ATI - COLOMBO

HNDCE – 2 year
nd

SEMESTER 2- 2020
Structural Design
CE 2212
BY
KVD Perera
Structure
• An assemblage of members arranged in a regular geometrical pattern, interact
through structural connections to support loads and maintain them in equilibrium
without excessive deformation
• Have capacity to carry loads in different ways
• May act in tension, compression, flexure, shear, torsion or combination of these
Structural design
• Determination of deposition and size of load bearing members and other
components in a structure
• Consists of application of theoretical analysis and design methods, validated by
experimental work and practical experience
Aim of structural design
• To provide, with due regard to economy , a structure capable of fulfilling its
intended function and sustaining the design loads for its intended life
• Design should facilitate safe fabrication, transport, handling, erection, future
maintenance, final demolition, recycling and reuse of materials.
Timber Design
BS 5268
 Timber
Timber is widely used for construction because its availability and
economy (doors, windows, roof construction, etc.). But today, it seems
timber is loosing its position due to arise of new technology substitutes
like steel, concrete, fiberglass, masonry, etc.
Any tree could be divided in to 3 sub systems
• Crown sub system
• Trunk sub system
• Root sub system
Functions of tree trunk
• A structural role – supports the crown of leaves, food production organs of the plant
• A conducting role – moves water and essential nutrients for food production from the
roots of the tree to the leaves
• A storage role – stores food
These functions are performed by different types of cells within the tree trunk
Properties of timber
• Fire resistant – high resistance to fire
• Permeability - low water permeability
• Workability - easy to work
• Durability
• Defects –free from defects

• Varies in different directions

• Depends on age, locality, growth related aspects
• Grain direction – direction, majority of the cells are aligned
Moisture in timber
A living tree contains a large amount of water (40-200%)
Importance
• Weight of timber
• Strength properties
• Tendency to shrink
• Susceptibility to attack by fungi and insects
• Ability to treat
2types
• Free moisture – moisture within the cell cavities
• Bound moisture – moisture within cell walls
Fibre saturation point (fsp)
When timber dries the free moisture is lost first and there comes a
hypothetical point when all the moisture is bound moisture. Occurs at a
moisture content of 25% to 30%.

• Timber is hygroscopic (Moisture content is varies with moisture

content of the environment )
Equilibrium moisture content (emc)
The moisture content which it is in equilibrium with environment. Occurs
at 8%-20% condition
Cross section of a tree
2 major parts
• Sap wood Both can be used for construction purpose after treated to acceptable
standards if it is not in acceptable condition
• Heart wood
Sap wood consist of living cells and food convey cells. So they may have
moisture (so liable to attack by insects and fungi) and less durable.
Heart wood consists of dead tissues and hence more dry and very
strong.
Classification of trees
• Outward growing trees Hard woods-broad leaves like Jak, Halmilla
Growth is visible by
means of leaves, Soft woods- conical shape trees
branches

• Inward growing trees

Growth is not visible. It occurs internally-
palmyra, coconut, bamboo
Strength determination of timber
• Density (heavy timber lasts long than light timber)
• Moisture content- shrinkage and expansion are associated with high
moisture content and also insects can easily attack.
• Defects (timber has many defects which could not be controlled by
man since it is a natural product)
• Grain pattern
Conversion of timber
This is preparing the felled tree to be used for constructional purposes. The tree in
Green state is not suitable & so its moisture content has to be brought down to a
suitable value
Green state moisture content-40%-200% (depending on the type of tree)

Seasoning
This is the process by which moisture is brought down to an acceptable limit. By
seasoning, moisture content is brought down to 10%-20%.
methods: Natural and artificial
Air seasoning (timber stacked in open sided sheds in such a way as to promote
drying without artificial assistant, Proper space should be maintained, Cheap)
Kiln seasoning (this method employs a heated, ventilated & humidified oven
controlled condition of drying should maintain, Expensive)
Defects in timber
• Reduce the strength of timber
• Infected by fungi and termites

