Download - Timber Lesson 2
-
8/10/2019 Timber Lesson 2
1/38
DESIGN IN TIMBERTO MS 544 PART2
-
8/10/2019 Timber Lesson 2
2/38
COURSE OUTCOME
At the end of the lecture student will be able to;
Understand design philosophy of permissiblestress design
Design flexural member
-
8/10/2019 Timber Lesson 2
3/38
INTRODUCTION
Timber can be used in a range ofstructural application (like piers),
heavy civil work (like bridge and piles)
or domestic housing (like roofs andfloors)
Timber structures are loaded with
different type of loadings ie flexural,compression, tension shear etc.
So it is important to know the
allowable and permissible stress of
-
8/10/2019 Timber Lesson 2
4/38
Different materials have differentstrength properties.
Permissible stress are the properties
of material under bending,compression, tension etc.
These stresses together with the
loading criteria will form the basis ofdesign for the various timber
components.
-
8/10/2019 Timber Lesson 2
5/38
Design of timber members
In accordance with
BS5268: The structural use of timber
or
MS544 Part2: Code of Practice for
Structural use of Timbers: Permissible
stress design of solid timber
-
8/10/2019 Timber Lesson 2
6/38
Design philosophy
Permissible stress design derived on a statistical basis and
deformations are limited.
Design based on the allowable and
permissible stress of the materials Elastic theory is used to analyse structures
under various loading conditions to give the
worst design case. Then timber sections are chosen so that
permissible stresses are not exceeded atany point of the structure.
-
8/10/2019 Timber Lesson 2
7/38
MS544 Part 2 and BS5268 is based on
permissible stress design (or elastic
design) rather than limit state design (as
in BS5950).
This means that in practice that a partial
safety factor is applied only to materialproperties not to the loading.
-
8/10/2019 Timber Lesson 2
8/38
Working stress design
Permissible stress design
Allowable stress design
Elastic method design
Has been used by designers and
engineers in timber construction and
this MS544 and BS5268 adopted this
design
-
8/10/2019 Timber Lesson 2
9/38
Working stress design
The adequacy of a structure ischecked by calculating the working
stress to maximum expected loads
and comparing them with thepermissible stresses.
The permissible stress is equal to the
failure stress design methods tosuccessful structures at that time
-
8/10/2019 Timber Lesson 2
10/38
The safety factors for the newmaterials were estimated in
comparison with those for traditional
materials by taking into account thenature for the new material and its
uncertainty or variability
Elastic method of design has formedthe basis of structural codes and
standards for most of the century.
-
8/10/2019 Timber Lesson 2
11/38
Structural timber design may be
based on either The grade stress for the individual
species for dry exposure or wet
condition given in Table 1 and 2 pg 5
16. The grade stresses for the strength
group SG for dry exposure condition
given in Table 4 pg 18
-
8/10/2019 Timber Lesson 2
12/38
Timber grade stresses
1. Since timber which is naturally occurringand has varying range of properties, its
grade has to be classified accordingly. Its
properties are affected by conditions of
growth(temperature, wet or dry season,
wind etc) and therefore its strength varies
2. Strength is also determined by the
process of various strength reducing
characteristics such as knots, slope of
grain, fissures and wane
-
8/10/2019 Timber Lesson 2
13/38
3. Strength Group, SG is defined as the
classification of timber based on
particular value of grade stresses Timber having similar strength and
stiffness properties have been group
together for simplicity in design procedure(Table 4 pg 18)
-
8/10/2019 Timber Lesson 2
14/38
Definition of stresses
a) Basic stress
The stress which can safely be
permanently sustained by timber
containing no strength reducing
characteristic
Includeduration of loading
- size and shape of actual
member
- factors of safety
- variabilit of stren th
-
8/10/2019 Timber Lesson 2
15/38
b) Grade stress
The stress which can safely be
permanently sustained by timber of aparticular grade.
c)Green stress
A stress applicable to timber having amoisture content exceeding 19%
d) Dry stress
A stress applicable to timber having amoisture content not exceeding 19%
-
8/10/2019 Timber Lesson 2
16/38
Derivation of Permissible
Stresses
Strength
properties ofsmall clear
specimens
Grade stress(individual species)
Characteristic stress
Basic stress
Permissible stress
Grade stress
(strength group)
-
8/10/2019 Timber Lesson 2
17/38
Derivation of Basic Stresses
Test results (for at least 30 samples)for each species were tabulated andthe average strength value (stress) is
calculated using the followingequation:
X =
X= strength value
N = number of samples
n
X
n
33.2X X
-
8/10/2019 Timber Lesson 2
18/38
-
8/10/2019 Timber Lesson 2
19/38
Basic stress
Basic stress: = X2.33
F.S
Where F.S is factor of safety
k
-
8/10/2019 Timber Lesson 2
20/38
Safety factor
A safety factor is applied and it is usually assumed thatthis factor will cover such items as accidentaloverloading, assumptions made during design anddesign accuracies together with errors in workmanship,etc.
