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

Superposition
Often, curved bars are loaded so that there is an axial force, as well as a moment, on the cross
section, as is illustrated in the next example. The normal stress given by Equation (8.28) may
then be algebraically added to the stress that is due to an internal axial force F. For this
simple case of superposition, the total stress at a point located a distance r from the center
of curvature, O, may be expressed as
313
bENdINg OF CuRVEd bARS
F Mr ( n − r)
σ x = −
A rAr ( − r )
c
n
(8.29)
A positive sign would be associated with a tensile internal force F, and a negative sign
would be used for compressive internal forces. Note that the internal bending moment M
produced by an external load P is calculated as the product of (the magnitude of) the force
and the perpendicular distance between the line of action of P and the centroid of the cross
section, not the neutral-axis location of the cross section.
ExAmpLE 8.14
The allowable stress for the clamp shown is 90 MPa. Use
the following dimensions for the clamp:
b
b
b
r
i
= 125 mm d = 25 mm
1 1
= 25 mm d = 75 mm
2 2
= 50 mm d = 25 mm
3 3
= 150 mm a = 20 mm
Determine the maximum permissible load P that the
member can resist.
B
A
r i
O
a
P
P
d3
d2
d1
b3
B (2) A b1
(3)
b 2 (1)
Section A-B
Plan the Solution
Begin by calculating the radial distances from the center of curvature, O, to the centroid
of the cross section and to the neutral-axis location. Next, determine an expression for
the bending moment in terms of the unknown load P and the perpendicular distance
between the line of action of P and the centroid of the curved-bar cross section. The
normal stresses at Section A–B will consist of both axial and bending stresses. Write an
equation, similar to Equation (8.29), for the sum of the stresses in terms of the unknown
load P, and use this equation to investigate the stresses at points A and B. At point A on
the inner surface of the curved bar, both the axial and bending stresses will be tensile
normal stresses. Set the sum of these two stresses equal to the specified allowable
stresses, and solve for the allowable magnitude of load P. At point B, the axial stress
will be a tensile normal stress while bending stresses will be compressive. Again, set
the sum of the axial and flexural normal stresses equal to the allowable stress and solve
for P. The smaller value obtained for the load P from these two calculations is the largest
load that can be applied to the clamp.