First, care should be taken to know the core and cover gum rubbers will be compatible between splice kit, original belt and saddle belt.

Second, the distance from steel cord center to pulley or bottom cover should not vary by more than 2 mm, hopefully, less, so as not to cause significant disturbance with the drive and splice dynamics.

Connecting belts, with different manufacturers has been done many times, by many who practice the splicing profession.

We have connected belts with very different belt cord constructions and cable pitch (by example, belts with 100 cables into 84 cables). I am sure there are many such stories.

In your case, I would guess the cable diameter is about 4mm, with a break strength of about 14.4 kN. The belt might have 64 cables. This will make a good single step splice.

You must take care not to have to large a disparity of cable diameter and count. Manufacturers keep different cord diameters in stock. I would guess you are ok.

If you need help to further understand, please email me the details. ■

Mr. Nordell is 100% right. As long as there is room within the splice area to lay out the cords, you should be OK. At this point it will be a matter of the level of skill of those splicing the belt. Splicing mix/match steel cord belting is as much of an art as it is a science in these situations. An experienced technician will save you a headache.

Buddy Wilson

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Applications Engineer / TSM-Southest

Fenner Dunlop Americas ■

Originally Posted by nordell

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We have connected belts with very different belt cord constructions and cable pitch (by example, belts with 100 cables into 84 cables). I am sure there are many such stories.

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Thank you Mr.Nordell,

In the above example, the efficiency of the joint is 184/200 = 92% ?

Am I right in assuming that if all the cords are present in the joints, the efficiency of the joint ( if done correct ) is 100% ?

Regards, ■

Define efficiency. ■

Dear Mr.Nordell,

For fabric ply, the joint efficiency is (n-1)/n where n is the number of plies in the fabric belts. i.e., at the joint area 5 ply belts joint will have 80% of the strength of the belt, 4 ply belts joint will have 75% and so on.

I was told that the steel cord belt joints are about 98% since all the cords are used in the joint.

So if two steel cord belts having 100 & 84 cables are used, the joint area will have 184 cables. So I thought the efficiency would be 184/200 of the belt strength . ( I have taken the belt having 100 cables into account and asumed that all the cords in both the belts are having same tensile strength, pull out strength etc., )

Waiting for opinions please.

Regards, ■

Dear experts,

I was not correct in my earlier post that steel cord joints would have 98% efficiency.

For single stage they are said to have 100% efficiency. It would go down as the number of steps increases and go below up to 70%.

Experts are requested to explain how the above were defined.

So I feel that it is important to have same numbers of cords on both ends for not loosing the joints efficiency.

Also request to explain how the minimum gap between the adjacent cords in the joint would be calculated if the ends are having different numbers of cords.

This makes me to remember my earlier thread on standardization of steel cord belts ( including the rip protection cords though they are not deciding the stages of the vulcanizing joints )

Thanks and regards, ■

Dear Sganesh,

You are incorrect in assessing the splice efficiency. Splice efficiency is a term used to identify the residual strength after cyclically loading the splice over a range between 6% and the nominate fully applied force the belt is expected to see in operation. A nominated repeat load pattern is applied. See the definition in DIN 22101 (earlier version), was considered about 36%. This value is how the steel cord safety factor 6.7:1 is derived.

You divide the necessary carrying loads of running (1.0 or unity), starting and stopping (0.4 or 40% of running) together with allowance for age, construction errors, all other forms of anomolies (1.0 also unity) by the residual 36% splice strength after 10,000 repeated load cycles. Summing the applied loads and dividing by the remaining splice endurance (1.0+0.4+1.0)/.36 = 6.7:1.

This process has been repeated in this forum a number of times. Please read available literature such as DIN 22101. ■