Principal Stresses in Beams
Let us consider the beam in Figure 3-1 Sc. The bending and shear stress
distributions over an arbitrary section a-a are shown in Figures.
In general, at an arbitrary section a-a along the beam's axis, the internal
shear V and moment M are developed by a parabolic shear- stress
distribution, and a linear normal-stress distribution. If these results are
applied to specific elements located at points I through 5 along the section,
the stresses acting on these elements will be as shown in Fig. In particular,
elements I and 5 are subjected only to the maximum normal stress, whereas
element 3, which is on the neutral axis, is subjected only to the maximum
shear stress.
1
2
3
4
5
4
4
5
5
4
4
3
3
2
2
1
1
The intermediate elements 2 and 4 resist both normal and shear stress. In each
case the state of stress can be transformed into principal stresses, using either
the stress-transformation equations or Mohr's circle. The results are shown in
Fig. If this analysis is extended to many vertical sections along the beam other
an a-a, a profile of the results can be represented by curves called stress
trajectories.
Pattern of stress trajectories of a beam in bending derived optically,
Stress Trajectories:
A knowledge of the directions of the principal stresses may aid in the prediction of the
directions of cracks in a brittle material (concrete, for example) and thus may aid in
the design of reinforcement to carry the tensile stresses. Curves drawn with their
tangents at each point in the directions of the principal stresses are called stress
trajectories.
The lines intersect the neutral axis at 45° angles, and the solid and dashed lines always
intersect at 90°. Knowing the direction of these lines can help engineers decide where
to reinforce a beam so that it does not crack or become unstable.
Since there are, in general, two nonzero principal stresses at each point (plane
stress), there are two stress trajectories passing through each point.
These curves will be perpendicular since the principal stresses are orthogonal;
one set of curves will represent the maximum stresses, whereas the other
represents the minimum stresses.
Fig 6.26 Stress Trajectories for Rectangular Beam
Shear failure of reinforced concrete beam: (a) overall view; (b) detail near right support
The trajectories for a simply supported rectangular beam carrying a concentrated
load at the midpoint are shown in Fig. 6-26, with dashed lines representing the
directions of the compressive stresses and solid lines showing tensile stress
directions. In the vicinities of the load and reactions there are stress concentrations,
and the trajectories become much more complicated. Figure 6-26 neglects all stress
concentrations.
In order to determine the principal stresses and the maximum shearing stresses
at a particular point in a beam, it is necessary to calculate the fiber stresses and
transverse (or longitudinal) shearing stresses at the point.
With the stresses on orthogonal planes through the point known, the methods \can be
used to calculate the maximum stresses at the point. The following example illustrates
the procedure and provides an example-in which the principal stresses at some interior
point are greater than the maximum fiber stresses.
Stress Trajectories for cantilever Beam
Problem:
A simply supported beam of 2ft. span in subjected to a load of 60,000 lb at its centre
the cross-section is show in Fig. Find the maximum normal and shearing stresses at
the junction of the flange and web.
P = 60,000 lb
1’
1’
R2 30,000 lb
R1 30,000 lb
+30,000
+
SFD
¯
-30,000
+30,000 lb-ft
BMD
+
 5  0.75 


 2 
C1 = 6"
C2 = 6"
 5  0.75 
4
 0.18  (5  .75)
  119.04 in
 2 
2
I1 N . A
I 2 N . A  48.23  0  48.23 in 4
I 3 N . A  I1N . A  119.04 in (due to symmetry)
4
I N . A  119.04  48.23  119.04
I N . A  286.31 in 4
Vmax  30,000 lb
M max  30,000 lb  ft
Flexural Stress or Bending Stress  x at upper or lower fiber
 max
MC1 30,000 12  6


 7544.27 Psi
I
286.33
 x the at the junction
My
x 
I
30000  12  5.25
x 
286.33
 x  6601 Psi
5.625
5.25/2
5.25
Shear Stress by shear formula
VQ
V 
bI
 NA

30,000
5.25


 5  0.755.625  0.55.25
286.31 0.5 
2 
 NA  5871.17 Psi
At Junction
30,000
 xy 
 5  0.755.625
286.31 0.5
 4424 psi
4424 Psi
5871.17 Psi
Then the principal stress is
1 
As
 x  y
2
  x  y 
   xy 2
 
 2 
2
y 0
2
6601
 6601 
2
1 
 
  4424
2
 2 
 1  8820 psi
Ans.
 max  5519.52 psi
Ans.
It is seen that (1) the principal stress at the junction between flange and web is larger than the
tensile stress in the most remote fiber (1 > max) and therefore should be considered in
design.
Problem:
In the previous problem Find the span length (l) such that the maximum Principal
stress at the junction between flange and the web will be equal to the maximum
flexural stress in the extreme fiber.
 1   x.max
As by Condition:
 max
 max
MC Plc Pl 6
Pl


   1. 5
I
4I
4 I
I
6
1.5 PL

6I
x

5.25
x
5.25
Pl
 x  1.312
I
VQ P 3.75  5.625
P
 xy 
 
 21.1
Ib 2 I
(0.5)
I
(1)
max
(5.25)"
x
6"
6"
1 
 x  y
2
  x  y 
   xy 2
 
 2 
2
y 0
2
Pl
 Pl 
P
 1  0.656  0.430   445.2 
I
 I 
I
2
Equating (1) and (2)
2
2
Pl
Pl
 Pl 
P
0.656  0.430   445.2   1.5
I
I
 I 
I
0.28l 2  445.2;
l 2  1580;
l  39.8 in.
(2)
Assignment
A cantilever beam loaded as shown in Figure a. Find The maximum
normal stresses and shearing stress in the beam at the junction of
the flange and the web.