• Natural defects- checks, shakes, splits

• Chemical defects-when contact with metals
• Seasoning defects –excessive , uneven drying, exposure to wind and
rain, poor seasoning, bad spacing during seasoning.
(end splitting, wash boarding, honey combing, bowing, twisting..)
Preservation of timber
• Moisture resisting coatings
• Use on fresh surface a lead based fiber paint and then an undercoating & 2
finishing coats of a glass paint
• Impregnation treatment
• Some synthetic resins are applied on timber. Expensive method, use only in
special situations
• Chemical preservatives
• The chemicals used are toxic to organisms which attack wood. Widely used in
Sri Lanka.
• Use proper seasoning method
Methods of applying preservatives
• Brushing or spraying-preservatives should be flooded over the timber,
particularly over the joints and cracks
• Immersion- submerge on a bath of preservative liquid
• Pressure – timber is placed in a closed cylinder and preservative fluid
is forced in to wood by high pressure
• Deluging-timber is fed through a tunnel conveyor as organic solvent
are sprayed on it
Uses of timber
• Marine works (structures on sea)
• Heavy construction works (bridges, piles)
• Medium-light construction works (roof trusses, partitions, floors & walls, falsework)
• Finishing works (flooring, skirting, door, window)

• Renewable resource
• Can erect entire buildings, towers, bridges, various roof structures
• Easy to work and joints can be fabricated with simple tools
• Has high strength to weight ratio
• Has reasonable strength in both tension and compression
• Durable in some conditions
• Attractive and decorative
Growth characteristics of a tree
• Slope of grain- can occur as a result of the method of conversion or issue of
the manner of growth
• Knots – reduce the strength
• Wane – caused by cutting from near the perimeter of the tree
• Fissures – these are separations between or across growth rings
• Duration - load duration of members
• Size and shape – bending strength is less for larger sections

Grading of timber
• This is a form of quality control
• Should be done before it put into use
• Growth characteristics and defects can affect
Clear timber – timber with no defects
Strength grouping of timber (page 16 -code)
By specifying a strength class, the supplier is able to supply a grade on
the basis of cost, availability, durability, apply preservatives treatment
and painting, workability and aesthetic considerations
Effect of moisture content
• Service classes (Page 10- code)

Service class Description

1 Timber use inside a heated

building. MC is below 12%
Dry exposure condition
We can neglect K2 value
2 Timber used in a covered
building, MC is below 20%

3 Exposed timber, MC exceeds 20%

Wet exposure condition – K2
only
Modification factors (MF)
• MF K1 and K2 refers to the service exposure condition. Environmental
conditions or service conditions are affected to the moisture content
variations in timber. High moisture content will reduce the strength.
• 2 exposure conditions
• Dry exposure conditions- air temperature and humidity result in solid
timber attaining an equilibrium moisture content less than 18%
• Wet exposure condition - air temperature and humidity result in solid
timber attaining an equilibrium moisture content more than 18%
• Then ; in dry exposure condition , use K1
• wet exposure condition K2
Page 28 –code
MF K2- wet exposure condition
Duration of load –K3
(code- page 28)
• Stresses are listed for long term durations. For other durations, the
given stresses are multiplied by the modification factor K3.
Load sharing system (K8)
(Page 29 -code)
• If 4 or more members spaced at a distance less than 610 mm , such as
joists in floors or compression members in stud walls act together to
support a common load, the grade stress must be multiplied by K8
equal to 1.1. 4 or more members =
K8 = 1.1
Design of axially loaded members
Tension members
Applications
• Bridges, roof trusses, towers, ceiling ties…,etc
Types of sections
• Circular cross sections
• Rectangular sections
Tensile stress
• Applied tensile stress σ t,a,par must not exceed the permissible tensile
stress σ t, adm, par
• Permissible tensile stress = grade tensile stress x relevant modification
factors
σt,a,par < σ t,g,par x K factors
Applied tensile stress = Tensile force/Area

• K2 = modification factor for appropriate service class –wet exposure

condition
• K3 = modification factor corresponding to duration of loads
• K8 = modification factor for load sharing
• K14 = modification factor for width
Width factor K 14
(code page -39)
Members subject to axial Tension and bending (page 40-code)
• If a member is subjected to tension and bending, it should be so
proportioned that the following expression is satisfied

σ m, a, par σ t,a par

• + ≤ 1.0
σ m, adm, par σ t, adm, par
Example 01
Timber beam section 50 mm x 100 mm, belongs to D40 is to be used
as a tension member in a roof truss. Member is subjected to a dead
load of 10 kN (tension), imposed load of 5.0 kN (tension) and wind
load of 7.0 kN (tension / compression). Service class 3.
a) Determine design axial forces in the member
b) If truss is subjected to wet exposure conditions determine whether
the proposed size is satisfactory. It is assumed that a central bolt of
20 mm diameter is adequate as the end connection
a) Design axial force
Long term load = DL only = 10 kN
Medium term load = DL + IL = 10 + 5 = 15 kN
Short term load = DL + IL + WL = 10+5+7 = 22 kN
DL+IL-WL = 10+5-7 = 8 kN 50 mm
b) 100 mm

