ME31B: CHAPTER FOUR

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ME31B CHAPTER FOUR

• DESIGN OF STRUCTURAL MEMBERS

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DESIGN OF MEMBERS IN DIRECT STRESS

• Structural members under direct stress are
  mainly ties, cables, and short columns.

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Solution Concluded
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Design of Short Columns
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Example

• A square concrete column(pier) which is 0.5 m
  high is made of a nominal concrete mix of 124,
  with a permissible direct stress of 5.3 N/mm2.
  What is the required cross-sectional area if the
  column is required to carry an axial load of 300
  kN? 
• Solution
• Area Force/stress 300 x 103 N 56600
  mm2
• 5.3
  N/mm2
• ie. the column should be minimum 240 mm
  square. 

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Note

• Maximum compressive stress (fc) occurs in the
  section where the bending moment is maximum. In
  the design of simple beams, section modulus (Z)
  should be selected such that fc does not
  exceed the allowable value. Allowable working
  stress values can be found in building codes or
  Engineering handbooks.
• For safe bending, fw gt f Mmax/Z
  where fw is the allowable bending stress f is
  the actual stress and Mmax is the maximum bending
  moment.

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b)     Deflection of Beams

• Excessive deflections cause cracking of plaster
  in ceiling and can lead to jamming of doors and
  windows.
• Most building codes limit the amount of
  allowable deflection as a
• proportion of the member's length ie. 1/180 ,
  1/240, or 1/360 of the length.

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4.2.1 Steps in the Design of Simple beams

• a) Calculate loading on the beam
• b) Calculate the bending moment, shear forces,
  etc.
• c) For the maximum bending moment (Mmax),
  provide a suitable section
• Z Mmax/fw , fw is allowable stress
  (from tables).
• d) Check for shear stress, deflection and
  buckling of web if necessary.

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Example

• Consider a floor where beams are spaced at 1200
  mm and have a span of 4000 mm. The beams are
  seasoned cypress with the following properties
  fw 8 N/mm2 , E 8400 N/mm 2 ,
  density 500kg/m3 . Loading on floor and
  including floor is 2.5 kN/m2 . Allowable
  deflection is L/240. Design the beam. Allowable
  shear stress is 0.7 N/mm2

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4.2.2 Bending Moment Caused By Askew Loads

• When the resulting bending moment on a beam is
  not about one of the axis, the moment need be
  resolved into components acting about the main
  axis.
• The stresses are then calculated separately
  relative to each axis and the total stress is
  found by adding the stresses caused by the
  components of the moment.

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Roof Truss Showing Purlin
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Example

• Design a timber purlin, which will span rafters
  2.4 m on centre. The angle of the roof slope is
  30 and the purlin will support a vertical dead
  load of 250 N/m and a wind load of 200 N/m acting
  normal to the roof. The allowable bending
  stress(fw ) for the timber used is 8 N/mm 2 .
  The timber density is 600 kg/m3

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Forces in the Purlin
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Solution

• Assume a purlin cross sectional size of 50 x 125
  mm
• i) Find an estimated self load
• w 0.05 m x 0.125 m x 600 kg/m3 x 9.81
  37 N/m
• Total dead load 250 37 287 N/m
• ii) Find the components of the loads relative to
  the main axes
• wx 200 N/m 287 N/m cos 30 448.5
  N/m
• wy 287 N/m sin 30 143.5 N/m

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Solution Contd.

• iii) Calculate the BM about each axis for a udl.
  The purlin is assumed to be a simple beam
• Mmax w L 2 /8
• Mmax x wx L2 448.5 x 2.42
• 8 8
• 323 x 103 N mm
• Mmax y wy L2 143.5 x 2.42
  
• 8
  8
• 103 x 10 3 N mm

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Solution Contd.

• iv) The actual stress in the timber must be less
  than the allowable stress.
• f Mmax x Mmax y lt fw
• Zx Zy
• v) Try the assumed purlin size of 50 x 125 mm
• Zx b d2 50 x 1252 130 x
  10 3 mm3
• 6 6

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Solution Concluded
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4.2.3 Design Using the Universal Steel Beams
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Example

• A steel beam used as a lintel over a door opening
  is required to span 4.5 m between centres of
  simple supports. The beam will carry a 220 mm
  thick and 3.2 m high brick wall, weighing 20
  kN/m3 . Allowable bending stress is 165 N/mm2.
  Assume allowable shear stress of 100 N/mm2, E is
  2 x 105 N/mm2. Assume self weight of the
  beam as 1.5 kN.

