[A Brief Discussion/calculations of the One Side Power Drive Only for
the BTeV Vertex Magnet Track-moving System.]{.underline}

August 23, 2004 Edward Chi

*[Introductions]{.underline}*

The BTeV Vertex Magnet (VM) is considering to design a track-moving
system to move the 450 tons magnet along the beam direction ( +/- Z
direction, i.e.: north-south dir. Of the C0 collision hall). Currently,
the designated travel distance is about 150".

The VM track-moving system is composed of: Support structures; 4 Hilman
rollers with Accu-roll guidance features; Rail system (special steel
plates for rail, alignment devices, special anchors and guiding device);
Power drive system (actuator, reducer, couplers, motor and control
system to generate about 45 tons push/pull force to move the VM along
the rail within the designated tolerances (home position with 0.5 mm,
skewing angle in plan view (xz plane) is about +/- 0.02 degree). The
moving speed will be controlled within 5"/minute in Z direction.

![](media/image1.jpeg){width="5.988888888888889in"
height="4.682638888888889in"}

Figure 1: General layout of the BTeV experiment at C0 collision hall

(with one side driving force only for VM track-moving system)

Currently, we're considering to design the VM track-moving system with
only one side driving force as showing on figure 1. It is necessary to
analysis and to understand the mechanical behavior of some related parts
resulting from the additional unbalance force, especially it is true for
the rail anchor bolts as showing on figure 2.

*[Assumptions and Calculations:]{.underline}*

![](media/image2.jpeg){width="5.999305555555556in"
height="4.490972222222222in"}

Figure 2: Partial side elevation view of the Vertex Magnet with moving
rollers and

Rail system.

Figure 3 on page 3 is a force distribute diagram based on the input from
Figure 1 & 2, it is also necessary to make some assumptions to simple
the calculation:

Four rollers will equally to take mass weight of the magnet, the
locations of the rail to

support (to contact?) the rollers are defined as A, B, C & D
respectively.

It is assuming that the two plates of the rail are rigidly to connect
together both on xy

& zy plane.

The coefficient of friction in the rolling direction between roller and
the top surface of

the rail is f~r~ = 0.10. so:

F~ay~ = F~by~ =F~cy~ =F~dy~ = F~y~ / 4 = 112.50 tons

Where: F~y~ = 450 tons, the mass weight of the Vertex Magnet,

F~ay~, F~by~, F~cy~ & F~dy~ are the vertical force to point A, B, C & D.

The lateral force caused by the allowable skew angle (+/- 0.02^o^) will
not discuss in

this section (It is also negligible to compare the lateral force F~ax~
or F~bx~).

F~az~ = F~bz~ = F~cz~ = F~dz~ = f~r~ . F~ay~ = 11.25 tons

Where, F~az~, F~bz~,F~cz~ & F~dz~ are the rolling forces subject to
point A,B,C & D.

So:

F~z~ = F~az~ + F~bz~ +F~cz~ + F~dz~ = 45 tons

F~ax~ = F~bx~

![](media/image3.jpeg){width="5.9944444444444445in"
height="4.838888888888889in"}

~Figure\ 3.\ Force\ distribution\ of\ the\ VM\ rail\ system\ with\ only\ one\ drive\ force~

For location B, where:

∑ M~By~ = 0,

So:

(F~cz~ + F~dz~) x 157.0 = F~ax~ x 95

F~ax~ = \[(F~cz~ + F~dz~) x 157.0\] ÷ 95.0 = \[(11.25 + 11.25) ton x
157.0 in\] ÷ 95.0 in

= 37.2 tons = F~bx~

The resultant shear load of the location A or B is:

F~rsa~ = F~rsb~ = (F~ax~^2^ + F~az~^2^)^1/2^ = 38.9 tons =77,800 lbs (1)

It is also necessary to find the tensile force of the location A or B of
figure 3.

![](media/image4.jpeg){width="5.988888888888889in" height="4.69375in"}

Figure 4: The enlarge xy view of location A or B of figure 3.

