design of Pressed steel tank pdf

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Winners academy

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Design of Pressed steel

tank

According to IS 804:1967

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Step 1: Dimensioning of tank:

Æ Dimensioning of tank is done in such a way that it fulfils the capacity criteria.

Æ You can use only pressed plate of size 1.25X1.25 m, so length breadth and height should (n X1.25m )where n is no. Of plate.

Æ Commonly two cases arises:

1. 2,5,5 plates along HLB respectively 2. 2,6,6 plates along HLB respectively

Step2: Thickness of plate:

Æ Thickness of plate of tank may be either 5mm or 6mm depending on height of tank and tier (For 2 plate one can say that bottom layer of plate is bottom tier and other will be top tier) Æ From IS 804:1967 clause : 6.3 and table 5, thickness of plates can be taken

Step 3: Provision of stay:

Æ As it is known that water will exerts the force in outside direction so it may be counteract by providing stay inside the tank.

Æ

Æ By above diagram we can see the use of stay.

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L1 L2 L3 L4 L5 L6 L7

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A

B

15.328

C 30.656

#Pressure distribution diagram and reactions#

#Force In stay# Æ Calculation:

∑ Above figure shows pressure distribution diagram; Pressure on each unit of plate for 1.25m width:

At 1.25m below top = 9.81X1.25X1.25 = 15.328kN/m2 At 2.50m below top = 9.81X2.50X1.25 = 30.656kN/m2 ∑ Reaction:

For top

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For bottom tier – P2= .5X(30.656+15.328)X 1.25 = 28.74kN Y= ( ) ( ) = ( . . ) . ( . . )

=

0.556m R2’ X1.25= 28.74X0.556 R2’=12.78kN Total reaction at A = 3.19kN At B = R1’+R2’ = 19.17kN At C= 15.96kN ∑ Force in stay: See the diagram— F1cos45 = 3.19 So F1= 4.512Kn F2cos45= 19.17kN F2= 27.11kN ∑ Design of stay—

Allowable stress in axial tension= 0.8X0.6X250 = 120N/mm2(0.8 factor is from IS 804)

Area required for long stay = 4.512E3 /120 = 37.6mm2 Area required for short stay= 27.11E3/120= 225.9mm2 Provide 30mm wide X 6mm thick mild steel flats for top stay. 60mm wide X 6mm thick flat

Step 4 : Design of longitudinal

beam:-Æ The tank is directly supported on seven longitudinal beam L1L1...L7L7. Æ Refer first arrangement of stay diagram.

Æ Weight of water per unit area = 9.81X1X1X2.5= 24.52kN/m2 Æ Weight of bottom plate = 0.006X1X1X78.5= 0.471kN/m2 Æ Vertical force component V1 of F1= F1sin45= 3.19kN Æ Vertical force component V2 of F2= F2sin45= 19.17kN

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∑ Longitudinal beam L4L4:

Æ Load on this beam is due to water column and bottom plate of one pair of HALF PLATE means half load from left and other from right as load is uniform that’s why there no problem.

Æ Another load is due to stay .Stay in between the span always create tension and at end it create compression.

Æ UDL :

24.52X1.25 + 0.471X1.25 + 1(self wt.)= 32.2kN/m =~ 32kN/m Plate weight = wt. Of upper plate= 1.34kN

Udl is due to wt of water and bottom plate

At end 1.34kN is force due wt. Of plate of upper and lower tier acting downward. At V2 (down)is due to small stay.

In fig see, only two small stay is arranged which have vertical component on beam at 1.25m from end.

Æ Now calculate reaction, moment at support , moment at centre.

∑ Longitudinal beam L3L3 and L5L5:

Again loading

Æ Udl is due to water and bottom plate

Æ V1 V2 and wt of plate at end(ref diagram , in diagram end is associated with two stay.) Æ At support vertical component of shorter stay.

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∑ Longitudinal beam L3L3 and L5L5:

Loading:

Æ Udl is due to water and bottom plate

Æ V2 and wt of plate at end(ref diagram , in diagram end is associated with one stay.) Æ At support two vertical component of shorter stay.

Æ At middle points vertical components of short stay because at all point shorter stay ia attached (See diagram)

Æ Again calculate reaction, moment at support , moment at centre. ∑ Longitudinal beam L1L1 and L7L7:

ÆAt end beam, the loading is slightly changed.

ÆLoad due to water= 0.5X 9.81X 1.25X2.5X1 = 15.325kN/m (half is because this beam carries on half column of water of end plate)

ÆSimilarly load of bottom plate= 0.5X0.006X1.25X1X78.5= 0.295kN/m ÆEnd beam also have to carry whole load of end wall so it also adds in udl Æ= 1.25X1X(0.006+0.005)X78.5= 1.07893kN/m

ÆTotal udl = 16.68 + self wt= 18kN/m(assume) ÆPoint load of wall at end= 1.34/2 = 0.675 Æ

Æ As it is a end beam all vertical component will act downward Æ At support one vertical of short stay

Æ At middle (at 2.5m from end) V1 and V2 both Again calculate reaction and moment

∑ Design:

Æ Determine from above analysis maximum Moment.

Æ These beams are laterally unsupported and take effective length =4X1.25 = 5m Æ For trial section assume σb= 165N/mm2

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Æ From table 6.1B of IS 800:1984 calculate exact value of σband fine true moment carrying capacity.

Æ If is greater than Mmax then Design is safe.

Step5- Design of cross beam:

Æ Cross beams are provides under longitudinal beams hence it carry’s all the reaction forces due to longitudinal beam.

Æ R1=R7; R2=R6; R3=R5; R4 ... These all are calculated in previous steps. Æ Calculate moment and reaction

Æ Design the beam by using same method as discussed in previous step.

Reference:

1. Design of steel structure by BC punmia 2. IS 804 and IS 800:1984 steel table.

Any query contact: Harsh Mahajan +91-9753351408

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