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With Z 221 cm3 and I
2070 cm4 , From SDM, choose a UB 254 x 102
x 22 with Z 225 .4 cm3 and I 2863
cm4Check for shear stress Shear stress
Qmax
D. tFrom Table 4.4.
Q max W /2 32.43 kN Shear stress
32.43 x 103 N 22.01
N/mm2 254 mm x 5.8 mmSince 22.01 N/mm2 lt
the allowable shear stress of 100 N/mm2 ,
ie. Beam is very satisfactory.
USE UB 254 x 102 x 22
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Composite Beams
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4.3 DESIGN OF COLUMNS

• Columns are compression members but the manner in
  which they tend to fail and the amount of load
  which causes failure depend on
• i) The material of which the column is made eg.
  a steel column can carry a greater load than
  timber column of similar cross-sectional size.
• ii) The shape of the cross-section of the
  column. A column having high c/s area compared
  to the height is likely to fail by crushing
  rather than by buckling.

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Columns Contd.

• iii) The end conditions of the column.
• To account for buckling of slender columns, the
  allowable compressive strength is reduced by a
  factor k , which depends on the slenderness
  ratio and the material used.
• Pbw k . cw . A where Pbw is the
  allowable load wrt buckling
• k is the reduction factor which depends on the
  slenderness ratio and
• A is the cross-sectional area of the column.

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4.3.2 Slenderness Ratio

• Slenderness ratio can be defined as
• k L l
• r r 
• Where is slenderness ratio
• k is effective length factor whose value depends
  on how the ends of the column are fixed
• L is the length of the column r is the radius of
  gyration (r I/A and l is the effective
  length of the columns (k. l)

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4.3.3 Types of End Conditions of a Column

• Columns can either be
• (a) fixed in position nor direction (the weakest
  condition)
• (b) fixed in position but not in direction
  (pinned)
• (c) fixed in direction but not in position
• (d) fixed in position and in direction

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4.3.4 Design of Axially Loaded Timber Columns

• Timber columns are designed with the following
  formulae
• k L and Pbw k . cw.
  A
• r
• NB In some building codes, a value of
  slenderness ratio in case of sawn timber is taken
  as l/b instead of l/r
•  

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Example

• Design a timber column which is 3 m long with a
  compressive load of 15 kN. Allowable compressive
  stress ( cw) for the timber is 5.2 N/mm2, kp
  is 1.00

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Note

• Actual load/allowable load
• 15 kN/16.9 kN 0.89
• This ratio is all right and shows that the
  section is economical.
• A ratio of 0.85 to 1.00 is acceptable.

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4.3.5 Design of Axially Loaded Steel Columns

• The allowable loads for steel with respect to
  buckling can be calculated in the same manner as
  for timber.
• The relationship between the slenderness ratio
  and the reduction factor ( k ) is slightly
  different (see Table 4.6).

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Solution Concluded
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4.3.7 Axially Loaded Concrete Columns

• Most building codes allow the use of plain
  concrete (no reinforcement) only in short columns
  ie. where the l/r ratio is lt 15 .
• For l/r between 10 and 15, the allowable
  compressive strength must be reduced.
• The tables of figures relating to l/b in place
  of a true slenderness ratio are only approximate,
  since radii of gyration depend on both b and d
  values in the cross section and must be used with
  caution.

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Example

• A concrete column, with an effective length of 4
  m has a cross section of 300 x 400 mm. Calculate
  the allowable axial load, if a nominal concrete
  mix is 124 is to be used.
• Solution Slenderness ratio, l/b 4000/300
  13.3
• Hence Table 4.7, gives Pcc 3.47 N/mm2 by
  interpolation.
• Pw Pcc . A 3.47 x 300 x 400
• 416.4 kN

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4.3.8 Plain Centrally Reinforced Concrete
Walls

• Walls are designed in the same manner as columns,
  but there are a few differences.
• A wall is different from a column by having a
  length which is more than five times its
  thickness.
• Plain concrete walls should have a minimum
  thickness of 100 mm.

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Plain and Reinforced Concrete Walls Contd.

• Where the load is eccentric, the wall must have
  centrally placed reinforcement of at least 0.2
  of the cross section if the eccentricity gt 0.20.
• This reinforcement may not be included in the
  load carrying capacity of the wall.

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Example

• Determine the maximum allowable load per metre of
  a 120 mm thick wall, with an effective height of
  2.8 m and made from concrete grade C 15 (a)
  when the load is central and (b) when the load
  is eccentric by 20 mm.

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4.4 DESIGN OF ROOF TRUSSES
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Roof Trusses Contd.

• Steel and timber trusses are usually designed
  assuming pin-jointed members. In practice,
  timber trusses are assembled with bolts, nails
• or special connectors and steel trusses are
  bolted, riveted or welded.
• These rigid joint impose secondary stresses which
  are negligible and therefore not used in design.

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4.4.2 Steps in Designing a Truss

• a) Select general layout of truss members and
  truss spacing
• b) Estimate external loads to be applied
  including self weight of truss, purlins and roof
  covering, together with wind loads.
• c) Determine the critical (worst combinations)
  loading. It is usual to consider dead loads
  alone, and then dead and imposed load combined.

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Steps in Designing a Truss Contd.

• d) Analyse framework to find forces in members.
• e) Select material and section to provide in
  each member a stress value which does not exceed
  the permissible value.

