technical paper 1.pdf

Technical paper number1

Bolts and Pins; An explanation of their differences

David Scott June 2011

Technical paper number 1 page 1

Introduction

This study was prompted by a question regarding the design of a joint between RHS beams and SHS columns. The joint was loaded in shear and torsion and employed endplates and through-bolts thus:

The forces involved were relatively modest and the bolt shear resulting from the combined torsion and direct shear was in the range 30-50 kN. A question was raised regarding this design requesting a check on the bending in the long bolts. This prompted the present paper to explore the nature of bending in bolts and pins, provide an explanation as to the reason bending in bolts is generally ignored in design calculations and to determine whether this unusual situation warrants a different approach from the conventional calculation methods set out in BS5950 and elsewhere.

Forces on a bolt

Considering a bolt in singe shear, the forces from the plates bearing on the fastener are as shown opposite:

These forces are not in equilibrium, as there is an overall unbalanced moment. More on this later; first drawing the shear force diagram, we get the following:





As the net area under the shear force diagram is not zero, it is clear that a net change of bending moment must occur from one end of the bolt to the other. The maximum rate of change of the bending moment occurs in the middle of the bolt (where the shear force is at a maximum). Furthermore, the forces and geometry have rotational symmetry; hence so must the bending moment diagram. These observations are sufficient to allow the bending moment diagram to be sketched as above.

The first point of interest in this diagram is the location of the point of contra-flexure, which is coincident with the shear plane in the bolt. This forms a large part of the explanation as to why bolts are designed for shear, not bending: The two forces are not coincident, where shear is applied, bending is zero.

A second point to note is that equal and opposite moments are applied at the bolt head and nut. These moments provide the overall equilibrium missing in the first diagram.

They are generated by direct bearing forces at the edges of the head and nut as shown opposite:





This observation has several practical implications:

i) Bolts must have a connection at the head and nut sufficiently rigid and strong to transmit the moment. Thus the essential difference between a bolt (or rivet) and a pin is the presence or absence of such moment resisting joints at its ends

ii) In a situation where pure shear is applied to a bolt, there remains a prying force applied to the nut. Thus any nuts incorrectly engaged on the threads carry a risk of catastrophic failure by prying of the nut, even where direct bolt tension is zero.

iii) There is a compressive force passing through the connected pieces. For bolts passing through hollow sections, it is therefore vital that this compressive force can be resisted or else the connection will not be effective as the side wall of the hollow section will crush. Thus either a CHS tube must be used as a sleeve as is shown below or else the wall of the section must be capable of resisting the resulting force, which is calculated later in this paper.

Illustration of reinforcing of thin-walled SHS section to resist clamping forces (ref 4)





Examining the magnitude of the forces involved let, let us now consider the case of grade 8.8 fasteners. The strength of these components is given in BS 5950 as follows:

Bearing on S275 material 460 N/sq mm

Shear 375 N/sq mm

Tension 560 N/sq mm

Shear capacity = Tensile stress area of bolt at thread x 375 N

Bearing capacity = Bolt diameter x bearing length x 460 N

Equating shear and bearing capacity, 375.pi.Dr2/4 = D.L.460

Where D= bolt diameter, Dr = reduced diameter at threads and L= bearing length

Dr is approx 0.88D, hence L may be calculated and is approximately D/2

Thus the full shear capacity of the bolt is utilised when the width of steel plates connected is around half the bolt diameter. The moment is given by

pv x bolt cross sectional area x diameter/4.

The tension in the bolt and compression in the joint is given by Moment at the end of the bolt divided by half the width across flats of the head or nut.

Thus the moment and tension in the bolt can both be calculated. In the case of grade 8.8 fasteners in the commonly used diameters, the results are tabulated below. It can be seen that, when the bolt is fully loaded in shear, around 20% of the bolt tension capacity and around 90 % of the moment capacity of the threaded section of the bolt is used.





For an M20 fully threaded fastener loaded in shear at the design limit of 92 kN, the stresses are as follows:

At the shear plane: fv = 375 N/sq mm, ft=128 N/sq mm

At the head or nut: fb=500 N/sq mm, ft = 128 N/sq mm

It should be remembered that even although the tension capacity of the fastener is quoted as 560 N/sq mm, the yield for the bolt is 80% of 800N/sq mm by definition of the grade, that is 640 N/sq mm.

