• Determine the horizontal force per unit length or shear flow q on the lower surface of the upper plank. • Calculate the corresponding shear[r]
(1)MECHANICS OF MATERIALS
Ferdinand P Beer
E Russell Johnston, Jr. John T DeWolf
Lecture Notes: J Walt Oler
Texas Tech University
CHAPTER
(2)Shearing Stresses in Beams and Thin-Walled Members
Introduction
Shear on the Horizontal Face of a Beam Element Example 6.01
Determination of the Shearing Stress in a Beam Shearing Stresses τxy in Common Types of Beams
Further Discussion of the Distribution of Stresses in a Sample Problem 6.2
Longitudinal Shear on a Beam Element of Arbitrary Shape Example 6.04
Shearing Stresses in Thin-Walled Members Plastic Deformations
Sample Problem 6.3
Unsymmetric Loading of Thin-Walled Members Example 6.05
(3)Introduction
( )
( )
0
0
0
= ∫ −
= =
∫ =
= ∫
= −
= ∫
=
=
∫ −
= =
∫ =
x z
xz z
x y
xy y
xy xz
x x
x
y M
dA F
dA z
M V
dA F
dA z
y M
dA F
σ τ
σ τ
τ τ
σ
• Distribution of normal and shearing stresses satisfies
• Transverse loading applied to a beam results in normal and shearing stresses in transverse sections
• When shearing stresses are exerted on the vertical faces of an element, equal stresses must be exerted on the horizontal faces • Longitudinal shearing stresses must exist
(4)Shear on the Horizontal Face of a Beam Element
• Consider prismatic beam
• For equilibrium of beam element
( )
∫ −
= ∆
∑ = = ∆ + ∫ −
A C D
A
D D
x
dA y I
M M
H
dA H
F σ σ
x V x dx dM M
M
dA y Q
C D
A
∆ = ∆ =
− ∫ =
• Note,
flow shear
I VQ x
H q
x I VQ H
= =
∆ ∆ =
∆ =
∆
(5)Shear on the Horizontal Face of a Beam Element flow shear I VQ x H
q = =
∆ ∆ =
• Shear flow,
• where section cross full of moment second above area of moment first ' = ∫ = = ∫ = +A A A dA y I y dA y Q
• Same result found for lower area
(6)Example 6.01
SOLUTION:
• Determine the horizontal force per unit length or shear flow q on the lower surface of the upper plank • Calculate the corresponding shear
force in each nail A beam is made of three planks,
nailed together Knowing that the spacing between nails is 25 mm and that the vertical shear in the beam is
(7)Example 6.01 ( )( ) ( )( ) ( )( ) ( )( ) 12 12 m 10 20 16 ] m 060 m 100 m 020 m 020 m 100 [ m 100 m 020 m 10 120 m 060 m 100 m 020 − − × = × + + = × = × = = I y A Q SOLUTION:
• Determine the horizontal force per unit length or shear flow q on the lower surface of the upper plank
m N 3704 m 10 16.20 ) m 10 120 )( N 500 ( -3 = × × = = − I VQ q
• Calculate the corresponding shear force in each nail for a nail spacing of 25 mm
m N q
F =(0.025m) = (0.025m)(3704
(8)Determination of the Shearing Stress in a Beam
• The average shearing stress on the horizontal face of the element is obtained by dividing the shearing force on the element by the area of the face
It VQ
x t
x I
VQ A
x q A H
ave
=
∆ ∆ =
∆ ∆ = ∆ ∆ =
τ
• On the upper and lower surfaces of the beam,
τyx= It follows that τxy= on the upper and lower edges of the transverse sections
• If the width of the beam is comparable or large relative to its depth, the shearing stresses at D1
(9)Shearing Stresses τxy in Common Types of Beams
• For a narrow rectangular beam,
A V
c y A
V Ib
VQ
xy
2
1
3
max
2
=
⎟ ⎟ ⎠ ⎞ ⎜
⎜ ⎝ ⎛
− =
= τ τ
• For American Standard (S-beam) and wide-flange (W-beam) beams
web ave
A V It VQ
= =
max
(10)Further Discussion of the Distribution of
Stresses in a Narrow Rectangular Beam
⎟ ⎟ ⎠ ⎞ ⎜
⎜ ⎝ ⎛
− =
2
1
3
c y A
P
xy τ
I Pxy
x = + σ
• Consider a narrow rectangular cantilever beam subjected to load P at its free end:
• Shearing stresses are independent of the distance from the point of application of the load
• Normal strains and normal stresses are unaffected by the shearing stresses
• From Saint-Venant’s principle, effects of the load application mode are negligible except in immediate vicinity of load application points