For long term duration

Applied tensile stress = 10 x 1000 N/ (100x50-50x20) mm2
Stress = load/effective area
= 2.5 N/mm2
Medium term = 3.75 N/mm2
Short term = 5.5 N/mm2
Permissible tensile stress = grade tensile stress x K2 x K3 x K8 x K14
= 7.5 x 0.8 x 1.0 x 1.0 x 1.17
= 7.02 N/mm2
Applied tensile stress ˂ permissible tensile stress
2.5 7.02
Member is satisfied for long term duration

Medium = 8.78 N/mm2

Short term = 10.53 N/mm2
Compression members
Applications
• Bridges, roof trusses, towers, props in formwork…,etc
Types of sections
• Circular cross sections
• Rectangular sections

Maximum applied compressive stress σ c,a par must not exceed the permissible compressive stress
σ c, adm,par

Permissible compressive stress = grade compressive stress x relevant modification factors

• σc,a,par < σ c,g,par x K factors
• K2
• K3
• K8
• K12 = buckling factor σ c,a par = compressive force/ effective cross sectional area
Buckling factor (K12)- code -page 36
This is the sudden change in shape (deformation) of a structural
component under load
b
Factor depends on
• Slenderness ratio h

• Ratio of E/σc,par Z = bh2/6

Section modulus (Z)

E = minimum modulus of elasticity

σc,par = grade compression stress modified for moisture content and
duration of load only
E is the tendency of an object to deform along an axis
when opposing forces are applied along that axis
 Slenderness ratio (code -page 34)
• Slenderness ratio λ = effective length/radius of gyration = Le/i
• Ratio may differ for two principle axes of the cross section. Larger
value of 2 ratios is critical hence it indicates a greater instability and
tendency to buckle under a lower axial load.

The slenderness ratio should not exceed 180 for:

a) any compression member carrying dead and imposed loads other than loads resulting from wind;
b) any compression member, however loaded, which by its deformation will adversely affect the stress in
another member carrying d

The slenderness ratio should not exceed 250 for:

c) any member normally subject to tension or combined tension and bending arising from dead and imposed
loads, but subject to a reversal of axial stress solely from the effect of wind;
d) any compression member carrying self weight and wind loads only (e.g. wind bracing).
Effective length (Le) (page 34-code)

1 2
1)
Le =0.7 L

L Le
2) Le =0.85 L
Radius of gyration – (i)
Radius of gyration is used to describe the distribution of cross sectional
area in a column around its centroidal axis

𝑠𝑒𝑐𝑜𝑛𝑑 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑎𝑟𝑒𝑎

• Radius of gyration =
𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
𝐼
• i= b
𝐴 Ix =bh3/12

h Iy =hb3/12
Members subject to axial compression and bending (page 35-code)

𝜋 2𝐸
𝜎𝑒 = 2
Stress = load/area 𝐿𝑒
𝑖
Example 02
A timber section 50 mm x 100 mm belonging to D50 is to be used as a
continuous compression chord in a roof truss. It is subjected to a long
term axial load of 20 kN and a bending moment of 1.0 kNm under wet
exposure conditions. The distance between effective lateral restraints is
600 mm and that between node points is 1.2 m. Check whether the
proposed section is satisfactory. Service class is 3
Answer
Applied compressive stress = load/effective area
= 20 x 1000 N/50x100 mm2 = 4 N/mm2
Permissible compressive stress = grade comp. stre. X K2 x K3 x K8 x K12
= 15.2 x 0.6 x 1.0 x 1.0 x K12

Assume Le= 1.0L

Radius of gyration ix = 28.87 mm
iy = 14.43 mm
λx = Le/ix = (1.0 x 1200)/ 28.87 = 41.57
λy = Le/iy = (1.0 x 1200)/ 14.43 = 83.16
Ratio E/σc,//
• = 12600 x K2/15.2 x K2 = (12600 x 0.8)/ (15.2 x 0.6) = 1105.26

• K12 = 0.556
Permissible compressive stress = grade comp. stre. X K2 x K3 x K8 x K12
= 15.2 x 0.6 x 1.0 x 1.0 x 0.54 = 4.92 N/mm2
• σc,a,par < σ c,adm,par