To further consider the implications of this state of stress, let us compare them to the Von Mises yield criteria. At the shear plane, the Von Mises criteria indicate that yielding is occurring.





However, out-with the shear zone, where the direct stress is 128 +/- 500 N/sq mm, no yielding occurs. This again illustrates that the bending stresses are not critical and may be safely ignored for design purposes.





Using the Von Mises criteria, the BS5950 interaction formula for combined shear and tension, which allows full shear capacity to be utilised in combination with 40% of the tensile capacity, produces a result out-with the yield criteria thus:

Although this perhaps says more about the unknowns associated with predicting the onset of yield in a multidimensional stress state, it nevertheless suggests that a closer examination of the interaction between shear and tension in bolts would be worthwhile at this juncture.





Reference 3, citing testing evidence from the University of Illinois, shows a good correlation between an elliptical interaction curve and test data. This is based on shear strength of 0.62 x Tensile strength. It is reproduced opposite:

BS 5950 (ref 1) has the following interaction formula:

And the Eurocode (ref 2) uses the following formula.





As each alternative design method also uses different strengths, this is best understood by plotting the various interaction curves on a graph:

From this it can be seen that the BS 5950 design method is in good agreement with the Univ of Illinois test results, albeit using a simplified and conservative formula. The Eurocode approach, in contrast, gives much lower tension capacity in the presence of high shear. Looking at the tension generated in a bolt when loaded to the maximum shear permitted under BS5950, we see that both The BS5950 and Univ of Illinois methods show this to be acceptable, but that it falls out-with the Eurocode criteria. This is clear evidence of over-conservatism in the Eurocode, as the tension





associated with maximum shear (about 20% of Pt) must exist and clearly should be reflected in the interaction formula.

Thus we may conclude that consideration should be given to the use of an elliptical interaction formula:

Ft2/Pt2 + Fv2/Pv2





Pins

In contrast to bolts, pins have no rigid connection to heads/nuts at their ends, instead relying on thin plates or split pins to prevent the pin being displaced from the joint.

Thus the bending moment at the free end of the pin must be zero. Hence the following applied load and shear force distribution applies:

This gives rise to a bending moment distribution such as:





It can be seen that a significant moment applies at the shear zone and thus the interaction of moment and shear must be considered in the design, with a reduced moment capacity being calculated in situations where applied shear exceeds 60% of the shear capacity of the pin. Shear capacity is taken as 0.9A x 0.6 py, where A is the cross sectional area of the pin. A second check is also required further along the pin where bending is a maximum although shear is zero.

It has become common to assume a uniform loading on a pin from the connected plies as shown opposite. However, in many cases, this will overestimate the bending in the pin. In accordance with the basic safe theorem for plastic design (which states that if a set of internal forces can be found which are in equilibrium with external loads, and which are everywhere less than the capacity of the material, then the structure is safe), it is perfectly appropriate to apply the load as close to the shear plane as is permitted by the bearing strength of

the materials employed.





As a final illustration of the importance of the clamping force in bolt performance and the absence of such forces in pins, it is useful to look back to the silver bridge disaster which occurred in 1967 in America. This major structure, which spanned the Ohio River, was a chain-link suspension bridge utilising suspension cables made from twin high tensile steel eye-bars, joined with pins. One cold December night, one eye-bar, affected by stress corrosion cracking, suffered a sudden brittle failure. Although the second eye-bar could have safely carried the whole load, the loss of one bar left the load on the pin unbalanced. The pin twisted around and the other eye bar slipped off, resulting in the loss of the whole suspension cable and consequently the collapse of the entire bridge.

I





Conclusion

Although simple design methods may be used in checking the capacity of bolted joints, it is important to bear in mind the full range of forces employed. In particular the presence of tension in the fastener, clamping the joint together is an essential aspect of bolt performance. Where detailing does not ensure that such forces can be generated, either due to the use of very long bolts or the presence of thin-walled, unreinforced hollow section members then normal bolt design methods will need to be modified.

References:

1. BS 5950: The Structural Use of Steelwork in Building; part 1, code of practice for design, welded and rolled sections.

2. BS EN 1993-1-8; Eurocode 3: Design of Steel Structures Part 1-8: Design of Joints

3. Guide to Design Criteria for bolted and riveted joints, 2nd edition, Kulak et al

4. SHS Jointing, Corus Tubes

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