Scaffold Basic Design Example

(1)

1

SCAFFOLD BASIC DESIGN EXAMPLE

Figure 1. Basic scaffolding working platform.

Let us design the components of a medium type scaffold with the configuration1_{as shown in}

Figure 1 and for the general building works (brickwork, window and mullion fixing, rendering and plastering)2_.

Table 1. Loading conditions.

Nominal load: 0.5 kNm-2 _{Imposed load: 2.0 kNm}-2

Wind load: Lateral load:

1. Platform

The platform is of 1.8 m wide, and may be considered to be made up of 4 planks of 450 mm width3_.

1_{Refer reg. 94(b). Construction of tubular scaffold.}

2_{Refer Table 1. Service loads for working platform, BS 1139-5:1990, Guidelines for Approval of}

Design Scaffolding.

3_{Refer reg. 87(1)(b). Planks must not less than 200 mm, but if thickness > 50 mm, width must ≥ 150}

(2)

2

1.1. Plank thickness

The putlogs4_{are the secondary beams for the platform.}

Span for platform = Spacing of putlogs, i.e. 1.2 m. The planks extends over three putlogs.

Figure 2. Platform rests on three putlogs.

Given the bending strength of timber platform5_{is 16 Nmm}-2_{, and a safety factor of 2,}

allowable bending stress6 max

 is 8 Nmm-2_{and allowable shear stress} allow



is 4 Nmm-2_.

For simply supported beam carrying a uniformly distributed load, maximum bending moment

M is at mid-span and equal to

8

2

wL

M 

(1) where

w



2 .

5 kNm

-2



1 .

8 m



4 .

5 kNm

-17_and

_L



₁

_.

₂

_m





kNm 81 . 0 8 m 2 . 1 kNm 5 . 4 -1 2    M (2)

Section modulus or elastic modulus S is functions of geometry only8_{and relates stress and}

internal moment during elastic or recoverable bending.

max max y I M S  



(3) where _-₂ 3 3 3 mm 250 , 101 Nmm 8 Nmm 10 10 81 . 0     S and

2

max

d

y



4_{Putlog or bearer means that part of the scaffold upon which the platform rests (reg.2).} 5_{Refer Table A.5, BS EN 12811-2:2004.}

6_{Table 2 of BS 5975:2008+A1:2011 specifies 7.20 Nmm}-2_{for D30 and 10.0 Nmm}-2_{for D40}

hardwoods. Table 5 of the same standards classify keruing, karri. opepe, merbau, teak, jarrah and iroko and typical tropical hardwoods.

7_{This is a load per unit width of the plank.}

(3)

3

Figure 3. Cross-section of a timber plank, where y is the distance from the neutral plane, and

y is maximum at the surface.

Given for rectangular cross-section, moment of inertia I

12

3

bd

I 

(4)

Substituting eqn. (4) into eqn. (3), and solving for d gives

b

S

d



6

(5)

mm

37 .

18 mm

800 ,

1 mm

250 ,

101

6

3







d

(6) USE 450 mm x 20 mm PLANK

Note: Width of the plank can be varied, but thickness must not be less than 20 mm. For example, if 450 mm x 20 mm is not commercially available, can use 300 mm x 25 mm (12’ x 1’) plank, but would require 6 planks.

1.2. Plank deflection

Assume, for wood9_{, Young’s Modulus} -2 6 -2

kNm 10 10 Nmm 10,000 GPa 10     E

Now, consider plank is 300 mm x 25 mm, and applying eqn. (4) gives





6 4 6 4 3

m

10

344 .

2 mm

10

344 .

2

12 mm

25 mm

800 ,

1 

_

_

_

_





I

(7)

9_{Table 2 of BS 5975:2008+A1:2011 specifies modulus of elasticity of hardwoods ranges from 7.6}

(4)

4 Figure 4. Deflection under cases (a) and (b).

Under case Figure 4(a), deflection is given by

EI wL4 009150 . 0 



(8)

Under case Figure 4(b), deflection is given by

EI wL4 005416 . 0 



(9)

Assume only one span is loaded with imposed load, but nominal load over both spans. Applying eqns. (8) and (9) gives,









mm 7285 . 0 m 10 285 . 7 m 10 344 . 2 kNm 10 10 m 2 . 1 m 8 . 1 kNm 5 . 0 009150 . 0 ₆ _-₂ ₆ ₄ 4 4 -2 dead           



(10)









mm 7278 . 1 m 10 73 . 1 m 10 344 . 2 kNm 10 10 m 2 . 1 m 8 . 1 kNm 0 . 2 005416 . 0 ₆ _-₂ ₆ ₄ 3 4 -2 live           



(11) mm 46 . 2 7278 . 1 7285 . 0 live dead total 







  



The maximum deflection for platform units shall not exceed 1/100 of the span length when suppporting the intended loads10_.

i.e.

