Lacing, battening

GUIDED BY: Prof. Sunil Jaganiya
Prof. Pritesh Rathod
Sub : Elementary Structural Design
NAME ENROLL NO.
Patel Jimi 131100106029
Patel Milind 131100106035
Patel Nirmal 131100106036
Patel Viraj 131100106040
Patel Yash 131100106042
Shah Ashit 131100106051

There are two types of lacing system.
1. Single lacing system
2. Double lacing system

 The compression member comprising two main components laced and tied should,
where practicable, have a radius of gyration about the axis perpendicular to the plane
of lacing not less than the radius of gyration at right angles to that axis.
 The lacing system should not be varied throughout the length of the strut as far as
practicable.
 Cross (except tie plates) should not be provided along the length of the column with
lacing system, unless all forces resulting from deformation of column members are
calculated and provided for in the lacing and its fastening.
 The single-laced systems on opposite sides of the main components should preferably
be in the same direction so that one system is the shadow of the other.
 Laced compression members should be provided with tie plates at the ends of the
lacing system and at points where the lacing system are interrupted. The tie plates
should be designed by the same method as followed for battens.

(1) Angle of inclination(θ): (cl. 7.6.4)
For single or double lacing system,
θ = 40 ͦͦ to 70 ͦ To the axis of the built up member
normally,=45 is taken
(2) Slendernes ratio(kL/r) : (cl. 7.6.5.1)
KL/r for each component of column, should not be gretear than 50.
or
kL/r not greater than 0.7 *most favourable slenderness ratio of the member as
a whole
The slenderness ratio of lacing shall not exceed 145 (cl. 7.6.6.3)

(3) effective length of lacing (le) :
For bolted connection :
For single lacing, le = L
For double lacing, le = 0.7 l
Where, L = distance between the inner end fastner
In welded connection :
Le = 0.7 * distance between the inner ends of welds
(4)width of lacing bars(b) :
minimum width of lacing bar, b = 3d
Where,
D = nominal diameter of bolt

(5) Thickness of lacing (t) : (cl. 7.6.3)
For single lacing, t > Le/40
For double lacing, t > Le/60
(6) Transvers shear (Vt) : (cl.
7.6.6.1)
Vt= 2.5% of the axial force in the
column.
This force shall be divided equally
among the lacing systems in
parallel
Planes.
For double lacing
F=Vt/4 sin
Where,

(7) Check for compressive strength
For lacing using Le/r min and fy = 250 Mpa
Find Fcd from IS: 800, table -9 (c)
For rectangular section buckling class is “c”.
Compressive load carrying capacity of lacing
Pd = (b * t) * fcd
If (b *t )* fcd > F(axial force n lacing) …. OK
b*t = area of lacing
i.e. pd > F …. OK

(8) check for tensile strength :
tensile strength of lacing flat is
Td = 0.9 (b-d)t fu /ϒ or fy.Ag/ ϒmo Which ever is less.
If Td > F…….Ok { Is: 800 cl.
6.3.1 pg 32 }
(9) End connection :
For case (a) : Resultant on force on bolt = R = F
No of bolt required = F/bolt value
For case (b) : Resultant on force on bolt = R =2Fcosθ
No of bolt required =
2𝐹 𝑐𝑜𝑠θ
𝑏𝑜𝑙𝑡 𝑣𝑎𝑙𝑢𝑒
For 16 dia. Bolt strength is single shear= 29 kN
For 20 dia. Bolt strength is single shear= 45.3 kN
Strength of bolt in bearing =2.5 kb.d.t.fu (cl.
10.3.4)

(10) Overlap:
In case of welded connection, the amount of overlap measured along either
edge of lacing bar shall not be less than , four times the thickness of the lacing
bar or the
thickness of the element of main member, whichever is less.

 Compression member can also be built up intermediate
horizontal connecting plates or angle connecting two or four
elements of column .these horizontal connecting plates are called
battens
 The battens shall be placed opposite to each other at each end
of the member and at point where the member is stayed in it
length and as for as practicable , be spaced and proportioned
uniformly throughout.
 The number of battens shall be such that the member is devided
into not less than three bays within its actual length

(IS : 800, cl. 7.2.2, P.51)
(1)The number of battens shall be
such that the member is divided into
not less than three bays.
(2) Battens shall be designed to resist
, simultaneous

 Longitudinal shear
Vb = Vt. C/Ns
And
 Moment
M=Vt.C/2N
Where,
Vt = transverse shear force
C = distance between centre to centre of battens longitudinally .
N = number of parallel planes of battens (2 usually)
S= Minimum transverse distance between the centroid of the bolt/
rivet group / welding.

