Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

At point A, r = 50 mm. Thus, the bending stress at point A is
Mr ( n − r)
(4,500 Nm)(79.957167 ⋅
mm − 50 mm)(1, 000 mm/m)
σ x =−
=−
2
rAr ( − r ) (50mm)(2,450 mm )(85 mm − 79.957167 mm)
c
n
=− 218.224 MPa = 218 MPa(C)
Ans.
At point B, r = 120 mm, making the bending stress at that point
Mr ( n − r)
(4,500 Nm)(79.957167 ⋅
mm −120 mm)(1, 000 mm/m)
σ x =−
=−
2
rAr ( − r ) (120 mm)(2450 mm )(85 mm − 79.957167 mm)
c
n
= 121.539 MPa = 121.5 MPa(T)
Ans.
(b) Flexural Stress Distribution
O
A
−218.2 MPa
79.9572 mm
85 mm
M
x
N.A.
Centroid
+
B
121.5 MPa
The bending stress created by M = 4,500 N ⋅ m in the curved rectangular bar is shown
in the accompanying figure. Note that the distribution is nonlinear and that the neutral
axis (N.A.) is not located at the centroid of the cross section.
(c) Comparison with Stresses from Straight-Beam Flexure Formula
For comparison, let’s consider the bending stresses that we would have obtained from the
flexure formula for initially straight beams. The section modulus of the rectangular shape is
2 2
bd (35mm)(70mm)
S = = = 28,583.333 mm
6
6
If the beam were initially straight, the stresses at A and B for M = 4,500 N ⋅ m would be
M (4500 Nm)(1000 ⋅ mm/m)
σ x =± =±
=± 157.434 MPa
3
S 28,583.333 mm
Therefore, the errors between the actual stress (determined from the curved-bar formula)
and the stress obtained by using the flexure formula for straight beams is
3
−157.434 MPa − ( −218.224 MPa)
−218.224 MPa
157.434 MPa − (121.539 MPa)
121.539 MPa
(100%) = 27.9% lowat A Ans.
(100%) = 29.5% high at B Ans.
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