written 8.7 years ago by | modified 3.1 years ago by |
Mumbai university > MECH > SEM 3 > Strength Of Materials
Marks: 10M
Year: Dec 2013
written 8.7 years ago by | modified 3.1 years ago by |
Mumbai university > MECH > SEM 3 > Strength Of Materials
Marks: 10M
Year: Dec 2013
written 8.7 years ago by | • modified 8.7 years ago |
First we determine the moment of inertia
Here, the neutral axis would be at
YNA=Σni=1AiYiΣni=1Ai
In the I-section,
No. | Section | Area(Ai) | Centroid Distance from the base of the section (Yi) | AiYi |
---|---|---|---|---|
1 | Upper flange | 240 × 40 = 9600mm2 | 40 + 240 + 402=300mm | 2880000 |
2 | Web | 240 × 40 = 9600mm2 | 40 + 2402=1600mm | 1536000 |
3 | Lower flange | 120 × 40 = 4800mm2 | 402=20 | 96000 |
- | - | Σni=1Ai=24000 | - | Σni=1AiYi=4512000 |
Thus,
YNA=Σni=1AiYiΣni=1Ai=451200024000=188mmfrom the base
From moment of inertia
I=∑ni=1(Ixi+Aidi2)
Where, Ixi=112(bd3)
di=Yi−YNA(Distance of the CG of the section from NA)
NO | Inertia Ixi in mm4 | Area Ai in mm2 | di | Aid2i |
---|---|---|---|---|
1 | 1.28 × 106 | 9600 | 112 | 120.442 × 106 |
2 | 46.08 × 106 | 9600 | -28 | 7.5264 × 106 |
3 | 0.64 × 106 | 4800 | -168 | 135.4752 × 106 |
- | Σni=1Ixi=48×106 | - | - | Σni=1Aid2i=263.4436×106 |
Thus,
I=Σni=1(Ixi+Aid2i)=48×106+263.4436×106
I=311.4436×106mm4
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=F×AdI×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 × 106
b = Width of the upper flange =240mm
Thus ,
τ=120×103×1.0752×106311.4436×106×240=1.726N/mm2
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=10.356N/mm2
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 × 106
b = Width of the web=40mm
Thus ,
τ=120×103×1.09088×106311.4436×106×40=10.508N/mm2
v) Shear stresses in the lower flange just at the junction of lower flange and web is given by,
τ=F×AdI×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 × 106
b = Width of the upper flange =240mm
Thus ,
τ=120×103×0.8064×106311.4436×106×240=1.295N/mm2
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