10_{In BS EN 12811-1:2003, the elastic deflection of platform unit shall not be exceed 1/100 of its span.}

In Scaffold Safety Handbook, Saudi Aramco, 2001, pp. 29, and OSHA US pp. 3, the limit is 1/60 of the span length.

(5)

5

L

100

1

total





(12)

Substituting L = 1,200 mm into eqn. (12), gives

mm

12 or

mm

200 ,

1

100

1

total total











(13) mm 12 mm 46 . 2 total  



USE 300 mm x 25 mm PLANK

Please note that for every bay, imposed load is only allowed on one span! Re-calculation is needed if imposed load is expected to be subjected on both spans.

1.3. Plank shear

Cross-sectional area of decking A = 1,800 mm x 25 mm = 45 x 103_mm2_.

Given for case Figure 4(a), maximum shear force V

wL

V



0 .

6250

(14)

Substituting

w



2 .

5 kNm

-2



1 .

8 m



4 .

5 kNm

-111_{and L = 1.2 m into eqn. (14) gives}

kN

3.375 m

2 .

1 kNm

5 .

4 6250

.

0 

-1





V

(15)

For rectangular cross-section, maximum shear stress



_max occurs at the neutral axis and is given by

Ib

VQ



max



(16)

By substituting

Q 

Ay

, it can be shown that

A

V





2

3

max



(17)

Maximum shear stress

2 -2 -2 3 max

112 .

5 kNm

0 .

1125

Nmm

m

10

45 kN

375 .

3

2

3 _

_





_



(18)

Therefore, the maximum shear stress is much lower than the allowable shear stress, i.e. 4 Nmm-2_.

OK TO USE 300 mm x 25 mm PLANK

(6)

6

2. Putlogs

Each putlog supports the reactions from the platform’s plank.

The worst reaction occurs in the middle putlog, receiving loads from both of its sides. The middle putlog carries the load from 1.2 m width of platform.

Load intensity12_{= 1.2 m x 2.5 kNm}-2_{= 3.0 kNm}-1_{of putlog span}

Span of putlog = spacing of primary beams (ledgers) = 1.8 m

Figure 5. Putlog rests on two ledgers.

2.1. Putlog size

Applying eqn. (1) gives,





Nmm

000 ,

215 ,

1 kNm

215 .

1

8 m

8 .

1 kNm

3

-1 2 max







M

(19)

Applying eqn. (3), section modulus, S

3 2 -

151 ,

875 mm

Nmm

8 Nmm

000 ,

215 ,

1 



S

(20)

Assume b = 100 mm, and applying eqn. (5) gives,

mm 46 . 95 mm 100 mm 875 , 151 6 3    d (21)

USE 100 mm x 100 mm (4’ x 4’) TIMBER PUTLOG

Alternatively, can try use the steel tube as ledger with the following properties13

Table 2. Option 1.

Nominal diameter = 48.3 mm Nominal wall thickness = 3.2 mm Nominal yield strength



_y = 235 Nmm-2 _{Allowable bending stress}

max



= 211.5 Nmm -2

12_{Load intensity is equals to total load per unit span of putlog. See footnote 6.} 13_{Refer BS EN 12811-1:2003 clause 4.2.1.2.}

(7)

7

Minimum required S, based on applied loads and strength of tube,

3 2 -

5 ,

744 .

68 mm

Nmm

5 .

211 Nmm

000 ,

215 ,

1 



L

S

(22)

Second moments of area I of hollow tube is given by



4 4



64 D

d

I









(23)

Applying eqn. (23), with D = 48.3 mm and d = 41.9 mm, gives



4 4



4

mm

5 .

856 ,

115

9 .

41

3 .

48

64 







I

(24)

Section modulus S, based on geometry of tube,

3 4 max mm 4 . 797 , 4 2 mm 3 . 48 mm 5 . 856 , 115 _   y I S_G (25)

SG is less than SL, means have to use tube of higher strength and/or thicker tube.

Try use tube with higher yield strength



_y



275 Nmm

-2. Table 3. Option 2.



_y = 27514_Nmm-2 _{Allowable bending stress}

max



= 247.5 Nmm -2 3 2 -

4 ,

909 .

1 mm

Nmm

247.5 Nmm

000 ,

215 ,

1 _



L

S

(26)

SG (= 4,797.4 mm3) is slightly higher than SL, can use tube with this properties, but with little

safety factor.

Then, try use tube with higher yield strength



_y



355 Nmm



max



= 319.5 Nmm -2

14_{Refer Table A.1 of BS EN 12811-2:2004 (E).} 15_{Refer Table A.1 of BS EN 12811-2:2004 (E).}

(8)

8 3 2 -

3 ,

802 .