Property Reduction factor Formula
Bending and shear 2.5 X2.33
2.5
Compression
parallel to grain
1.5 X2.33
1.5
Compression
perpendicular tograin
1.3 X2.33
1.3
Mean modulus of
elasticity
1.0 X
Minimum modulus
of elasticity
1.0 X2.33
-
8/10/2019 Timber Lesson 2
21/38
Grade stress
The stress can be safely permanentlysustained by timber at a particular grade
They are derived from individual species
and are governed by the effect of visiblegross features of defect such as knots,
sloping of grains, fissures, etc.
Reduction strength expressed in terms of
strength ratio i.e of strength of piece oftimber with defects to the strength of the
same piece without defect.
-
8/10/2019 Timber Lesson 2
22/38
Grade stress
Grade stress = Basic stress x reductionfactor
Reduction factor for grade stress taking into account of varying
amounts of defects based on basic stress and this gives three stress
grades of timbers namely Select, Standard and Common
Property Select
%
Standard
%
Common
%
Bending, tension and
compression parallel tograin
80 63 50
Compression
perpendicular to grain
85 80 75
Shear 72 56 45
Modulus of elasticity Same as the basic value for all grades
Table 2.0 Strength ratio
-
8/10/2019 Timber Lesson 2
23/38
Permissible stress
The stress which can be safely besustained by a structural component
under the particular condition of service
and loading Permissible stress is the final stage at
which all allowances are made for the
particular condition of services and
loadingPermissible stress = grade stress x modification factor
= grade stress x K1xK2xK3xK4xK5xK6
-
8/10/2019 Timber Lesson 2
24/38
Modification factors
-
8/10/2019 Timber Lesson 2
25/38
DURATION OF LOADING
Modification factor K1Table 5 MS 544:2001 Part 1
Duration of loading Value of K1
Long term (dead + permanent imposed)
Medium term (dead + temporary imposed)
Short term (dead + imposed + wind)
Very short term (dead + imposed + wind)
1.00
1.25
1.50
1.75
-
8/10/2019 Timber Lesson 2
26/38
LOAD SHARING FACTOR (Clause 10 of the MS544:2001 Part 2)
Load sharing means a system whereby a number of members act together tosupport a common load.
For load sharing , modification factor K2is 1.1 and the mean value of modulus ofelasticity may be used.
(For non-load sharing, there is no increase in the grade stress and the minimum Eis used).
For load sharing system to be applicable the following must be satisfied:
I. There must be at least four or more members
II. The spacing of members must not be more than 610 mm apart with adequate
provision of lateral distribution of loadIII. The stresses due to dead load or permanent load are not more than 60% of
stresses due to the total design load.
Members include rafters, joists, trusses or wall studs with adequate provision forlateral distribution of loads by means of purlins, binders, boarding etc.
Emeanshould be used to calculate deflections and displacements.
-
8/10/2019 Timber Lesson 2
27/38
LENGTH AND POSITION OF BEARING
For bearing < 150mm in length located 75mm or more from theend of member, grade stress should be multiplied by
modification factor K3.
75mm Bearing less than
or more 150mm or more
Length of bearing (mm) 10 15 25 40 50 75 100
-
8/10/2019 Timber Lesson 2
28/38
NOTCHING FACTOR
Beam with notch on the top edge Beam with notch on the underside
For a he
K4=
For a >he
K4= 1.0
K4= he/h
Beam with notch on the top edge Beam with notch in the underside
2
)(
e
e
h
ahahh e
-
8/10/2019 Timber Lesson 2
29/38
FORM FACTOR
K5= 1.8 for solid circular sectionK5= 1.41 for solid square sections loaded on a diagonal
DEPTH FACTOR
For depth of beams > 300mm, the grade bending stressesshould be multiplied by modification factor K6
For solid and glue laminated beams,
K6= 0.81
-
8/10/2019 Timber Lesson 2
30/38
Design of bending members
e.g, girders, stringers, bearers, purlinsand joist
In design, all follow the same design
principles as member subjected tobending forces
Beam may be supported as freely
supported, supported by intermediatesupports, fixed end with other end free
(cantilever) etc.