(3) Slenderness ratio : (cl. 7.7.1.4)
the effective slenderness ratio (
𝑘𝐿
𝑟
)e of battenced column shall be taken as 1.1 times
the (
𝑘𝐿
𝑟
)o, the maximum actual slenderness ratio of the column, to account for shear
deformation effects.
(4) Spacing of battens (C) : (cl. 7.7.3)
For any component of column
(i)
𝑐
𝑟 𝑚𝑖𝑛
should not greater than 50
(ii)
𝑐
𝑟 𝑚𝑖𝑛
should not greater than 0.7 * kL/r of built up column (about z-z axis)
(5) Thickness of battens (t) : (cl. 7.7.2.4)
t >
𝐿𝑏
50
where Lb = Distance between the inner most connecting line of bolts, perpendicular
to the main member

(6) Effective Depth of battens (de) : (cl 7.7.2.3)
 de > 3/4 *a ……… for intermediate battens
 de > a,……. For end batten
 de > 2b , ………. For any battens
where
de = effective depth of battens
= distance between outermost bolts longitudinally
a = distance between centroid of the main member
b = width of one member
Overall depth of battens
 D = de + (2 * end distance)

(7) transverse shear (Vt) : (cl. 7.7.2.1)
Vt = 2.5 % of the factored axial column load
(8) Ovrlap (cl. 7.7.4.1)
for welded connection, the overlap shall be not less than four
times the thickness of the battens
It should be noted that the battens columns have least
resistance to shear compared to column with lacings

the minimum thickness of rectangular slab bases , supporting columns
under axial compression shall be
 ts =√(2.5 w (a2 - 0.3b2) ϒmo/fy) > tf
Where
ts = thickness of slab base
w = uniform pressure below the base
a,b = larger and smaller projection, respectively of slab base beyond
the column
tf = flange thickness of compression member

 Design a slab base foundation for a column ISHB 350 to carry a factored
axial load of 1200 KN. Assume fe 410 grade steel and M25 concrete. take
safe bearing capacity of soil as 200 kN/m2
Solution :
For steel fe 410 fy = 250 N/mm2
For m 25 concrete, fck = 25 N/mm2
FOR ISHB 350 COLUMN
h = 350 mm
Bf =250 mm
Tf = 11.6mm
Tw= 8.3 mm

(a) Area of base plate : {IS 800 -2007 CL. 7.4.1 P.46 }
pu = 120 kn ( factored load )
 bearing strength of concrete = 0.6 fck
= 0.6 * 25
= 15 N/mm
2
 area of base plate :
=
p
u
𝑏𝑒𝑎𝑟𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒
A =
1200∗103
15
= 80,000 mm2
size of built up column
b = 350 mm
d = bf =250 mm

( B ) THICKNESS OF BASE PLATE :
 a =larger projection
= 50 mm
 b = smaller projection
= 50 mm
 W = uniform pressure on base plate
=
1200 ∗103
450 ∗350
= 7.62 n/ mm2
thickness of base plate =t
provide 50 mm equal projection all around the column
 width of plate
 Bp = 350 + 50 + 50 = 450 mm
 Dp = 250 + 50 +50 = 350 mm
Use base plate of size 450 mm* 350 mm
 Gross area of base plate provided = 450 * 350 = 157500 mm2

(C) WELD CONNECTING COLUM TO BASE
PLATE :
 Use a 6 mm fillet weld all around the colum section to hold the base
plate in position
 total length available for welding along the periphery of ISHB 350 ,
there are 12 ends for ISHB
 DEDUCTION = 12* 2S
=12 * 2 * 6
= 144 mm
 effective length of weld available
= 1683.4 – 144
= 1539.4

 capacity of weld per mm length
= 0.7 s * fwd
= 0.7 * 6 * 189
= 793.8 n/mm
= 0.7938 KN/mm
 required length of weld
= 1200
0.7938
= 1512 mm < 1539.4 mm
6 mm weld is adequate .

(D) SIZE OF CONCRETE BLOCK :
 Axial load on column =120 kN(factored load)
 Working load =1200/1.5=800kN
 Add 10% as self weight of concrete block =80KN
 Total load =800+80=880 kN
 Area of concrete block required
=Total load /S.B.C. of soil
=880/200
=4.4m2

 Concrete block is designed for working
load
 Consider rectangular concrete block
with equal projection beyond base plate.
 Let, X= projection of concrete block
 Area of concrete block =L*B
4.4=(0.45+2x) *(0.35 + 2x)
4.4=0.1575 + 0.7x +0.9x + 4x2
4x2 + 1.6 x – 4.2425 = 0
Solving it, x=0.849 m
Using calculator , say x= 0.85 m

 L=0.45 + 2 * 0.85 = 2.15m
 B=0.35 + 2*0.85 = 2.05m
 Area of concrete block
provide = 2.15 * 2.05
=4.407m2
> 4..4 m2……OK
 Assumme angle of dispersion
=45°
 Depth of concrete block = d =
x
= 0.85 m

column splice:
A joint when provided in the length of column to get to required length
it I called column splice.
If a column is loaded axially, theoretically no splice is required.
Compression will be transmitted by direct bearing, and column sections
could be rested one on top of each other.
How ever , In practice the load on column is never truely axial and the
real column has to resist bending due to this eccentrically applied load.
In addition , the column may be subjected to bending moments.
Also, the bearing surface of the adjacent sections can never be
machined to perfection.

