procedure of steel design.pdf

8/14/2019 Procedure of Steel Design.pdf

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Design of Tension Member as per IS: 800-1984:-

1. Allowable permissible tensile stress Where,= yield stress=250 MPa

2. Area required =

, increase 30%ofabove area for rivet connection and 10% of above area

for welding connection.3. Choose a section for above area from steel table.4. Calculated tensile stress 5. Check for slenderness ratio , effective length for truss , =for

single angle and for double angle or . =180 for only tension and =350 for reversible oftension and compression.

6. Design of end connection:for welding- assume size of the weld and equate with the strength of weld i.e. 108(0.7

s)

l=Load

for rivet-rivet value minimum of below two-a. Bearing= b. Shearing=n

No. of rivet = Load/ rivet value.

Calculation:-1. In case of single angle connected through one leg the net effective sectional area shall be taken as

below -

For riveting of for welding of

ISA one side ISA one side

Where,


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2. In case of a pair of angles back to back (or a single Tee) connected by one leg of eachangle (or by flange of the Tee) to the same side of a gusset the net area shall be taken as

below-

For rivet of 2 ISA on same side A single T section

With tack rivetFor rivet: For rivet: For welding: For welding: Where, 3. For double angles or Tees placed back to back and connected to each side of the gusset

plate, net effective area shall be taken as below-

4. a. Where the angles are back to back but are not tack riveted or welded, then 2 and 3 shall not apply

and each angle shall be designed as a single angle connected through one leg only i.e. 1.

b. When two tees are placed back to back but are not tack riveted or welded, then 3 shall not apply

and each Tee shall be designed as a single tee connected to one side of the gusset only i.e. 2.


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If no tack rivet is being provided then,

of each angle is calculated separately.

Note- The area of the leg of an angle shall be taken as the product of the thickness and the length from

the outer corner minus half the thickness and the area of the leg of a tee as the product of the thickness

and the depth minus the thickness of the flange.


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DESIGN OF COMPRESSION MEMBER AS PER IS: 800-1984-

1. Assume allowable permissible compressive stress 2. Area required =

3. Choose a section for above area from the steel stable.

4. Calculated compressive stress 5.calculation-a. Slenderness ratio , effective length for truss ,=for single angle and for double angle or .

=180 for only tension and =350 for reversible of tension and compression and =250 for

compression due to earthquake.

b. Elastic critical stress in compression, , Where, E= 200x103MPa.c. permissible stress in axial compression

If ,then ok.1. End connection-

for welding- assume size of the weld and equate with the strength of weld i.e.

108x(0.7xs)xl=Load

for rivet-rivet value minimum of below two-

a. Bearing= b. Shearing=n x


Note- when a section choose, we can use plate along with the section, if the section is not sufficient

then following procedure is adopted

1. Assume allowable permissible compressive stress 2. Area required =

3. Choose a section below the above area from the steel stable.

4. Plate is provided for remaining area.

5. Calculated compressive stress 6.calculation-a. Slenderness ratio , effective length for column as per support conditions.

b. Elastic critical stress in compression,

, Where, E= 200x103MPa.

c. permissible stress in axial compression , If ,then ok. Find the shear stress at the junction per unit length and find the size of weld.


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DESIGN OF BEAM OR BENDING MEMBER AS PER IS: 800-1984

1. Calculate the maximumbending moment M

2. Assume 3. Section modulus required 4. Choose a section by a single Iand

For built up section, Area of plate =

5. Assume a thickness and width of plate.

6. Calculate and of whole section.7.

For restrained-8. For unrestrained-

8. 9. ( )

10. Elastic compressive stress in bending Where, = reduction coefficient=1,=coefficient for the inequality of flanges=

c2, c1are the greater and lesser distances from N.A.

T= thickness of flanges of the section and D= overall depth of section.

11. Permissible bending stress > Hence, ok.12. Check for Shear-

a. calculate SFmax.=V

b. For single I < For plate with I < , where d1=depth of web of I section.


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13. Check for Deflection-

a. Calculate maximum deflection according to load condition and it should be less than

14. Check for web crippling-

Bearing stress for point load =

Bearing stress for reaction=

Where a= width of point load or reaction.