8 mm

Nmm

5 .

319 Nmm

000 ,

215 ,

1 _



L

S

(27)

SG (= 4,797.4 mm3) is significantly higher than SL, can use tube with this properties, and with

bigger safety factor.

And, try use thicker tube

t



4 .

05 mm,



_y



275 Nmm

Nominal diameter = 48.3 mm Nominal wall thickness = 4.0516_mm

Nominal yield strength



_y = 275 Nmm-2 _{Allowable bending stress} max



= 247.5 Nmm -2

Second moments of area of hollow tube is given by eqn. (23), with D = 48.3 mm and d = 41.9 mm, gives



4 4



4 6 4

m

10

139 .

0 mm

1 .

956 ,

138

2 .

40

3 .

48

64















I

(28)

Section modulus S, based on geometry of tube,

3 4 max mm 88 . 753 , 5 2 mm 3 . 48 mm 1 . 956 , 138    y I SG ₍₂₉₎

Section modulus S, based on applied loads

3 2 -

4 ,

909 .

1 mm

Nmm

247.5 Nmm

000 ,

215 ,

1 



L

S

(30)

SG is higher than SL, can use tube with this properties, but with higher safety factor.

THEREFORE, EITHER USE TUBES PROPERTIES AS IN OPTION 3 OR OPTION 4.

Note that, the allowable bending stress is assumed as 0.9 x



_y.

Also note that, it is preferable to use tube as putlog (diameter 48.3 mm), to a bigger timber section (100 mm x 100 mm).

2.2. Putlog deflection

Choose Option 4, and for loading conditions as shown in Figure 5, the deflection is given by

(9)

9 EI wL 384 5 4 



(31)

For steel, assume17_{modulus of elasticity, E = 210,000 MPa = 210,000 Nmm}-2_{, shear}

modulus, G = 81,000 MPa = 81,000 Nmm-2_{and density,}_{ρ = 7,850 kgm}-3





mm 36.42 m 03642 . 0 m 10 139 . 0 Nm 10 000 , 81 384 m 8 . 1 Nm 10 3 5 4 6 -2 -6 4 -1 3          



(32)

mm

18 or

mm

800 ,

1

100

1 







(33)

Since



36.42mm18mm, therefore try double up the tube for putlog, so that









0.01821m 18.21mm m 10 139 . 0 2 Nm 10 000 , 81 384 m 8 . 1 Nm 10 3 5 4 6 -2 -6 4 -1 3          



(34)

Now, the



18.21mm18mm, therefore OK.

Note:

1. Putlog is made up of two steel tubes, coupled together (side by side). 2. This is a case whereby deflection governs the design.

2.3. Putlog shear

Maximum shear force for loading condition in Figure 5 is given by

wL

V



0 .

5

(35)

Substituting

w



3 .

0 kNm

-118_{and L = 1.8 m into eqn. (35) gives}

kN

7 .

2 m

8 .

1 kNm

0 .

3

5 .

0 

-1





V

(36)

For hollow tube cross-section, area A is given by



2 2



2 4 2

m

10

63 .

5 mm

01 .

563

2 .

40

3 .

48

4















A

(37)

Applying eqn. (17) gives the maximum shear stress

2 -2 -3 2 4 max

3 .

6

10 kNm

3 .

6 Nmm

m

10

63 .

5

2 kN

.7

2

3 









_



(38) 17_{Refer Table 1, BS EN 12811-2:2004.} 18_{See footnote 6.}

(10)

10

Therefore, the maximum shear stress is much lower than the allowable shear stress19_{, i.e.}

123.75 Nmm-2_.

OK TO USE DOUBLE STEEL TUBE (YIELD STRENGTH 275 Nmm-2_{) DIAMETER 48.3}

mm, THICKNESS 4.05 mm AS PUTLOGS

3. Primary Beams (Ledgers)

Ledgers are subjected to concentrated loads from putlogs above them, and supported by the vertical standards (poles).

Span for ledgers = spacing of standards = 2.4 m

The three putlogs exert concentrated loads on each ledger, and two of these putlogs are coupled directly to the standards.

Total load on the middle putlog = Distributed load (2.5 kNm-2_{) over platform area (0.6 m x 1.8}

m) on both sides = 5.4 kN

This load is transferred onto two ledger, and therefore one ledger takes 2.7 kN. The two side putlogs exert point loads of half of this value, as shown in Figure 6.

Figure 6. Ledger.

3.1. Ledger size

Maximum moment

kNm

62 .

1 m

4 .

2 m

1.2 m

1.2 kN

7 .

2 .

.

_



_



L

b

a

F

M

(39)

Using allowable bending strength



_max as 247.5 Nmm-2_{, the required section modulus S}

based on applied loads

3 2 -6

mm

5 .

