A cast iron bracket subjected to bending has a c/s of I shape with unequal flanges. If the section is subjected to shear force of 120kN, draw shear stress distribution diagram over the depth of the se

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written 8.7 years ago by

teamques10 ★ 69k

• modified 8.7 years ago

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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