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

I ci (mm 4 ) | d i | (mm) d i 2 A i (mm 4 ) I z (mm 4 )
(1) 2,500,000 15.0 675,000 3,175,000
(2) 337,500 20 1,800,000 2,137,500
(3) 2,500,000 15.0 675,000 3,175,000
8,487,500
The moment of inertia of the U-shaped cross section about its z
centroidal axis is I z = 8,487,500 mm 4 .
For the positive bending moments M A and M B acting on
the beam segment as shown, compressive normal stresses will
be produced above the z centroidal axis and tensile normal
stresses will occur below the z centroidal axis. The flexure formula
[Equation (8.7)] is used to compute the bending stress at
any coordinate location y. (Recall that the y coordinate axis has
its origin at the centroid.) For example, the bending stress at the
top of area (1) at section A is calculated with y = 65 mm:
My (11kNm)(65 ⋅ mm)(1, 000 N/kN)(1, 000 mm/m)
σ x =- =-
4
I
8,487,500 mm
z
=- 84.2 MPa = 84.2 MPa (C)
30 mm 150 mm
30 mm
65 mm
35 mm
A
(1)
M A = 11.0 kN·m
B
Tensile bending stress
z
z
y
(1)
(2)
y
(3)
(2)
(3)
30 mm
Compressive bending stress
100 mm
x
M B = 16.5 kN·m
The maximum tensile and compressive bending
stresses at sections A and B are shown in the accompanying
figure.
(b) Of particular interest in this example are the bending
stresses acting on area (2) of the U-shaped cross
section. The normal stresses acting on area (2) are
shown in the following figure:
84.2 MPa (C)
45.4 MPa (T)
A
B
126.4 MPa (C)
x
68.0 MPa (T)
84.2 MPa (C)
y
126.4 MPa (C)
6.48 MPa (T)
45.4 MPa (T)
A
x
(2)
B
9.72 MPa (T)
68.0 MPa (T)
The resultant force of the bending stresses acting on area (2) must be determined
at section A and at section B. The normal stresses acting on area (2) are all of the same
sense (i.e., tension), and since these stresses are linearly distributed in the y direction,
we need only determine the average stress intensity. The stress distribution is uniformly
distributed across the z dimension of area (2). Therefore, the resultant force
acting on area (2) can be determined from the product of the average normal stress
and the area upon which it acts. Area (2) is 150 mm wide and 30 mm deep; therefore,
A 2 = 4,500 mm 2 . On section A, the resultant force in the x direction is
F
A
1
2
= (6.48 MPa + 45.4 MPa)(4,500 mm ) = 116,730 N = 116.7 kN
2
and on section B, the horizontal resultant force is
1
2
FB
= (9.72 MPa + 68.0 MPa)(4,500 mm ) = 174,870 N = 174.9 kN
2
as shown in the following figure:
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