545 ,

6 Nmm

247.5 Nmm

10

62 .

1 





L

S

(40) 19_{Taken as}

2

max



.

(11)

11

From eqn. (29), the S_G 5,753.88mm3, therefore need to use bigger diameter and thicker tube as ledger, or use metal with higher strength.

Try use steel as shown in Table 6.

Table 6. Ledger.



max



= 319.5 Nmm -2

The section modulus S based on applied loads

3 2 -6

mm

4 .

563 ,

4 Nmm

355 Nmm

10

62 .

1 

_



L

S

(41)

SG is higher than SL, can use tube with properties as shown in Table 6.

3.2. Ledger deflection

Deflection due to concentrated load,













27.

2 .

3

2 .

.

E.I.L

b

a

b

a

b

a

F







(42)

For steel, assume21_{modulus of elasticity, E = 210,000 MPa = 210,000 Nmm}-2_{, shear}

modulus, G = 81,000 MPa = 81,000 Nmm-2_{and density,}_{ρ = 7,850 kgm}-3_{, and} 4 6 4

m

10

139 .

0 mm

1 .

956 ,

138 







I

Substituting into eqn. (41) gives,













m

4 .

2

10

139 .

0 Nm

10

210

27 m

2 .

1

2 m

2 .

1 .

m

2 .

1

3 m

2 .

1

2 m

2 .

1 m.

1.2 m

2 .

1 kN

7 .

2

6 2 -9













_



(43)

mm

2.66 m

10

2.66 

-3





(44)

mm

24 or

mm

400 ,

2

100

1 _

_





(45)

The





2 .

66 mm



24 mm

, therefore OK.

20_{Refer Table A.1 of BS EN 12811-2:2004 (E).} 21_{Refer Table 1, BS EN 12811-2:2004.}

(12)

12

3.3. Ledger shear

Maximum shear force for loading conditions in Table 6 is given by,

L

a

F

V





(46)

kN

35 .

1 m

4 .

2 m

2 .

1 kN

7 .

2 

_



V

(47) From eqn. (37),

A



563 .

01 mm

2



5 .

63 

10

4

m

2 Average shear stress

2 -2 3

Nmm

4 .

2 mm

01 .

563 N

10

1.35 







which is significantly lower than the allowable shear stress22_,

i.e. 159.75 Nmm-2_.

USE STEEL TUBE (YIELD STRENGTH 355 Nmm-2_{) DIAMETER 48.3 mm, THICKNESS}

4.05 mm AS LEDGERS

4. Standards

The standards is designed to take the compression due to the loads exerted by ledgers. The worst loaded is the middle standard, with 1.35 kN of force is applied from each of the two mid-span putlogs on either side, in addition to the 1.35 kN force applied by the two putlogs it supports directly.

Total axial load, P = 2 x 1.35 kN + 2 x 1.35 kN = 5.4 kN

Unbraced length is assumed as, H = 1.5 m (in both transverse (vertical) and longitudinal (vertical) planes)

4.1. Standard size

Assume the allowable stress 159.75 Nmm-2_.

Required area, A 2 2 -

33 .

8 mm

Nmm

159.75 kN

4 .

5 



A

(48) With

I



138 ,

956 .

1 mm

4



0 .

139 

10

6

m

4 and

A



563 .

01 mm

2



5 .

63 

10

4

m

2 Radius of gyration, r is given by

A

I

r 

(49) 22_{Taken as}

2

max



.

(13)

13

mm

71 .

15 mm

01 .

563 mm

1 .

956 ,

138

2 4



r

(50)

Slenderness ratio is given by

48 .

95 mm

71 .

15 mm

500 ,

1 _



r

L

(51) Based on

(14)

14

Table 7, for slenderness ratio of 95.48, and nominal yield strength



_y = 355 Nmm-2_{, the}

permissible stress 83 MPa = 83 Nmm-2_{(the lowest estimate)}

Actual capacity of the tube = 207 mm2_{x 83 Nmm}-2_{= 17.2 kN}

THEREFORE, THE ACTUAL CAPACITY OF THE TUBE = 17.2 KN > TOTAL AXIAL LOAD = 5.4 KN

4.2. Braces

Assume lateral load23_{is 3% of the vertical load, i.e. F}

b = 5.4 kN x 0.03 = 162 N

Assume the allowable stress 159.75 Nmm-2_{, the area required for steel brace is 1 mm}2_.

Very small.

THUS, PROVIDE 25 MM OUTSIDE DIAMETER, 3.2 MM THICKNESS STEEL TUBE AS BRACES, BOTH DIAGONALS.

REFER Figure 7 FOR SUMMARY OF DESIGN.

(15)

15 Table 7. Maximum permissible stress24_.

Figure 7. Summary of design.