Shearing stresses in Beams & Thin-walled Members .

CH. 6

6.1 Transverse loading applied to abeam will result in normal and shearing stresses in any given transverse section of the beam .The normal stresses are created by the bending couple “M” and the shearing stresses by the shear “V” .

x

y

z

𝜏 (𝑥𝑦 )𝑑𝐴

𝜏 (𝑥𝑧 )𝑑𝐴

𝜎 (𝑥 )𝑑𝐴x

y

z

V

M

Y components: = -VZ components: = 0

The first of these equations shows that vertical shearing stresses must exit in a transverse section of a beam under transverse loading .The second equation shows that the average horizontal shear stresses in the section = 0So ,we conclude that shearing stresses in the horizontal plan = 0 = 0 (in the horizontal plan) .

• A force P is applied .So ,planks are observed to slide with respect to each other .

• M is applied ,no shear happens ,no slide planks .

P

(b)

(a)

M

• We call shearing force in horizontal face

in the direction shown before in x .

• q is the shear per unit length “shear

flow” .

6.2 Shear in the horizontal face of a beam element .

If we took an element

w

P1 P2

x

y

z

y1 c

y

z

Δ𝑥y1

• Vertical shearing forces and ,a horizontal shearing force

= 0 + - )dA = 0, dA is the sheared area

w

Δ𝐻𝜎𝐷 𝑑𝐴𝜎𝐶𝑑𝐴

𝑉 ′𝐶❑𝑉 ′ 𝐷

❑

, y = From the former equation . = , is the first moment with respect to the neutral axis of the portion located at “y” . = Q, = = () = V = , q (shear flow) = I is the moment of inertia , Q = A y’

6.3 Determination of the shearing

stresses in a beam .

= =

=

The average shearing stresses in

the horizontal stress .

∆𝐻 dA

∆ 𝑥

6.4 Shearing stresses in common types of beams .• In common types at which bWhere b is the width of the beams , h is the depth . = 0.8 % . = t = L

b or t

h

• For a rectangular cross section area ,shear stresses in x-y plane (horizontal plane) .

First we get Q (the 1st moment of the section) .Q = Ay’ = b(c-y)() = b(), = = = = = = , A = 2bc (total area) = (1 -When y = 0 =

Zy𝑦

b

hC=

• For beam .

=

, = tb

t bZ

y