4.19]
Eliminating qfrom Eqs [4.18] and [4.19], we have
[4.20]
At this instant, let the centre of the mid-hinge cross-section O2move side- ways by an increment dw, the corresponding rotational increment of the segment is then dw/aand the corresponding increment of deflection dis
[4.21]
But by differentiating Eq. [4.17], we have
[4.22]
The negative sign means that when b increases,wdecreases. Substituting Eqs [4.20] and [4.22] into Eq. [4.21], we obtain after some rearrangement
[4.23]
Therefore
[4.24]
d d b b
b
b b b
b
b
b b
D D
h D D
h
o
= = -
+ +
Ê
ËÁ ˆ
¯˜ Ê
Ë ˆ
¯Ê - Ë
ˆ
¯-Ê - Ë
ˆ
¯ È
ÎÍ ˘
˚˙
Ú
Ú
d d1 2
1
2 8
1 4
2 1
2 4
3
2 1 2
dd b b d
b
b b b
b D b
h D D
h
= -
+ +
Ê
ËÁ ˆ
¯˜ Ê
Ë ˆ
¯Ê - Ë
ˆ
¯-Ê - Ë
ˆ
¯ È
ÎÍ ˘
˚˙ 1
2 1
2 8
1 4
2 1
2 4
3
2 1 2
dw h d
= - Ê + ËÁ ˆ
¯˜ 4
22 1
b b
d d d
d b
b
= Ê + b Ë
ˆ
¯ = Ê +
Ë ˆ 2 ¯
2
1 w h w 2
a h w
a
a wR w Dh h
= - = Ê -
Ë ˆ
¯-Ê Ë
ˆ
¯ Ê - Ë
ˆ 2 ¯
4 2
4
2 2
2 2
b b
b b a=Rcosq
w=R(1 sinq- )
h w
4 2 b -b Ê Ë
ˆ
¯= e1+e2=w
Thus for a given value of D/h,bocan be determined from Eq. [4.14] and then Eq. [4.24] gives the value of d/Dcorresponding to any value of b. The tension force can then be determined from Eq. [4.10]. This result is shown in Fig. 4.2 as a solid line for the case of D/h =5. This tension force is, of course, lower than the case where the effect of axial force was not consid- ered. Also, maximum non-dimensional displacement,d/D, before full mem- brane yielding develops is 0.61, which is higher than the 0.57 given by Eq.
[4.8] for q=p/2. This clearly demonstrates stretching of the ring as a result of the axial force. The maximum tensile force is limited to the full axial yielding force (=2Np), and the ring then behaves like a bar under simple tension.