When designing a weldment, determining the proper weld size is important, but sometimes, this isn’t enough. Load path and weld orientation can also be critical to safe and effective weld design.

An application that illustrates this concept was a “heave compensator” used by an oil company working offshore.

A barge supported the equipment that extended to ocean floor where the drilling took place. Large waves would cause the barge to oscillate up and down with respect to the ocean’s floor. Some movement of the barge was necessary, but excessive movement could damage the drilling equipment. The heave compensator, depicted in Figure 1, dampened the effect of large waves. The assembly was big — over six feet long. While for the sake of simplicity, Figure 1 shows three pulleys, the actual unit had many more.

The weldment portion was composed of two “inner uprights,” two “outer supports,” and a semi-cylindrical wraparound shell. Two rectangular cutouts with rounded corners permitted the steel rope to pass through the shell.

My task was simple: The engineer who designed the assembly asked me to double-check the calculations for the weld sizes.

The forces involved were huge: The vertical forces that the cables resisted totaled 1.3 million lbs — see Figure 2. Two sets of pulley sheaves were wrapped with steel cables and each pulley supplied 650 kilopounds (kips) of resistance in the opposite direction.

The sheaves were mounted to an axle, and the two supports for each axle transferred 325 kips of load through the bearings.

The engineer’s original calculations went something like this:

Weld length: (Circumference/2) + 2 length of 21 in. (62π/2) + 2(21 in.) = 139.34 in. Four welds: 4 X 139.34 in. = 557.36 in. Total load: 1,300,000 lb Force per unit length of weld: 1300000/557.36 = 2330 lb/in. Weld size: 2330 in./(0.707 in. X 0.3 in. X 70,000 in.) = 0.157 in. Use 3/16 in. (0.1875 in. > 0.157 in.) okay.

These calculations were accompanied by a sketch that looked something like Figure 3.

The small 3/16 in. fillet didn’t seem right, and therefore the engineer asked for my opinion.

I’m glad he did, because there were a number of problems in the analysis.

Before we get into those, notice that the above calculations assumed that only four welds would be made, even though welds could be placed on one side of the inner upright and the outer support.

If that were considered, his computations would have led to an even harder to accept weld size of 3/32 in.! To unravel this problem, let’s first look at the loads and the load path: The cable loads were transferred to the shaft and then to the two bearings. In the case of the bearings in the “inner uprights,” the load was transferred directly from the bearing support. No welds were required to transfer this load. For the “outer support,” however, the load would have to be transferred from the support, to the shell, then from the shell to the “inner upright.”

In the original analysis, it was incorrectly assumed that all of the 1.3 million lb load would be transferred through the welds. In fact, only half the load was transferred through welds.

However, the four welds could not be considered as all resisting the load, since the force to the outer support first had to go through the weld on the outboard side, then through the weld on the inboard size. So, while the engineer considered twice the force involved, he also used twice the weld length. Coincidentally, the two errors cancelled each other out.

But, the problems did not end there.

The engineer assumed the full length of the weld would be effective in transferring the load.

Two details of this application, however, challenged the validity of that assumption.

First, there is the basic principle of weld design that states “Provide a path so the force can enter into the section that lies parallel.” [See “Provide for Transverse Forces,” Welding Design & Fabrication, February 2004, p. 18-19.]

In this case, a significant portion of the curved weld is nearly perpendicular to the direction of force. Clearly, the straight portions of the weld were effective since they were parallel to the applied force. The curved portion was divided into four segments, and only the two portions that adjoined the straight portion were considered as being effective for this application.

The second complicating factor in this particular case was the presence of the slots in the shell.

Forces could not be transferred through these openings. The load path of force that entered into the outer support would be primarily transferred through the sides of the shell. From the shell to the inner upright, the majority of the force would be transferred through the portions of the weld that were nearly vertical.

For computation purposes, the weld that would resist the applied loads would consist of two straight segments (21 in. each) and two curved quarters of weld (243/8 in. each) for a total length of 90.75 in. The average force per length therefore was assumed to be 325 kips/90.75 in. or 3.58 kips/in. This increased the load per length by slightly more than 50 percent.

The next issue to consider was the nature of the load. The engineer had assumed a weld allowable of 21 kips per sq. in. (ksi), which is appropriate for E70 filler metal in statically loaded conditions.

However, this application was subject to cyclic loading. In use, the stress ranged from no load to the full 1.3 million lbs.

The stress ratio (minimum stress/ maximum stress) was zero, and the stress range (maximum stress – minimum stress) was 3.58 kips/ in. of weld.

For infinite life, the fatigue threshold value would apply. The current AWS D1.1 category F stress range allowable for this condition is 8 ksi. Thus, the required weld size was 3.58/(0.707 in. X 8) or 0.63 in. Rounding to the next even weld size, a ¾ in. weld would be used. This is four times the size originally calculated.

Finally, we return to the fact that two-sided welds were possible, and instead of a single-sided, ¾ in. fillet, a pair of 3/8 in. fillets could have been used as well.

There is nothing wrong with applying a weld along the lower quadrant of the compensator, as is shown in Figure 1. However, this portion of the weld should not be considered effective for transferring load.

At the end of the day, I imagine most people would call for a pair of ¾ in. welds, just to be safe.

After all, for one-of-a-kind weldments such as this, and given the critical nature of this compensator for the operation of the drilling barge, it is better to be safe than sorry.

The takeaway lesson?

Dividing the load by the total length of the weld, then determining the required weld to resist that force per length, isn’t always good enough.

Considering the load path, as well as the orientation of the weld to the force, will help avoid mistakes like the ones the original engineer nearly made.

Omer W. Blodgett, Sc.D., P.E., senior design consultant with The Lincoln Electric Co., struck his first arc on his grandfather’s welder at the age of ten. He is the author of Design of Welded Structures and Design of Weldments, and an internationally recognized expert in the field of weld design. In 1999, Blodgett was named one of the “Top 125 People of the Past 125 Years” by Engineering News Record. Blodgett may be reached at (216) 383-2225.