Design of column spices:
The steps inn the design of splices are:
1. Determine the nature of loads to which the splice is subjected. The splice
may be subjected to axial compressive load, bending moment and shear
force.
2. For axial compressive load the splice plates are provided on the flanges of
the two columns.
if the ends of columns are milled/machined, the splice is designed only to
keep the column in position and to carry tension due to the bending
moment. In this case splice plate is designed to carry 50% of the axial load
and tension due to B.M.
if the ends of column are not milled/machined, the splice and connections
are designed to resist the total axial load and tension, if any.

3. Load due to axial load for machined ends of column,
 Pul= load on splice due to axial factored load Pu on the column.
=
𝑃𝑢
4
(total load on splice plates =
𝑃𝑢
2
but load on each splice plate =
𝑃𝑢
2
)
For non-machined ends of column,
Pul =
𝑃𝑢
2
4. Load due to bending moment Pu2 =
𝑀𝑢
𝑙𝑒𝑣𝑒𝑟 𝑎𝑟𝑚
=
𝑀𝑢
𝑎
Where,
a = lever arm
= c/c distance of two splice plates.

5. Column splice plates are assumed to act as short column (with zero
slenderness). Hence, the plates will be subjected to yield stress (fy).
 fcd=
𝑓𝑦
1.10
6. The cross-sectional area of splice plaate(A)
 A=
𝑃𝑢
𝑓𝑐𝑑
Pu= Pul + Pu2
7. The width of splice plate is kept equal to the width of the column
flange.
 thickness of splice plate=
𝐴
𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑠𝑝𝑙𝑖𝑐𝑒 𝑝𝑙𝑎𝑡𝑒
For column exposed to weather , the thickness of splice should not
Be less than 6 mm.

8. Nominal diameter of bolts for connection is assumed.
 No. of bolts=
𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑎𝑑 𝑜𝑛 𝑠𝑝𝑙𝑖𝑐𝑒 𝑝𝑙𝑎𝑡𝑒
𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑜𝑛𝑒 𝑏𝑜𝑙𝑡
9. When the bearing plates are to provided to join two columns of
unequal
sizes:
- The bearing plate may be assumed as short beam to transmit the axial load
to the lower column.
- Axial load of the column is assumed to be taken by flanges only.
shown in figure
Maximum B.M in bearing plate:
 M=
𝑃𝑢
2
*a1

The length and width of the bearing plates are kept
equal to the size of the lower
storey column.
Thickness of bearing plate,
 M= fbs * Z
Where ,
 fbs= design bending stress
=
𝑓𝑦
1.10
=
250
1.10
= 227.27 N/𝑚𝑚2
 Z =
𝑏𝑡2
6

10. The web splice plates are designed to resist maximum shear force.
11. If packing are provided between the splice plate and column flange
and more than 6mm in thickness, the design shear capacity of the
bolts is reduced as per cl. 10.3.3.3 of IS : 800-2007.

 A column section ISHB 250@ 500.3 N/m is carrying a factored load of 600
kN. Design a suitable column splice. Use 16 Ø 4.6 grade bolts and steel of
grade Fe 410.
Solution..
For 4.6 grade bolts,
Fub =400 N/mm2
For – fe 410 plate fu = 410 N/mm2
fy = 250 N/mm2
 For column ISHB 250 @ 50.3 N/m
bf = 250 mm
tf = 9.7 mm

 Assume ends of columns are miled /machined for complete bearing.
Therefore , splice plate are designed for 50 % of axial load of column .
load on each splice plate ,
 pu1 = 𝑝 𝑢
4
= 600
4
= 150 KN
 Fcd
𝑓 𝑦
ɣ 𝑚0
= 250
1.10
= 227.27 N/mm2

Area of splice plate requride = 150 ∗103
227.27
= 660 mm2
 width of splice plate should be equal to the width of the column flange .
b = 250 mm
 thickness of splice plate,
t = 𝑎𝑟𝑒𝑎
𝑏
= 660
250
= 2.54 mm
 provide 6 mm thick splice plate as colum may be exposed to weather .
For 16 mm dia , 4.6 grade bolts
strength of bolt in single shear = 29 KN

 Stength of bolt in bearing ( on 6 mm
plate )
= 2.5 kb . D .t .fu /ɣ𝑚𝑏
= 2.5 * 1* 16 * 6 * *400/1.25
= 76800 N
= 76.8 KN
bolt value = 29 KN
 N0 0f bolt required =150/29=5.17
say 6 nos.
 Provide 16 mm dia , 6 bolt on each
side of the splice (joint) in two
vertical raws to connect splice plate
with column flangers.
Minimum pitch = 2.5 d = 2.5* 16
= 40 mm

 Provide pitch = 50 mm
 Edge distance = 1.5 d0 =1.5 *18 =27 mm provide 30 mm
 Depth of splice plate
=(4 * 50) +(4*30)
=320 mm
 Provide splice plate 320*250*6mm 0n column flanges.

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