section subjected to a concentrated load P at its free end A (Fig. 6.32). The

largest bending moment occurs at the fixed end B and is equal to M = PL.

As long as this value does not exceed the maximum elastic moment MY

(i.e., PL ≤ MY), the normal stress σx will not exceed the yield strength σY

anywhere in the beam. However, as P is increased beyond MY∕L, yield is

initiated at points B and B′ and spreads toward the free end of the beam.

Assuming the material is elastoplastic and considering a cross section CC′

located a distance x from the free end A of the beam (Fig. 6.33), the

half-thickness yY of the elastic core in that section is obtained by making

M = Px in Eq. (4.38). Thus,

Px =

32

MY(1 - 1 3 yc2 Y2) (6.14)

where c is the half-depth of the beam. Plotting yY against x gives the boundary between the elastic and plastic zones.

As long as PL < 3 2MY, the parabola from Eq. (6.14) intersects the

line BB′, as shown in Fig. 6.33. However, when PL reaches the value 3 2MY

(PL = Mp) where Mp is the plastic moment, Eq. (6.14) yields yY = 0 for

x = L, which shows that the vertex of the parabola is now located in section BB′ and that this section has become fully plastic (Fig. 6.34). Recalling Eq. (4.40), the radius of curvature ρ of the neutral surface at that point

is equal to zero, indicating the presence of a sharp bend in the beam at

its fixed end. Thus, a plastic hinge has developed at that point. The load

P = M

p∕L is the largest load that can be supported by the beam.

The above discussion is based only on the analysis of the normal

stresses in the beam as it exhibits plastic behavior. Now examine the distribution of the shearing stresses in a section that has become partly plastic. Consider the

portion of beam CC″D″D located between the transverse sections CC′ and DD′

and above the horizontal plane D″C″ (Fig. 6.35a). If this portion is located

entirely in the plastic zone, the normal stresses exerted on the faces CC″ and

DD″ will be uniformly distributed and equal to the yield strength σY (Fig. 6.35b).

The equilibrium of the free body CC″D″D requires that the horizontal shearing

force ΔH exerted on its lower face is equal to zero. The average value of the

horizontal shearing stress τyx across the beam at C″ is also zero, as well as the

average value of the vertical shearing stress τxy. Thus the vertical shear V = P

in section CC′ must be distributed entirely over the portion EE′ of the section

located within the elastic zone (Fig. 6.36). The distribution of the shearing

stresses over EE′ is the same as that in an elastic rectangular beam with the

same width b as beam AB and depth equal to the thickness 2yY of the elastic

zone.† The area 2byY of the elastic portion of the cross section A′ gives

τ xy = 32 P A′ (1 - yy2 Y2) |
(6.15) |

τ

max =

32

P A′

(6.16)

As the area A′ of the elastic portion of the section decreases, τmax

increases and eventually reaches the yield strength in shear τY. Thus, shear

contributes to the ultimate failure of the beam. A more exact analysis of this

mode of failure should take into account the combined effect of the normal

and shearing stresses.