First we determine the moment of inertia
Here, the neutral axis would be at
$Y_{NA} = \frac{Σ_{i=1}^nA_iY_i}{Σ_{i=1}^n A_i}$
In the I-section,
| No. |
Section |
Area$(A_i)$ |
Centroid Distance from the base of the section $(Y_i)$ |
$A_iY_i$ |
| 1 |
Upper flange |
240 × 40 = 9600$mm^2$ |
40 + 240 + $\frac{40}{2} = 300mm$ |
2880000 |
| 2 |
Web |
240 × 40 = 9600$mm^2$ |
40 + $\frac{240}{2} = 1600mm$ |
1536000 |
| 3 |
Lower flange |
120 × 40 = 4800$mm^2$ |
$\frac{40}{2} = 20$ |
96000 |
| - |
- |
$Σ_{i=1}^nA_i = 24000$ |
- |
$Σ_{i=1}^n A_iY_i = 4512000$ |
Thus,
$Y_{NA} = \frac{Σ_{i=1}^n A_iY_i}{Σ_{i=1}^nA_i} = \frac{4512000}{24000} = 188 mm from \ \ the \ \ base$
From moment of inertia
$I = \sum_{i=1}^{n}(I_{xi} + {A_id_i}^2)$
Where, $I_{xi} = \frac{1}{12}(bd^3)$
$d_i = Y_i - Y_{NA} (Distance \ \ of \ \ the \ \ CG \ \ of \ \ the \ \ section \ \ from \ \ \ NA)$
| NO |
Inertia $I_{xi}$ in $mm^4$ |
Area $A_i$ in $mm^2$ |
$d_i$ |
$A_id_i^2$ |
| 1 |
1.28 × $10^6$ |
9600 |
112 |
120.442 × $10^6$ |
| 2 |
46.08 × $10^6$ |
9600 |
-28 |
7.5264 × $10^6$ |
| 3 |
0.64 × $10^6$ |
4800 |
-168 |
135.4752 × $10^6$ |
| - |
$Σ_{i=1}^{n} I_{xi} = 48 × 10^6$ |
- |
- |
$Σ_{i=1}^{n} A_id_i^2 = 263.4436 × 10^6$ |
Thus,
$I = Σ_{i=1}^{n} (I_{xi} + A_id_i^2) = 48 × 10^6 + 263.4436 × 10^6$
$I = 311.4436 × 10^6 mm^4$
Shear stress distribution
i) Shear stresses at the extreme edges of the flanges is zero
ii) Shear stresses in the upper flange just at the junction of upper flange and web is given by,
$τ = \frac{F × Ad}{I × b}$
where,
Ad = Moment of area of upper flange about NA
= A(upper flange) × Distance of the CG of upper flange from NA
= 9600 × 112 = 1.0752 × $10^6$
b = Width of the upper flange =240mm
Thus ,
$τ = \frac{120 × 10^3 × 1.0752 × 10^6}{311.4436 × 10^6 × 240} = 1.726 N/mm^2$
iii) The shear stress in the web at the junction of web and upper flange will suddenly increase from $1.726 \ \ t0 \ \ 1.726 × 240/40 = 10.356 N/mm^2$
iv) The shear stress will be maximum at the NA
here,
Ad = Moment of total area about NA
=Moment of area of upper flange and of web about NA
= 9600 × 112 + 28 × 40 × 28/2 = 1.09088 × $10^6$
b = Width of the web=40mm
Thus ,
$τ = \frac{120 × 10^3 × 1.09088 × 10^6}{311.4436 × 10^6 × 40} = 10.508 N/mm^2$
v) Shear stresses in the lower flange just at the junction of lower flange and web is given by,
$τ = \frac{F × Ad}{I × b}$
where,
Ad = Moment of area of lower flange about NA
= A(upper flange) × Distance of the CG of lower flange from NA
= 4800 × 168 = 0.8064 × $10^6$
b = Width of the upper flange =240mm
Thus ,
$τ = \frac{120 × 10^3 × 0.8064 × 10^6}{311.4436 × 10^6 × 240} = 1.295 N/mm^2$
vi) The shear stress in the web at the junction of web and upper flange will suddenly increase from 1.295 to 1.295× 120⁄40=3.885 N/mm^2
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