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4.4.2.1 Spacing and Layout

• Roof trusses should as far as possible be spaced
  to achieve a minimum of weight and economy of
  materials used in the total roof structure.
• For spans up to 20 m, the spacing of steel
  trusses is likely to be about 4 m, and in case of
  timber, 2m.

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Spacing and Layout Contd.

• Short spans up to 8 m should have pitched timber
  rafters or light weight trusses either pitched or
  flat. Medium spans of 7 to 15 m or 16 m require
  truss frames designed of timber or steel.
• Long spans of over 16 m should if possible be
  broken into small units or the roofs should be
  designed by specialists.

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Pitch or Slope of a Roof
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Estimation of Roof Loads
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Example

• Determine the critical forces for design for each
  of the members in the left hand of the truss
  shown above. Assume the following Trusses are
  spaced 4 m on centres. The roof deck is of
  galvanised sheet weighing 6 kg/m2 . All purlins
  weigh 22kg/m. Openings constitute 20 of the
  wall surface. Calculate the panel loads.

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Solution

• (a) Dead load
• i) Determine the panel area supported by one
  purlin
• Distance between the purlins x spacing of
  trusses 3.2 m x 4 m 12.8 m2
• ii) Calculate the roof deck load supported by
  one purlin
• 12.8 m 2 x 6 kg/m 2 x 9.81 753.4
  N 0.75 kN.
• iii) Weight of each purlin 22 kg/m x 4 m x
  9.81 863.28 N 0.86 kN

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Solution Contd.

• Total dead load to be carried by truss
• 0.75 0.86 kN 1.61 kN
• iv)Estimate total truss weight per panel as 10
  of the total load to be carried 0.16 kN
• v) Total dead load per panel point , P 1.61
  0.16 1.77 kN

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Solution Contd.
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Final Design Diagram
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4.5 FOUNDATIONS AND FOOTINGS DESIGN

• 4.5.1 Introduction Foundation is the part of
  the structure through which the load of the
  structure is transmitted to the ground.
• A combination of footing and foundation
  distributes the load on the bearing surface (the
  soil) and keeps the building level and plumb and
  reduces settling to a minimum.
• Footing and foundation are normally made of
  concrete, no matter the construction material.

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Footing and Foundation Contd.

• Before design of footing and foundation can be
  made, the total load from the building as well as
  soil bearing characteristics should be
  determined.

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Soil Bearing Capacity

• Strength of the soil required to resist the loads
  resting in it.
• This is after the top soil has been removed (See
  Table 5.6, FAO book for values of soil bearing
  capacities). Detailed investigations of the soil
  is not usually required for small scale buildings

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Soil Bearing Capacities
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4.5.3 Foundation Footings

• 4.5.3.1 Description A footing is an
  enlarged base for a foundation designed to
  distribute the building load over a larger area
  of soil and to provide a firm, level surface for
  constructing the foundation wall.

Foundation wall
Footing
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Footings and Foundation Contd.

• The footing is wider than the foundation wall
  because the soil's bearing stress is less than
  that of the material (concrete) of the wall e.g.
  concrete has a strength of about 1000 kN/m2 .
• A 1 3 5 ratio of cement, dry sand, and
  gravel is suggested for footings with 31 litres
  of water per 50 kg sack of cement.

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Note

• Unlike the continuous wall footings, the pier and
  column footings are heavily loaded.
• It is very important to correctly estimate the
  proportion of building load to be carried by each
  pier or column.
• If wall footings are very lightly loaded, it is
  advisable to design any pier or column footings
  required for the building with approximately the
  same load per unit area.
• If any settling occurs, it should be uniform
  throughout.

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Example

• For the building below, the loads are as follows
• a) The roof framing plus the expected wind load
  130 kN
• b) The wall above the foundation is 0.9 kN/m
• c) The floor will be used for grain storage and
  will support as much as 7.3 kN/m2 . The floor
  structure is an additional 0.5 kN/m2
• The foundation walls and piers are each 1 m high
  above the footing. The wall is 200 mm thick and
  the piers 300 mm square. The soil on the site is
  compacted clay in a well drained area. Find the
  size of the foundation and pier footing that will
  safely support the loads. Assume that the weight
  of the mass 1 kg approximately equals 10 N. The
  mass of concrete is 24 kN/m3.

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Solution

• Note that each pier(column) carries 4/32 ie.
  1/8 of total floor load. Each wall carries 7/32
  of floor load ie. (2 x 3) 1/2 1/2
• 1) The division of load on each foundation wall
  is as follows
• a) Roof load 50 on each wall i.e. 50 of
  130 kN 65 kN
• b) Wall load for each side 16 m x 0.9
  kN/m 14.4 kN
• c) Floor load each side of wall carries 7/32
  
• Total floor load (7.3 0.5) x 16 x 8
  998.4 kN
• i.e. floor load carried by each wall 7/32 x
  998 kN 218.4 kN

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Solution Contd.
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