• 4 N/mm2
Selected section is satisfied for long term duration without bending
effect
 • Permissible bending stress = grade bending stress x K2x K3x K6 x K7 x K8
300 0.11
= 16 x 0.8 x 1.0 x 1.0 x x 1.0 = 14.44 N/mm2
100
Applied bending stress = bending moment/section modulus
= (1.0 x 1000 x 1000)/(50 x 1002/6) = 12 N/mm2
𝜋2 𝐸 𝜋2 𝑥 12600𝑥𝐾2
𝜎𝑒 = 𝐿𝑒 2
= = 14.38 N/mm2
83.16 2
𝑖

1.87 > 1.0

Section is not
satisfied for both
bending and
compression
Example 01 …cont…
C) can the member withstand any bending for all load durations?
Determine its magnitude
Answer
• Permissible bending stress = grade bending stress x relevant
modification factors
• σm,a,par < σm,g,par x K2 x K3 x K6 x K7 x K8 K6 =not defined
K8 = not defined
For long term duration
300 0.11
Permissible bending stress = 12.5 x 0.8 x 1.0 x = 11.25N/mm2
100
 σ m, a, par σ t,a par
• + ≤ 1.0
σ m, adm, par σ t, adm, par

σ m, a, par 2.5
• + = 1.0
11.28 7.02

• σ m, a, par = 7.26 N/mm2

• σm,a,par < σm,adm,par

• Short term = applied bending stress = 8.09 N/mm2

• Permissible = 16.93 N/mm2
• Medium term –applied = 8.08 N/mm2
• Permissible = 14.11 N/mm2
Design of flexural members
• A member of a structure that is subjected to both tension and compression within its
depth can be called as a flexural member.
• Beams are considered as flexural members
Applications
• Purlins, rafters in roof
• Supporting members

Failure modes
• Bending
• Shear
• Bearing To check all design criteria, It is needed to consider
bending, shear, bearing, deflection and buckling
• Deflection
• Buckling
Bending
• Maximum applied bending stress σm,a,par must not exceed permissible
bending stress σm,adm,par
• Permissible bending stress = grade bending stress x relevant
modification factors
• σm,a,par < σm,g,par x K factors Bending stress = bending moment/section modulus
= M/Z

• K2 = wet exposure condition

• K3 = duration of load
• K6 = form factor (depends on shape of the timber)
• K7 = depth factor
• K8 = load sharing
 Form factor (K6) – code-page 31
• This depends on the shape of the timber
K6 = 1.18 for solid circular sections; and
K6 = 1.41 for solid square sections loaded on a diagonal

Depth factor (K7) – code page 31

K7 = 1.17 for solid timber beams having h ≤ 72 mm;
K7 = (300/h)0.11 for solid and glued laminated beams having 72 mm ≤ 300 mm; b
ℎ2 +92300
K7 = 0.81 for solid and glued laminated beams having h > 300 mm
ℎ2 +56800

h
Shear
• Maximum applied shear stress τ a, par must not exceed the permissible
shear stress τ adm, par
• Permissible shear stress = grade shear stress x relevant modification
factors
• τ a, par < τ g, par X K factors Load

• K2
• K3
• K5 = notched factor
• K8
Applied shear stress = shear
force/cross sectional area of shear
plane
• Notched factor (K5) – code –page 30
• Square cornered notches at the ends of a flexural member causes a stress concentration. Shear stress is
calculated by using effective depth (he).
Bearing
• Maximum applied bearing stress σ a, perp must not exceed the
permissible bearing stress σ c, adm,perp
• Permissible bearing stress = grade bearing stress x relevant
modification factors
• σc,a,perp < σ,g,perp x K factors
• K2
• K3
• K4 = bearing factor Bearing stress = bearing force/cross sectional area

• K8
 Bearing factor (K4) – code page – 29,30
This depends on the length and position of bearing.

