Question Details

Consider an elastic straight beam of length L = 10π m, with square cross-section of side a = 5 mm, and Young’s modulus E = 200 GPa. This straight beam was bent in such a way that the two ends meet, to form a circle of mean radius R. Assuming that Euler-Bernoulli beam theory is applicable to this bending problem, the maximum tensile bending stress in the bent beam is _________ MPa.

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Correct Answer :

100

Solution :

The correct answer is 100.

Step-by-step Explanation:

As shown in the provided illustration, a straight elastic beam of length L (left) is bent until its ends meet, forming a closed circle of mean radius R (right), with the point of contact labeled as "Ends of the beam".

First, we relate the length of the beam to the geometry of the circle. Since the beam is bent to form a complete circle, the length of the beam is equal to the circumference of the circle:
L=2πR

Given:
L=10π m
We can solve for the mean radius R:
2πR=10π
R=5 m

According to the Euler-Bernoulli beam theory, the bending stress σ at any distance y from the neutral axis is given by the bending relation:
σy=ER
where:
E is the Young's modulus of the material,
R is the radius of curvature (which is the mean radius of the circle, 5 m).

The maximum bending stress (maximum tensile stress on the outer fiber) occurs at the maximum distance from the neutral axis, ymax. For a square cross-section of side a:
ymax=a2

Given the side of the square cross-section a is 5 mm:
ymax=5 mm2=2.5 mm=2.5×103 m

The Young's modulus is given as E = 200\text{ GPa} = 200 \times 10^9\text{ Pa}.

Now, we calculate the maximum bending stress σb:
σb=E·ymaxR
σb=(200×109 Pa)·(2.5×103 m)5 m
σb=5×1085 Pa=108 Pa=100 MPa

Thus, the maximum tensile bending stress in the bent beam is 100 MPa.

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