Analysis and retrofitting design of the fastener group
for the spreader
bar of
SM3 disassembly
Background:
Try to re-use (or to modify) the existing Mino
project spreader bar for SM3
disassembly project. the capacity of the spreader bar for SM3 project
is:
Overall load Pt = 9.4
tons with span distance Ls = 201 in
Applicable codes:
ASME B30.20; “Below – the – Hook Lifting Devices”
ASD, AISC 9th edition
References:
ME – 397459
ME – 397426
MD – 397452
“Steel Structures Design and Behavior” by C. Salmon & J. Johnson, 3rd
edition
Assumptions:
Slip critical connection, single
shear
Figure 1 of
page 1 is showing that when the applying load P is eccentric to the
centroid of the bolt group, this physical configuration is the actual design of
the spread bar using for sm3 project.
Where:
P = 9,400 lbs, applying load
L = 47.625 in
Location A is the geometrical centroid of the bolt group, 6 bolts are located as
showing in Figure 1. currently, it is assuming: A325, ¾ - 10, UNC
n = 6
per Table I –D, part 4 of ASD, Rav
= 7.51 kip (allowable shear load)
Find out the local properties of the
fastener group:
∑x2 = 4 (4)2
= 64 in2
∑y2 = 6 (2)2 = 24 in2
∑x2 + ∑y2 = 88 in2
The primary shear load Rv of
each bolt subject to the applying load P:
Rv = P/n = 9,400 lbs / 6
= 1,567 lbs ↓
The secondary torsional shear load of the
bolt subject to the moment PL:
To
pick the bolt of the most right top one as showing in figure 1,
Where: Rx = PLy ÷ (∑x2 + ∑y2)
= (9,400 lbs) x (47.625 in) x
(2 in) ÷ 88 in2
= 10,174 lbs. →
Ry = PLx ÷ (∑x2 + ∑y2)
= (9,400 lbs) x (47.625 in) x (4 in) ÷ 88 in2
= 20,349 lbs ↓
The resultant force applying to the most
right top bolt:
R = [(Rv + Ry)2 + Rx2
]1/2
= [(1,567 +
20,349)2 + (10,174)2 ]1/2 lbs
= 23,259 lbs > Rav
= 7.51 kip
It is necessary to look for:
a. Different specifications of the fastener with the same fastener
group.
b. Another pattern of the fastener group
A.Different specification of the fastener
with the same fastener group:
A1.
If using 6 bolts with A325, 1 3/8”- 6 UNC,*
then Rav = 25.2 ksi > R
= 23.26 ksi,
*: The
hole ctr. to hole ctr. distance Le = 4.0 in < 3d = 4.13 in
A2.
If using same fastener group with bolt of A490, 1 ¼”- 7, UNC,
then Rav = 25.8 ksi > R =
23.26 ksi (per Table I-D, part 4 of ASD, 9th edition)
where: Le = 4.0 in > 3d = 3.75
in
B. Modify the current pattern of the fastener
group:
Figure 2 on page 3 is the new fastener group
with adding additional 10 fasteners to
the original group, it can be found the new properties of the fastener
group:
∑x2 = (4 (4)2
+ 4 (2)2 + 6 (6)2) in2
= (64 + 16 + 216) in2
= 296 in2
∑y2 = (6(2)2
+ 8(3.375)2) in2
= 115 in2
∑x2 + ∑y2
= (296 + 115) in2
= 411 in2
Also
n = 16
The primary shear load Rv of
each bolt subject to the applying load P:
Rv = P/n = 9,400 lbs / 16
= 588 lbs ↓
The secondary torsional shear load of the
bolt subject to the moment PL,
To pick the bolt of the most right top one
as denoted as bolt A of Figure 2:
Where: Rx = PLy ÷ (∑x2 + ∑y2)
= (9,400 lbs) x (47.625 in) x
(3.375 in) ÷ 411 in2
= 3,677 lbs. →
Ry = PLx ÷ (∑x2 + ∑y2)
= (9,400 lbs) x (47.625 in) x (6 in) ÷ 411 in2
= 6,536 lbs ↓
The resultant force R applying to the most
right top bolt A:
R = [(Rv + Ry)2 + Rx2
]1/2
= [(588 + 6,536)2
+ (3,677)2 ]1/2 lbs
= 8,017 lbs > Rav
= 7.51 kip
If the bolt material change to ASTM A490,
then Rav = 9.28 kip
R = 8.017 kip < Rav = 9.28 kip
Since only 2 bolts of the fastener group
will experience shear load of R ~ 8.017 kip, all the rest bolt shear load is
less than 7.51 kip, so there are two choices:
1.
The most top right and bottom
right bolts use A490, the rest bolts use A325. (3/4 – 10, UNC.)
2.
Or all of them use A490 bolts
(3/4 – 10, UNC)
The
conclusions:
There are two ways to modify the current
fastener group to meet the new design criteria of spreader bar for SM3
disassembly:
1.
Using (6) A490 high strength
structural bolts with spec. of 1¼ - 7, UNC (original is ¾ -10, UNC), or
2.
Using (16) A490 high strength
structural bolts with spec. of ¾ -10, UNC.