Note ;
According to figure 1,
K4 = 1.0 for bearing of any length at the ends of a member
K4 = 1.0 for bearings 150 mm or more in length at any position
K4 = Use Table 18- for bearings of less than 150 mm length located 75 mm or more the end of a member as shown in figure
Buckling (code – page 32)
• The depth to breadth ratio of solid and laminated beams of rectangular
section should be checked to ensure that there is no risk of buckling
under design load. Depth/breadth = d/b

d
Deflection (code –page 32)
• This is limited to 0.003 of the span
• Deflection can be caused due to combine effect of bending and shear.
L

• Δmax = 0.003L
Example 03
Timber beam section 50 mm x 100 mm, belongs to D60 is to be used to span
a distance of 3.0 m and carry a uniformly distributed load of 1.2 kN/m
(including self weight). Beam is subjective to wet exposure condition. Assume
the load duration as medium term.
a) Determine design bending moment, shear force and bearing force if the
ends of the beams are simply supported with no lateral support
b) check whether the maximum bending stress is within the permissible
stress
c) If maximum allowable deflection is 8.0 mm, check whether the total
deflection is satisfactory
d) A notch of 50 mm is cut on the underside of the beam, check whether the
maximum shear stress is within the permitted level
e) Check whether the bearing width (another beam) of 50 mm is adequate
to withstand the bearing force .
f) Check whether all design criteria are satisfied
Answer
a)
• Bending moment = WL2/8 = (1.2 x 3.02)/8 = 1.35 kNm
• Shear force = WL/2 = (1.2 x 3.0)/2 = 1.8 kN
• Bearing force = WL/2 = 1.8 kN
1.35 𝑥 1000 𝑥 1000 2
b) Applied Bending stress = M/Z = 100 2 = 16.2 𝑁/𝑚𝑚
50 𝑥
6
Permissible bending stress = grade bending stress x K2 x K3 x K6 X K7 x K8
300 0.11
= 18 x 0.8 x 1.25 x 100
𝑥 1.0 = 20.31 𝑁/𝑚𝑚2
σm,a,par < σm,adm,par

Applied maximum bending stress is within the permissible bending stress.

• C) total deflection = bending deflection + shear deflection
5𝑤𝐿4 𝐾 𝑤𝐿2
• = +
384 𝐸𝐼 8 𝐺𝐴
• Rectangular section K = 1.2
• G = E/16 = shear modulus

• = 5 𝑥 1.2𝑥 34 𝑥 1012 𝑥 1000

+
1.2 𝑥 1.2 𝑥 32 𝑥 106 𝑥1000
1003 15600 𝑥 0.8
1000 𝑥 384 𝑥15600 𝑥0.8 𝑥 50𝑥 8𝑥 𝑥 50𝑥 100𝑥1000
12 16
 d) Applied shear stress = (1.8 x 1000)/ 50 x 50 =
0.72 N/mm2

Permissible shear stress = grade shear stress x K2 x K3 X K5 x K8

50 = 2.4 x 0.9 x 1.25 x 0.5 = 1.35 N/mm2

100 K8= not

applicable

K5 =
50/100
= 0.5
τ a, par < τ adm, par
Maximum shear stress is within the permissible
shear stress
• e)

50 mm

50 mm
• Applied bearing stress = bearing force/ area = 1.8 x 1000/(50x 50 )
• = 0.72 N/mm2
• Permissible bearing stress = grade bearing stress x relevant
modification factors
• σc,a,perp < σ,g,perp x K2 x K3 x K4 x K8
• Permissible bearing stress = σ,g,perp x K2 x K3 x K4 x K8
• = 5.2 x 0.6 x 1.25 x 1.0 = 3.9 N/mm2
K8 = not applicable
σc,a,perp < σ,adm,perp
Applied bearing stress is adequate with permissible bearing stress value
f)
Bending stress is within the permissible limit
Deflection is not in permitted value
Shear stress is within the permissible limit
Bearing stress is within the permissible limit
Buckling
Depth/breadth = 100/50 = 2
Maximum value (Table 19 ) = 2
Example 01 …cont…
• C) can the member withstand any bending for all load durations?
Determine its magnitude
Long term duration
σ m, a, par σ t,a par
+ ≤ 1.0
σ m, adm, par σ t, adm, par

• σm,a,par < σm,g,par x K factors K7 = (300/100) 0.11

= 1.128

• σm,adm,par = σm,g,par x K2 x K3 x K6 x K7 x K8
• = 12.5 x 0.8 x 1.0 x 1.0 x 1.128 x 1.0 = 11.28 N /mm2
 σ m, a, par σ t,a par
+ = 1.0
σ m, adm, par σ t, adm, par
σ m, a, par 2.5
11.28
+ 7.02
= 1.0

