NAT+2 / -0MediumMoment of InertiaStatics & Rigid BodiesStructural Engineering
Structural Engineering → Statics & Rigid Bodies → Moment of Inertia
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Question
The cross-section of a girder is shown in the figure (not to scale). The section is symmetric about a vertical axis (Y-Y). The moment of inertia of the section about the horizontal axis (X-X) passing through the centroid is _______cm4 (round off to nearest integer).
Correct answer
464000 to 472000
Solution
1.Identify the components of the cross-section:
Top flange: 40 cm×10 cm
Web: 20 cm×50 cm
2. Calculate the area of each component:
A1=40×10=400 cm2
A2=20×50=1000 cm2
Total Area A=1400 cm2
3. Determine the centroid (yˉ) from the bottom edge:
y1=50+210=55 cm
y2=250=25 cm
yˉ=AA1y1+A2y2=1400400×55+1000×25=140022000+25000=140047000≈33.57 cm
4. Calculate the moment of inertia about the centroidal axis (X-X) using the parallel axis theorem (Ixx=∑(Igi+Aidi2)):
For the top flange: Ixx1=1240×103+400×(55−33.57)2=3333.33+400×459.24=187029.33 cm4
For the web: Ixx2=1220×503+1000×(25−33.57)2=208333.33+1000×73.44=281773.33 cm4
Total Ixx=187029.33+281773.33=468802.66 cm4
5. Final Answer: Rounding to the nearest integer, we get 468803 cm4, which falls within the range of 464000 to 472000.