Mechanical Engineering - Strength of Materials - Discussion
Discussion Forum : Strength of Materials - Section 6 (Q.No. 40)
40.
A beam of triangular section is placed with its base horizontal. The maximum shear stress occurs at
Discussion:
7 comments Page 1 of 1.
Shiva Shukla said:
4 years ago
Maximum Shear Stress in Triangular section = Above the N. A.
Virat said:
5 years ago
Centroid is at h/3.
Mid way means h/2. Clearly mid way is above centroid and neutral axis above centroid has higher stress.
Mid way means h/2. Clearly mid way is above centroid and neutral axis above centroid has higher stress.
(1)
SARAFRAJ ANSARI said:
8 years ago
It seams Above the neutral axis.
Shivani said:
8 years ago
@Neeraj.
What you said is for triangle. For rectangle it's h/2.
For shear apply formula.
τ = shear force*area* distance from neutral axis/(MI*Width).
Hence shear is inversely proportional to width.
What you said is for triangle. For rectangle it's h/2.
For shear apply formula.
τ = shear force*area* distance from neutral axis/(MI*Width).
Hence shear is inversely proportional to width.
Neeraj said:
9 years ago
For rectangular section h/6 distance from the neutral axis or above the neutral axis for all.
Milindsardar said:
9 years ago
Above the neutral axis.
Rakesh said:
1 decade ago
Any one please suggest a method to check shear for triangular section.
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