σ m, a, par = 7.26 N/mm2

• σm,a,par < σm,adm,par
• For long term duration, bending is satisfied
Medium term
Admissible value = 14.10 N/mm2
Applied = 8.07 N/mm2
Short term
Admissible value = 16.92 N/mm2
Applied = 8.1 N/mm2
Example 04
A timber section of 3 m length spans over an opening and supports on
220 mm thick brick wall at the ends which exerts 3 kN/m UDL. Check
whether D 70 , 100x 250 mm teak timber is satisfied in wet exposure
condition for medium term duration.
Check for bending
Maximum bending moment = wL2/8 = (3x 32)/8 = 3.375 kNm
Applied bending stress = max bending moment/ section modulus
= 3.375/(100 x 2502 /6) = 3.24 N/mm2
σm,adm,par = σm,g,par x K2 x K3 x K6 x K7 x K8
= 23 x 0.8 x 1.25 x 1.0 x 1.02 x 1.0
=23.46 N/mm2

σm,a,par < σm,adm,par K7 = (300/250) 0.11

bending is satisfied = 1.02
Check for shear
Shear force = wL/2 = (3 x 3)/2 = 4.5 kN
Applied shear stress = 4.5 x 10 3/ 100x 250 = 0.18 N/mm2
τ a, par < τ g, par X K factors
Permissible shear stress = τ g, par X K2 xK3 xK5 x K8
= 2.6 x 0.9 x 1.25 x1.0 x 1.0 = 2.93 N/mm2
τ a, par < τ adm, par
Shear is satisfied
Check for bearing
Bearing force = wL/2 = 4.5 kN
Applied bearing stress = 4.5x 1000/ (100 x 220) = 0.2 N/mm2
Permissible bearing stress = σ,g,perp x K2 x K3 x K4 x K8
= 6 x 0.6 x 1.25 x 1.0
= 4.5 N/mm2 K8 = not use
σc,a,perp < σ,adm,perp
Bearing is satisfied

Check for buckling

d/b = 250/100 = 2.5
Maximum value 3 (Table 19)
Buckling is satisfied
Check for deflection
5𝑤𝐿4 𝐾 𝑤𝐿2
Total deflection = +
384 𝐸𝐼 8 𝐺𝐴

= 1.86 mm
Maximum value = 0.003 x L = 0.003 x 3000 = 9 mm
Deflection is satisfied.
Example 05
Timber beam section 50 mm x 100 mm, belongs to D40 is to be used to
span a distance of 2.0 m and carry a uniformly distributed load of 1.2
kN/m (including self weight). Beam is subjective to dry exposure
condition. Check the adequacy of the member for long term duration.
Bearing width of an another beam is 50 mm.
Check for bending
Maximum bending moment = wL2/8 = (1.2x 22)/8 = 0.6 kNm
Applied bending stress = max bending moment/ section modulus
= 0.6 x 1000x1000/(50 x 1002 /6) = 7.2 N/mm2
σm,adm,par = σm,g,par x K2 x K3 x K6 x K7 x K8 0.11
K7 = (300/100)
= 12.5x 1.0 x 1.0 x 1.0 x 1.128 x 1.0
= 1.128
=14.1 N/mm2
K2 = not use for dry exposure condition
σm,a,par < σm,adm,par
bending is satisfied
Check for shear
Shear force = wL/2 = (1.2 x 2)/2 = 1.2 kN
Applied shear stress = 1.2 x 10 3/ 100x 50 = 0.24 N/mm2
τ a, par < τ g, par X K factors
Permissible shear stress = τ g, par X K2 xK3 xK5 x K8
= 2.0 x 1.0 x1.0 = 2.0 N/mm2
τ a, par < τ adm, par K2 = not use for dry exposure condition
K8 = not use
Shear is satisfied
Check for bearing
Bearing force = wL/2 = 1.2 kN
Applied bearing stress = 1.2x 1000/ (50 x 50) = 0.48 N/mm2
Permissible bearing stress = σ,g,perp x K2 x K3 x K4 x K8
= 3.9 x 1.0 x 1.0
= 3.9 N/mm2 K8 = not use
σc,a,perp < σ,adm,perp K2 = not use for dry exposure condition

Bearing is satisfied

Check for buckling (Assume ends held in position condition)

d/b = 100/50 = 2
Maximum value 3 (Table 19)
Buckling is satisfied
Check for deflection
5𝑤𝐿4 𝐾 𝑤𝐿2
Total deflection = +
384 𝐸𝐼 8 𝐺𝐴

= 8.3 mm
Maximum value = 0.003 x L = 0.003 x 2000 = 6 mm
Deflection is not satisfied.