From Figure 3 & Figure 4, it is found that:

F~ax~ x 2.625 in = F~ay1~ x 19.75 in

F~ay1~ = (F~ax~ x 2.625 in) / 19.75 in

= 4.95 tons

Where: F~ay1~ is the tensional force of the one side of the anchor bolts

subject to the shear force F~ax~ @ location A.

From Figure 3 and Figure 5, it is found that:

F~az~ x 2.625 in = f~ay2~ x 9.75 in

F~ay2~ = (F~az~ x 2.625 in) / 9.75 in

= 3.03 tons

Where: F~ay2~ is the tensional force of the one side anchor bolts

subject to the shear force F~az~ @ location A.

> ![](media/image5.jpeg){width="5.995138888888889in"
> height="4.975694444444445in"}

Figure 5: The enlarge zy view of location A or B of figure 3.

The resultant tensional force for one side of the anchor bolts:

F~ay~ = F~by~ = (F~ay1~^2^ + F~ay2~^2^)^1/2^ = 5.8 tons =11,600 lbs (2)

*[The discussion of the anchor bolt applying to the rail
system]{.underline}:*

It is assuming that the anchoring design is based on the design criteria
of the "HILTI Product Technical Guide", 2002 edition; and ASD 9^th^
edition:

Per Figures 4 & 5, it is assuming there are number of anchor bolts
within 30.75" (z dir.) force applying area to each side of the roller.

If using Carbon Steel Kwik Bolt II in concrete (3,000 psi), for

¾" bolt dia. With embedment depth 4.75", the allowable loads are:

F~t~ = 4,130 lbs

F~t~: allowable load in tension

F~v~ = 5,120 lbs

F~v~: allowable load in shear

(Per page 156, section 4.3.3 of HILTI Product Technical Guide, 2002
edition)

Per eq. (1) of page 4, and figures 4 & 5, it is found that the working
shear load per bolt:

f~v~ = F~rsa~/ 10 = 77,800^\#^ / 10 = [7,780^\#^ \> F~v~
(5,120^\#^)]{.underline}

Per eq. (2) of page 6, and figure 4 & 5, it is found that the working
tensional load per bolt:

f~t~ = F~ay~ / 5 = 11,600^\#^ / 5 = 2,320^\#^ \< Ft (4,130^\#^)

Since the space distance 9.75" \> S~cr~ (9.50"), it can be assuming
that:

Recommended working tensional load F~rect~ = F~t~

Recommended working shear load F~recs~ = F~v~

To check the combined loading by formula:

(f~t~/F~t~)^5/3^ + (f~v~/F~v~)^5/3^ ≤ 1.0

(per Section 4.1.3 of HILTI Product Technical Guide, 2002 edition)

(2300/4130)^5/3^ + (7780/5120)^5/3^ = 0.376 + 2.009 = [2.385
\>1.0]{.underline}

Both **f~v~** and the **combined load** don't meet the design spec.
under current design assumptions.

If choose 1" dia. Anchor bolt:

Since the min. space dist. S~cr~ for 1" dia. Bolt \* will be 12.0", then
it will be 4 anchor bolts instead of 5 anchor bolts for each effect side
(See figure 5), so the working shear load for each anchor bolt will be:

f~v~ (1" dia.) = 77,800# / 8 = [9,725^\#^ \> 9,200^\#^]{.underline}

1" anchor bolt is not available for some anchors.

*[Conclusions:]{.underline}*

Under the current design assumptions, the applying working load (shear)
**will be larger** than the designated allowable shear load of HILTI
anchor bolt (3/4" & 1" dia.) when VM track-moving system is driven by
one side force only.

However, the results will be changed by:

The coefficient of friction is not as it was assumed as f~r~ = 0.10 (the
most important

factor).

The friction force between top surface of the rail and the roller in x
direction is lesser

than F~ax~ or F~bx~ (i.e: the VM rotating along y axis).

The change of the distance between two rollers in z or x direction.

References:

1.  "Product Technical Guide", HILTI North America, 2002 edition

2.  "ASD" 9^th^ edition, AISC.