-
8/10/2019 Timber Lesson 2
31/38
Design considerations
The main design considerations for flexural members are:1. Bending stress and prevention of lateral buckling
2. Deflection
3. Shear stress
4. Bearing stress.
The cross-sectional properties of all flexural members have to satisfyelastic strength and service load requirements. In general, bendingis the most critical criterion for medium-span beams, deflection forlong span beams and shear for heavily loaded short-span beams. Inpractice, design checks are carried out for all criteria listed above.
The permissible stress value is calculated as the product of the
grade stress and the appropriate modification factors for particularservice and loading conditions, and is usually compared with theapplied stress in a member or part of a component in structuraldesign calculations. In general:
Permissible stress ( = grade stress K-factors ) appliedstress
-
8/10/2019 Timber Lesson 2
32/38
Bending stress and prevention of
lateral buckling
The design of timber beams in flexure requires the application of the elastictheory of bending as express by:
=My
I
The termI/yis referred to as section modulus and is denoted by
Z. The applied bending stress about the major (x-x) axis ofthe beam (say) (see Fig. 4.1), is calculated from:
My M
Ixx Zxx=
h
y
x x
y
b
Where:
m,adm,//=applied bending stress (in N/mm2)
M= maximum bending moment (in Nmm)
Zxx= section modulus about its major (x-x) axis
(in mm3). For rectangular sections
-
8/10/2019 Timber Lesson 2
33/38
Design considerations
Includes all as mention above
Additions:
Depth factor (Sec 11.6 pg 23)
Effective span Le
Clause 11.3 pg 22 recommends that the span of flexural
members should be taken as the distance between thecentres of bearing
Beam or joist
Clear span
Effective span
Span to centres of actual bearings
-
8/10/2019 Timber Lesson 2
34/38
Depth factor, K6
The grade bending stress given in Tables 1,2 and4 of MS544 : Part 2 apply to beams having adepth, h, of 300 mm (Clause 11.6). For otherdepths of beams, the grade bending stressshould be multiplied by the depth modification
factor, K6 where:
for h 72 mm, K6 = 1.17
for 72 < h< 300 mm, K6
= (300 )0.11
h
for h> 300 mm, K6 = 0.81h2+ 92
300 h2+ 56800
-
8/10/2019 Timber Lesson 2
35/38
Depth to breath ratio
I xx/ I yy 1 4 9 16 25 36 49
d/b 1 2 3 4 5 6 7
Note:forasimplerectangular beamtheIxx/Iyyratioissimplythesquareof
thed/bratio.
ateral Stability of built in beams
Beamswithlargedepthtothicknessratiosareatriskofbucklingunderbendingforces.BS5268uses
theratioofIxx(2ndmomentofareaofsectionaboutneutralaxis)toI yy(2ndmomentofareaofthesection
perpendiculartotheneutralaxis)toidentifythesupportrequirementssuchthatthereisnoriskofbucking
-
8/10/2019 Timber Lesson 2
36/38
Stiffness and deflection (Sec 11.7 pg 24)
Stiffness is related to deflection. When a member issaid to be stiff, it means that it is able to resist
deflection to a certain extend depending on the
degree of stiffness.
Deflection limits are decided through practical
experience and arbitrarily fixed.
For floors when fully loaded should not exceed
0.003 x span
In purlins, deflection should not exceed 0.005 x
span
Members may be pre-cambered to account for the
deflection under full dead or permanent load and in
this case the deflection under live or intermittent
load should not exceed 0.003 of the span
-
8/10/2019 Timber Lesson 2
37/38
In calculating the deflection, either the meanvalue or the minimum value of modulus ofelasticity (E) is used.
The mean value Emeanis used for roof joist, floorjoist and other systems where transversedistribution of load is achieved and where thestress induced by the dead load or permanentload is not more than 60% of the permissible
stress induced by the full design load.( the latterstatement is included because where thepermanent load is a large proportion of the totalload, will induce creep.
The minimum value of Emin is used for principals,
binders and other components which actsalone.
-
8/10/2019 Timber Lesson 2
38/38
Procedure in beam design
Design steps:
i. Calculate the loads to be applied
ii. Effective span, Le (Clause 11.3 pg 22)
iii. Find m,g,// = bending stress (grade
stresses) parallel to grain from table 1-4
M < MRWhere M: design moment
MR: moment resistance
MR = m,adm,// , // Zxx
Z = section modulus
m,adm,// = permissible bending stress // to grain
difi ti f t ( K1 K8)