use physics to solve your weld design problems

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Use Physics to Solve Your Weld Design

Problems

Dec 7, 2006 12:00 PM

When addressing a weld design problem, the designer often will rifle through various engineering texts or

handbooks, looking for an applicable equation that will yield the desired result. Euler's column formula, for

example, quickly enables the designer to determine the required section properties to resist buckling of a

compression member of a certain length. But, what do you do when no such engineering formulas exist?

That's where we may need to dust off our physics books and return to basic principles.

Several years ago, a friend called me to discuss the design of a "bumper plate". This device was part of a

concentrating plant where iron ore was processed in a gyratory crusher, producing a concentrated ore

called taconite. The ore was discharged through a chute to the crusher (see Figure 1). However, the

impact of the ore falling directly into the crusher jammed the unit. To eliminate this problem, the designer

proposed to add a bumper plate. Ore falling from the chute would hit the plate, which would absorb much

of the energy associated with the falling material.

In order to size the welds that would hold the bumper plate in place, the designer had to estimate the loads

that would be imposed on the bumper plate by the falling ore. Intuitively, he knew that he wouldn't find any

equations for "forces on bumper plates" in his engineering texts or handbooks, so he called me for

assistance.

The basic purpose of the bumper plate was to slow the falling ore, thereby reducing the impact of the ore

on the crusher. The conveyor transported the ore to the chute, after which it fell through a distance of 120

in. During the freefall, the vertical velocity of the ore increased. It then hit the bumper plate, which slowed

the rate of fall. The impact of the ore on the bumper plate would create a force that had to be resisted by

the mechanical structure supporting the plate.

From physics, we know that

Impulseis change in momentum, and impulse is also force multiplied by time.

Momentumis mass times velocity. Mathematically, the above can be expressed as follows:

Impulse = change in momentum

Ft = mV2 - mV1 = m(V2 - V1)

Finally, the interval of time (t) can be set equal to D t, and F can be found as follows:

F(Dt) = m D t(V2 - V1)

This relationship is helpful, since F represents the force of the ore hitting the bumper plate. The problem

now consists of determining the mass of ore delivered in the increment of time, and the change in velocity.

The mass involved is that of the constantly flowing ore. The ore left the chute at a speed of 100 in. per

second. The ore in the chute had a cross section of 20 in. by 30 in. (see Figure 2). In any increment of time

D t, the mass of the ore (with a density of 0.0868 lb per cubic inch) can be found by the following: m = (area

of chute) x (ore feed rate) x (ore density) D t/ gravity



= (20 in. x 30 in.) x (100 in./sec) x (0.0868 lb/cu in.) D t/ 386 in./sec/sec

= (13.49 lb/sec) D t

Next, we need to find the velocities. While the velocity V1 of the falling ore was significant, after it hit the

bumper plate, V2 was essentially zero, simplifying the relationship to the following:

F D t = - mV1

The ore left the chute at 100 in./sec. Since the chute was at a 45-degree angle from the horizontal, the

horizontal velocity Vh was equal to 70.71 in./sec, and this value was constant, regardless of the horizontal

location of the ore. The initial vertical velocity Vv was also 70.71 in./sec, but increased due to the

acceleration of gravity. Basic physics was used to determine that the drop of 120 in. caused Vv to

increase to 312.53 in./sec, and that the fall took 0.626 sec. During this time, the ore moved forward 44.2

in. Therefore, the ore hit the bumper plate at an angle of about 12.74 degrees from vertical, as shown in

Figure 3.

The bumper plate was oriented at a 30-degree angle from vertical (Figure 4). Normal to the face of the

plate, the ore flowed at a velocity Vn of 217.53 in./sec, and the flow parallel to the face Vp was 235.41

in./sec. Vn is the value that is expected to be reduced to zero on impact. This is the velocity of interest,

since it is the force that must be resisted by the structure and the welds that support the assembly. And so

here, more physics came into play.

Since Vn = V1 = 217.53 in./sec, then

F D t = - mV1 = (13.49 lb/sec)Dt x 217.53 in./sec

F = 2934.0 lb

At this point, the force on the bumper plate is known, making it easy to size both the members and the

welds that join them.

This problem required us to apply basic physics and some trigonometry, but by methodically analyzing itstep by step, we were able to solve the problem.



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.

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use physics to solve your weld design problems

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