Question: The member having a rectangular cross section, Fig. 6 - 2 8 a , is designed to resist a moment of 4 0 N *

The member having a rectangular cross section, Fig. 6-28a, is
designed to resist a moment of 40N*m. In order to increase its
strength and rigidity, it is proposed that two small ribs be added at
its bottom, Fig. 6-28b. Determine the maximum normal stress in the
member for both cases.
SOLUTION
Without Ribs. Clearly the neutral axis is at the center of the cross
section, Fig. 6-28a, so ?bar(y)=c=15mm=0.015m. Thus,
I=112bh3=112(0.060m)(0.030m)3=0.135(10-6)m4
Therefore the maximum normal stress is
max=McI=(40(N)*m)(0.015(m))0.135(10-6)m4=4.44MPa
With Ribs. From Fig. 6-28b, segmenting the area into the large
main rectangle and the bottom two rectangles (ribs), the location ?bar(y) of
the centroid and the neutral axis is determined as follows:
?bar(y)=(tilde(y))AA
=[0.015(m)](0.030(m))(0.060(m))+2[0.0325(m)](0.005(m))(0.010(m))(0.03(m))(0.060(m))+2(0.005(m))(0.010(m))
=0.01592m
This value does not represent c. Instead
c=0.035m-0.01592m=0.01908m
Using the parallel-axis theorem, the moment of inertia about the
neutral axis is
I=[112(0.060(m))(0.030(m))3+(0.060(m))(0.030(m))(0.01592(m)-0.015(m))2]
+2[112(0.010(m))(0.005(m))3+(0.010(m))(0.005(m))(0.0325(m)-0.01592(m))2]
=0.1642(10-6)m4
Therefore, the maximum normal stress is
max=McI=40(N)*m(0.01908(m))0.1642(10-6)m4=4.65MPa
NOTE: This surprising result indicates that the addition of the ribs to
the cross section will increase the maximum normal stress rather than
decrease it, and for this reason they should be omitted.
The member having a rectangular cross section,

Step by Step Solution

There are 3 Steps involved in it

1 Expert Approved Answer
Step: 1 Unlock blur-text-image
Question Has Been Solved by an Expert!

Get step-by-step solutions from verified subject matter experts

Step: 2 Unlock
Step: 3 Unlock

Students Have Also Explored These Related Mechanical Engineering Questions!