Bridge Deck Slab Design Example as per IRC || Slab Bridge Design example || LSM Method-Part 4

Bridge Deck slab design example as per IRC || Slab Bridge Design Example || Bridge Design Using LSM Method-Part 4  

We get, 

Maximum bending moment for dead load=241.344 kN.m

Maximum bending moment for live load=163.967 kN-m

Total design bending moment for Dead Load and Live Load=

=1.35xBMDL + 1.5xBMLL

=1.35x241.334kN-m+1.5x163.967 kN-m

=571.751 kN-m/m-width

Maximum shear force for dead load=100.56 kN.

Maximum shear force for live load =71.66kN.

Total Design shear force for Dead Load and Live Load=

=1.35xVDL + 1.5xVLL

=1.35x100.56 kN + 1.5x71.66 kN

=243.246 kN.

 1.   Checking Depth Of Slab: 

We using materials: Concrete M30 and Steel: fy-415

We know moment of resistance by concrete

Mu=0.138fckbd2  (for fy: 415)

405.301x10⁶=0.138x30x1000xd2

d=312.888 mm<700mm (provided effective depth) ( Hence, ok)

 2.      Design of Main Steel Bars:  

Check section is under-reinforced or over-reinforced section.

 Maximum depth of neutral axis (as per IS 456: 2000, Clause-38.1, page 70)

Xu(max)=0.48xd

=0.48x700

=336mm

 The actual depth of neutral axis:

Ultimate moment of resistance by concrete=

Mu=0.36fckXub(d-0.42Xu)

571.751x10^6=0.36x30.Xu.1000(700-0.42xXu)

52939.91=700Xu-0.42Xu^2

0.42Xu^2-700Xu+52939.91=0

After calculating the equation

Xu=1587.25mm or 79.41mm

Here, Xu=1587.25mm is not acceptable because overall depth of slab is 750mm.

So, Xu=79.41mm

 Xu=79.41mm< Xu(max)=336mm

Hence, this section is under-reinforced section. LSM method is applicable to design under reinforced and balanced reinforced sections.

 

If the actual depth of the neutral axis (Xu) is greater than maximum depth of neutral axis (Xumax), the section shall be redesigned. LSM method is not applicable to design over reinforced section. (from IS 456: 2000, Clause-38.1 G.1.1(c), page-96)

Moment of resistance by steel reinforcements:

Mu=0.87fyAst(d-0.42Xu)

571.751x10^6=0.87x415xAst(700-0.42x79.41)

Ast=2375.435mm²

 No of bars= Area of steel/area of individual steel bar=(2375.435)/(π/4x20^2)= 7.56

Spacing =1000/7.564=132.197mm

Providing 20mm diameter bars @130mm spacing C/C.

So, Provided area of steel reinforcement, Ast=2416 mm²

2.      3. Check the percentage of steel:

(According to IS 456:2000, Clause 26.5.2.1, page-48)

For Slabs:  Minimum percentage of steel 0.15% of A-gross for Fe 250 and 0.12% of A-gross for Fe 450. Maximum percentage of steel 4%.

Minimum steel reinforcement=0.12% of bD

=0.12x1000x750/100

=900 mm² <Provided area of steel reinforcement (Ast=2416 mm²) (Hence safe)

3.      4. Check Shear Stress:

As per IS 456:  2000, Clause 40.1, page 72

Picture-1: Check Shear Stress for Slab Bridge Design Example

τv = Vu/ b.d

=243.246/(1000x700) kN/mm²

=0.000347494 kN/mm²

= 0.347494 N/mm²

 Percentage of steel

=Astx100/bd

=2416x100/(1000x700)

=0.345

We get permissible shear strength of concrete for grade M30 (as per IS 456 2000, Table-19, page 73)

Permissible shear stress=τc

= 0.419 N/mm²

Nominal shear stress (τv) is less than permissible shear stress (τc).

Hence, there is not required to provide shear reinforcement.

       5. Calculation for distribution bar:

Bending moment for distribution reinforcement:

M=0.2BMDL+0.3BMLL (As per IRC 21, Clause 305.18.1, page 56)

Design bending moment for distribution reinforcement

=1.35x(0.3x241.334)kN-m+1.5x(0.2x163.967) kN-m (multiplied by safety factor 1.35 for Live Load and 1.5 for Dead Load).

=146.930 kN-m

Find out effective depth for distribution bar=d’

d’=(oval all depth)-(clear cover)-(diameter of main bar)-(half diameter of distribution bar).

Let's use  12 mm diameter distribution bar.

d’=(750-40-20-12/2) mm

d’=684mm

We know,

Mu=0.87fy.Ast.d (1-Ast.fy/bdfck)

146.930x10^6=0.87x415xAstx684(1-Astx415/1000x684x30)

594.959= Ast(1-Astx2.022x10^^-5)

Ast²(2.022x10^^-5)-Ast+594.959=0

Ast=48854mm² or 603mm²

Take the acceptable value 603 mm²

But percentage of steel will be P(t)=603/ (1000x750)

=0.000804

=0.0804% of bD< Minimum steel reinforcement=0.12% of bD

So, provide minimum area of distribution bars.

=Area of distribution bars=0.12% of bD

=0.12x1000x750/100

=900 mm²

Spacing=1000x (π/4x12^2)/900

=125.71 mm

 So, Provide 12mm diameter distribution steel bars @ 120 mm Centre to centre distance.

Picture-2: Main bars and distribution bars details.

 

You may visit my youtube channel: 

1.  Slab-Bridge-Design-In-Midas-Civil-as Per Indian Code IRC 112

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