Showing posts with label "how to design simply supported beam manually". Show all posts
Showing posts with label "how to design simply supported beam manually". Show all posts

Sunday, July 11, 2021

How to design a simply supported beam manually

 Suppose we have a beam whose span is 30 Ft , and a Point Load "F"=10000 Pounds, which includes weight of beam and imposed load.



In this case we are going to design a RCC Beam using Load Factor Theory in following steps.

1-Find out Support or Reactions

Left and Right Support = F/2= 5000  Pounds

In given symmetrical loading case both reactions are equal

2-Find out Max shear force = Load /2 = 10000/2 =5000 Pounds

Shear force at support = +5000 Pounds

Shear force at Mid      = +5000 + (-10000) =   -5000 Pounds

3-Max Bending Moment at Center of  Beam =    Reaction x L/2

Max MB=            5000x (30/2)x12= 900000 Pound-Inch--------1

4-Moment of Resistance =  K.b.d1^2----------------EQ.2

5-b (width of beam ) = (Span x12)/30 =12"

Value of  "K" = 250  in Load Factor Method using (1000-20000) Concrete

Where fc =1000 PSI  and Fs = 20000 PSI

We know that Moment of Resistance =Bending Moment  so comparing both equations EQ----1 AND EQ-----2

6-       900000 = 250x12xd1^2

d1 (Effective Depth) = 17.33 inch Say 17.5"

7-   Area of Steel in Tension zone = BM /(FstxLever Arm)

                        Ast  =900000/(20000x0.75/17.5)= 3.42 Sq.in

 No of bars according to 3/4" dia bars =  Ast/area of one bar =8 nos.

 

8-Area of  steel in compression zone =(B.M - M.r)/fst.Lever Arm

               =((900000-(250x12x17.5^2))/20000x 0.75x17.5

               =(900000-918750)/20000x0.75x17.5= - (0.07)


Use (4+4) nos of 3/4" dia bars and 2 no of 1/2" dia bar at top sides.

Use  3/8" dia bar stirrups 2-Leg @ 10 " c/c

Minus sign shows that the M.r value is greater than the applied value of B.m so no need of Asc (Area of steel in compression zone ), Only two steel bars of 1/2" dia bars are enough to hold stirrups.

 

Use (4+4) nos of 3/4" dia bars and 2 no of 1/2" dia bar at top sides.

9- Spacing of stirrups = Aw.Fw.Lever arm/(Shear Force)

                                    =(0.22x18000x0.75x17.5) /5000 = 10" C/c

Here Aw = Area of 2 leg stirrup bars

      Fw    =   Permissible value of Tensile stress of stirrups

Overall depth of beam = Effective depth+ 2 concrete covers

concrete cover = 1"

Concrete ratio =1:2:4

Factor of safety in Fc = 3000/1000=3 Times (considering compressive strength of concrete)

 The factor of safety in Fst= 60000/20000=3 Times (considering Tensile strength of Steel).

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