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Derivation of stress block parameters

Basic Assumptions:

• Plane sections to axis remains plane even after bending.
• The maximum strain in concrete at the outermost compression fibre is 0.0035 in bending.
• The tensile strength of concrete is ignored.
• The relationship between the compressive stress distribution and strain in concrete may be assumed to be rectangle, trapezoid, parabola or any shape which result in substantial agreement with the test result.
• For design purposes, compressive strength of concrete = 0.67 x f_ck.
• Partial safety factor for concrete = 1.5.
• Partial safety factor for steel = 1.15.
• Maximum strain in tension reinforcement in section shall not be less than

\[{{\rm{\varepsilon }}_{\rm{s}}}{\rm{ = }}\frac{{{f_y}}}{{1.15x{E_s}}} + 0.002\] </ul>

Where,

• d = Effective depth of beam.
• D = Overall depth of beam.
• b = Width of beam.
• C = Compressive force of concrete.
• T = Tensile force of steel.
• x_u = Depth of neutral axis.
• ε_c = Maximum strain allowed in concrete.
• ε_y = Maximum strain allowed in steel.
• 0.002 = Strain at maximum characteristic compression strength of concrete is present.
• f_ck = characteristic compression strength of concrete.
• f_ck = characteristic yield strength of steel.

Calculation of stress block parameters:

Depth of parabolic part of stress block ( x¹ )

\[\frac{{{x_u}}}{{0.0035}} = \frac{{{x^1}}}{{0.002}}\]

Simplyfy the above equation :

\[{x^1} = \left( {\frac{{0.0035}}{{0.002}}} \right) \times {x_u}\]

Depth of the parabolic curve :

\[{x^1} = \left( {\frac{4}{7}} \right) \times {x_u}\]

Depth of the rectangular portion :

\[{x^{11}} = \left( {\frac{3}{7}} \right) \times {x_u}\]

Area of stress block = Area of rectangular portion + Area under parabolic curve

\[{A_{Total}} = \left( {\frac{3}{7} \times {x_u} \times 0.446 \times {f_{ck}}} \right) + \left( {\frac{2}{3} \times \frac{4}{7} \times {x_u} \times 0.446 \times {f_{ck}}} \right)\]

After simplyfying above equation :

\[{A_{Total}} = 0.36 \times {f_{ck}} \times {x_u}\]

The distance of stress block from the top fiber :

\[{x_c} = \frac{{\left( {\frac{3}{7} \times {x_u} \times 0.446 \times {f_{ck}} \times \left( {\frac{1}{2} \times \frac{3}{7} \times {x_u}} \right)} \right) + \left( {\frac{2}{3} \times \frac{4}{7} \times {x_u} \times 0.446 \times {f_{ck}} \times \left( {\frac{3}{8} \times \frac{4}{7} \times {x_u} + \frac{3}{7} \times {x_u}} \right)} \right)}}{{0.36 \times {f_{ck}} \times {x_u}}}\]

Centroid of stress diagram from top fiber:

\[{x_c} = 0.42 \times {x_u}\]

Calculation of depth of neutral axis:

Depth of neutral axis is obtained by considering equillibrium of internal forces of compression and tension.

Compression force of concrete, C = Average stress x compression area.

\[C = 0.36 \times {f_{ck}} \times b \times {x_u}\]

Tensile force of steel, T = Design yield stress x area of steel.

\[T = 0.87 \times {f_y} \times {A_{st}}\]

Tensile force of steel = Compressive force of concrete.

\[0.36 \times {f_{ck}} \times b \times {x_u} = 0.87 \times {f_y} \times {A_{st}}\]

simplyfying the above equation we get.

\[{x_u} = \left( {\frac{{0.87 \times {f_y} \times {A_{st}}}}{{0.36 \times {f_{ck}} \times b}}} \right)\]

Calculation of lever arm:

Lever arm is the distance between compressive and tensile force.

\[Z = \left( {d - 0.42 \times {x_u}} \right)\]

Calculation of moment of resistance:

Moment of resistance is calculated by using M_u = C x Lever arm.

\[{M_u} = 0.36 \times {f_{ck}} \times b \times {x_u} \times \left( {d - 0.42 \times {x_u}} \right)\]

Moment of resistance is calculated by using M_u = T x Lever arm.

\[{M_u} = 0.87 \times {f_y} \times {A_{st}} \times \left( {d - 0.42 \times {x_u}} \right)\]

Table:

Grade of steel	Xumax
Fe-250	0.53d
Fe-415	0.48d
Fe-500	0.46d

1. First video

2. Second video

3. Third video

Refrence

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1. Brooks, J.J & Neville A. M. (2019). Concrete Technology (2^nd ed.). Pearson Publishers Pvt Ltd.

2. Shetty, M. S & Jain, A. K. ( 2018). Concrete Technology: Theory And Practice (8^thed.).S Chand Publishers Pvt Ltd.