Engineering Mechanics

LESSON 25.

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

a \[\bar y\]  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