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

310
bENdINg
To remove θ from this expression, we once again use Equation (b), rewriting it,
however, as
Edθ
σ xr
=−
θ r − r
n
(f)
A positive bending moment
makes the curve tighter.
Then we substitute Equation (f) into Equation (e) to obtain
M
σ xr
=−
r − r
Next, we expand Equation (g) as follows:
n
∫
A
( rn
− r)
r
2
σ xr
⎛ n
M
r r
rr dA rr n
r dA rr n
r dA rdA ⎞
=− ∫ −
A ∫ −
A ∫ +
n − ⎝
⎜
A ∫A
⎠
⎟
σ xrrn
⎛
r r r dA ⎞ σ xr
=−
n ∫ − dA
rn
dA r dA
A r
∫
(
A r r
∫A
∫ )
− ⎝
⎜
⎠
⎟ − − + (h)
−
A
n
Notice that the first two integrals in Equation (h) are identical to the terms in parentheses
in Equation (c). Accordingly, these two integrals vanish, leaving
σ xr
M =−
r − r
n
n
2
dA
(g)
( − rn
dA+ rdA)
(i)
∫ A ∫ A
The first integral in Equation (i) is simply the area A. The second integral is the first
moment of area about the center of curvature. From the definition of a centroid, this integral
can be expressed as r c A, where r c is the radial distance from the center of curvature,
O, to the centroid of the cross section. Thus,
σ r
⎛
x
M
rA rA rA r c −
( )
r ⎞
n
=− − n + c =−σ
x ⎜ ⎟
r − r
⎝ r − r ⎠
n
Solving this equation for the circumferential normal stress created in a curved bar by a
bending moment M gives
Mr ( n − r)
σ x =−
(8.28)
rAr ( − r )
In this formula, the bending moment M is considered to be a positive value if it creates compressive
normal stress on the inner surface (i.e., at radius r i ) of the curved bar. A positive
bending moment decreases the radius of curvature of the bar.
c
n
n
ExAmpLE 8.13
M
r i
y
O
50 mm
A
A
35 mm
A curved bar with a rectangular cross section is subjected to a
bending moment M = 4,500 N ⋅ m, acting in the direction shown.
The bar has a width of 35 mm and a height of 70 mm. The inside
radius of the curved bar is r i = 50 mm.
r o
B
M
x
B
70 mm
(a) Determine the bending stresses in the curved bar at points
A and B.
(b) Sketch the distribution of flexural stresses in the curved
bar.
(c) Determine the percent error if the flexure formula for
straight beams were used for part (a).