15. Check for buckling- and and Elastic critical stress in compression, then ok.


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DESIGN OF TENSION MEMBER AS PER IS: 800-20071. Factor Load P =1.5x Load

2. Gross area of cross-section Where,=yield stress= 250 MPa. And =1.13. Choose a section for above area, by increasing area 10%

4. End connection- by equating strength of weld to factor load, i.e. and calculatel and for double angle length for each angle=l/2.Where,=ultimate stress, for Fe250, , for Fe300, ,S=size of weld, minimum is 3mm,

5. Calculate design strength, least of following-

a. Design strength due to yielding of gross area- b. Design strength due to rupture of critical section- Net area of the total cross-section,

= Gross area of connected leg, Where, w=outstanding leg width, = shear lag width, as shown fig.=length of the end connection that is the distance between the outermost bolts in the end jointmeasured along the load direction or length of the weld along the load direction.

c. Design strength due to block shear-

or

For, double angle Anc=2xAncfor each. minimum gross and net area in shear along line parallel to external forces, Minimum gross and net area in tension from the bolt hole to the toe of the angleperpendicular to external forces.


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DESIGN OF COMPRESSIN MEMBER AS PER IS: 800-2007

1. Area=

2. If one angle or two angle back to back on one side of gusset plate.

Effective slenderness ratio

Where, yield stress ratio and are the leg width and t= thickness of angle section.2.If double angle placed opposite side of the gusset plate-

and it should be greater than 4. End connection- by equating strength of weld to factor load, i.e.

and calculatel and for double angle length for each angle=l/2.

Where,=ultimate stress, for Fe250, , for Fe300, ,S=size of weld, minimum is 3mm,


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DESIGN OF BEAM COLUMN IS: 800-1984

Members subjected to axial compression and bending shall be proportional to satisfy the following

requirement-

Calculation-Calculated compressive stress a. Slenderness ratio , for I section or .

b. Elastic critical stress in compression, , and Where, E= 200x103MPa.c. permissible stress in axial compression

and For unrestrained- and

( ) ( )

Elastic compressive stress in bending Where, = reduction coefficient=1,=coefficient for the inequality of flanges= c2, c1are the greater and lesser distances from N.A.

T= thickness of flanges of the section and D= overall depth of section.

Permissible bending stress


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DESIGN OF LACING AS PER IS: 800-1984

1. Area A= where, (assume)

2. Choose a section of I or channel.

3. 4.

and For effective length of column as per support condition.

FIND c/c distance b/w the I section by equating

5. Calculate a, l, L

L=2a for single lacing, L=a for double lacing.

6. Dia. of rivet 20mm. and width of lacing 60mm.

7. Thickness of lacing

t=l/40 for single lacing, and l/60 for double lacing

8. and 9. =l for single lacing with rivet

=0.7xl for double lacing with rivet

=0.7xl for both single and double lacing with weld.

10. Transverse force

11. Force in lacing F = for single lacing

F= for double lacing

12. Check for axial compression and tension as below-Calculation-a. Slenderness ratio for lacing as calculated in 8.

b. Elastic critical stress in compression, Where, E= 200x103MPa.c. permissible stress in axial compression

d. Calculated compressive stress


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Calculation-Calculated tensile stress 13. Determine the rivet value and find out the no. of rivet=

DESIGN OF BATTEN AS PER IS: 800-19841.A= where, (assume)2.Choose a section of I or channel.

3. 4.

and for effective length of column as per support condition.

FIND c/c distance b/w the I section by equating i.e.s

5.Calculate by assuming gauge distance-

Where, B=width of I or channel section.

6. D is calculated by adding gauge distance.

7. Thickness of batten =


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8.

And find C9. Transverse force 10. Longitudinal force where, N= no. of parallel plane. S= c/c distance b/w rivet group, c=spacing of batten.

11. Moment M 12. Longitudinal shear stress =

13 .Bending stress =

14. Rivet connection-

Rivet value, then assume no. rivet.

15. Find the Load carried by rivet-

a. Force due to shear

b. Force due to bending Resultant = Rivet Value -

a. Bearing= b. Shearing=n


DESIGN OF PURLIN AS PER IS: 800-1984

1. Calculate Mxand My2. Zxrequired = Mx/1653. Choose a suitable section.4. Mx/Zx+ My/Zy165 Mpa

For angle, length =l/45 for max load side and l/60 for other side. Mx=My=(WL)/10.

procedure of steel design.pdf

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