## LESSON 25.

25.1 SHEAR STRESS DISTRIBUTION FOR BEAM SECTIONS OF VARIOUS SHAPES

Fig……… shows a rectangular section of width b and depth d. Let the section be subjected to shear force V.

Consider a section EF at a distance y from the neutral axis.

The intensity of shear stress at this section is given by

τ =  ${{Va\bar y} \over {Ib}}$

V = Shear Force

y = distance from Neutral axis

b = width of the beam

d = Depth of the beam

I = Moment of Inertia of the beam section about the neutral axis

a $\bar y$   Moment of the area about the neutral axis

${\tau _{max}}$  =   ${{3V} \over {2bd}}$

Example: A timber beam is simply supported at the ends and carries a concentrated load at mid span. The maximum longitudinal stress is σ and the maximum shearing stress is τ. Find the ratio of the span to the depth of the beam ignoring the self weight of the beam.

If σ = 10 N/mm2 and τ = 1 N/mm2

Solution:

Maximum Shear Force, V = ${W \over 2}$

Maximum shear stress, τ=  ${{Va\bar y} \over {Ib}}$

${\tau _{max}}$   =  ${{3V} \over {2bd}}$

${\tau _{max}}$   = ${{3W} \over {4bd}}$  ----------------------------------------------------------------(i)

Maximum Bending Moment =  ${{WL} \over {4}}$

Maximum Bending Stress =  ${M \over Z}$

${{WL} \over {4}}$   = σ ×  ${{b{d^2}} \over 6}$

$\sigma={{3WL} \over {b{d^2}}}$   -----------------------------------------------------------(ii)

Ratio of span to depth =  ${L \over D}$

Dividing equation (ii) by (i)

${\sigma \over \tau }$  =  ${{{{3WL} \over {b{d^2}}}{\rm{}}} \over {{{3W} \over {4bd}}}}$

${L \over d} = {\sigma\over {2\tau }}$

=  ${{10} \over {2 \times 1}}$  = 5    (σ = 10N/mm2 and τ = 1N/mm2)

${L \over d}$   = 5

Example: A beam of I-Section 550mm deep and 200mm wide has flanges 30mm thick. It carries a shear force of 450 kN at a section. Calculate the maximum intensity of shear stress in the section assuming the moment of inertia to be 6.45 × 108 mm4. Also calculate the total shear force carried by the web and sketch the shear stress distribution across the section.

Solution: Maximum shear stress occurs at the web (N.A)

${\tau _{max}}$  =  ${{Va\bar y} \over {Ib}}$

V = 450 kN

$\bar y$  Moment above and about the neutral axis

I = 6.45 × 108 mm4

b = breadth at the level of ${\tau _{max}}$  = 20mm

a  $\bar y = {a_1}\overline {{y_1}}+{a_2}\overline {{y_2}}$

= (200 × 30) × 260 + (20 × 245) × 122.5

= 2.16 × 106 mm3

${\tau _{max}}={{450 \times {{10}^3} \times 2.16 \times {{10}^6}} \over {6.45 \times {{10}^{8}} \times 20}}$   = 75.35 N/mm2

For ${\tau _2}$  , at the bottom and top flange

a $\bar y$  (200 × 30) × 260 = 1560000 mm3

b = 200mm

${\tau _1}={{Va\bar y} \over {Ib}} = {{450 \times {{10}^3} \times \left( {200 \times 30} \right) \times 260} \over {6.45 \times {{10}^{8}} \times 200}}$  = 5.442 N/mm2

For  ${\tau _2}$  , a  $\bar y = 1560000$    mm3

(area is always taken above the point of consideration)

${\tau _2}={{450 \times {{10}^3} \times 1560000} \over {6.45 \times {{10}^{8}} \times 20}}$  = 54.42 N/mm2

Example: The T-beam section is subjected to vertical shear force of 150 kN. Calculate the shear stress at the neutral axis and at the junction of the web and the flange. Moment of inertia about the horizontal neutral axis is 1.134 × 108 mm4.

Solution:  (from top flange) =  ${{{a_1}\overline {{y_1}}+{a_2}\overline {{y_2}} } \over {{a_1} + {a_2}}}$

=  ${{\left( {60 \times 250} \right) \times 30+\left( {250 \times 60} \right) \times 185} \over {\left( {250 \times 60} \right)+\left( {250 \times 60} \right)}}$  = 107.5 mm

${\tau _{max}}$  =  ${{Va\bar y} \over {Ib}}$

=  ${{150 \times 1000 \times \left( {250 \times 60} \right) \times \left( {107.5 - 30} \right) + \left( {47.5 \times 60} \right)(47.5/2} \over {1.134 \times {{10}^8} \times 60}}$  = 25.77 N/mm2

${\tau _1}$  =  ${{\left( {150 \times 1000} \right) \times \left( {250 \times 60} \right)\left( {107.5 - 30} \right)} \over {1.134 \times {{10}^8} \times 250}}$  = 6.15 N/mm2

${\tau _2}$  =  ${{\left( {150 \times 1000} \right) \times \left( {250 \times 60} \right)\left( {107.5 - 30} \right)} \over {1.134 \times {{10}^8} \times 60}}$  = 25.63 N/mm2