Centroid of Area (Center of Gravity of an area): Point...
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Centroid Centroid or center of gravity is the point within an object from which the force of gravity appears to act. Centroid of 3D objects often (but not always) lies somewhere along the lines of symmetry. Centroidal axis or Neutral axis A A x x A A y y A A ∫ ∫ = = δ δ Q x A y A y A ⋅ = = ∫ δ Ib VQ = τ Centroid of Area (Center of Gravity of an area): Point that defines the geometric center of the area ( ) y x , First Moment of an Area with respect to the x-axis First Moment of an Area with respect to the y-axis Q y A x A x A ⋅ = = ∫ δ
Transcript of Centroid of Area (Center of Gravity of an area): Point...
CentroidCentroid or center of gravity is the point within an object from which the force of gravity appears to act.Centroid of 3D objects often (but not always) lies somewhere along the lines of symmetry.
Centroidal axisor Neutral axis
A
Axx
A
Ayy AA
∫∫==
δδ
Q x AyAyA
⋅== ∫ δ
IbVQ
=τ
Centroid of Area (Center of Gravity of an area):Point that defines the geometric center of the area
( )yx,
First Moment of an Area with respect to the x-axis
First Moment of an Area with respect to the y-axis
Qy AxAxA
⋅== ∫ δ
Calculate the center of gravity of the rectangle:(a)Without a hole(b)With a hole of dimensions c and d
a) Without a hole
22
22
20
20
aabba
ab
xax
A
Axx
bab
abab
yay
A
Ayy
a
A
b
A
====
====
∫∫
∫∫
δδ
δδ
b) With a hole
dcba
dccebaa
xA
Axx
A
Axx
dcba
dcdfbab
yA
Ayy
A
Ayy
n
iiA
n
iiA
⋅−⋅
⋅⋅−−⋅⋅⎟⎠⎞
⎜⎝⎛
===
⋅−⋅
⋅⋅+−⋅⋅⎟⎠⎞
⎜⎝⎛
===
∑∫
∑∫
)2
(2
)2
(2
1
1
δ
δ
The centroid of any area can be found by taking moments of identifiable areas (such as rectangles or triangles) about any axis. The moment of an area about any axis is equal to the algebraic sum of the moments of its component areas.
Moment of Inertia
Second Moment or Moment of Inertia of an Area
∫∫ ==A
yA
x AxIAyI δδ 22
∫ +==A
yxO IIAJ δρ 2
Rectangular Moments of Inertia: Moments of inertia with respect to an axis
Polar moments of inertia: Moment of inertia with respect to a point
also known as the Second Moment of the Area is a term used to describe the capacity of a cross-section to resist bending. It is a mathematical property of a section concerned with a surface area and how that area is distributed about the reference axis. The reference axis is usually a centroidal
Simple rectangular shape
⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−−===
−−∫ 8833
332
2
32
2
2 hhbybbdyyI
h
h
h
hx
2 bdydAdAyIA
x == ∫
2 hdxdAdAxIA
y == ∫
⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−−===
−−∫ 8833
332
2
32
2
2 bbhxhhdxxI
b
b
b
by
12
3bhIx = 12
3hbI y =
The polar moment of inertia is an important parameter in problems involving torsion of cylindrical shafts and rotations of slabs.
∫= dArJ 20
The polar moment of inertia is related to the rectangular moments of inertia,
( ) ∫∫∫∫ +=+== dAydAxdAyxdArJ 222220
Polar Moment of Inertia
Simple rectangular shape
12
3bhIx =
12
3hbI y =( )22
33
12
1212
bhbhI
hbbhIII
P
yxPolar
+=
+=+=
xy IIJ +=0
Moment of Inertia: Parallel Axis Theorem for an AreaIf the moment of inertia of an area is known about its neutral axis (centroidaxis), we can determine the moment of inertia of area about a corresponding parallel axis using the parallel axis theorem
Consider moment of inertia of the shaded area
( )
∫∫∫
∫
++=
+=
dAdAdAx
dAx
dAdydAddAyI
dAydI
211
2
21
2
First integral represents the moment of inertia of the area about the centroidal axisSecond integral is equal to zero, since xc passes through the area’s centroid CThird integral represents the total area A
21dAII xcx ⋅+=
22dAII ycy ⋅+=Similarly:
Simple rectangular shape
12
3bhIx =3
41223
332
bhI
bhbhhAII
BB
xBB
=
+=⎟⎠⎞
⎜⎝⎛+=
Note:
Radius of Gyration2222
22
yxOOO
yyxx
rrrArJ
ArIArI
+==
==
Moment of Inertia of Composite Areas
Determine the moment of inertia of the area shown with respect to the x-axis
X-axis
The moment of inertia of the area shown can be obtained by sustracting the circle from the rectangle
( ) ( ) ( ) ( ) 46224
2'
104.1175252541 mm
AdII yxx
=+=
+=
ππ
( )( ) ( )( )( ) ( ) 4623
2'
105.11275150100150100121 mm
AdII yxx
=+=
+=
Circle
Rectangle
4666 10101104.11105.112 mmI x ×=×−×=
Determine the moment of inertia of the area shown with respect to the centroids.
( )( ) ( )( )( )
( )( ) ( )( )( ) 4923
2'
4923
2'
1090.1250300100100300121
10425.1200300100300100121
mm
AdII
mm
AdII
xyy
yxx
×=+=
+=
×=+=
+=
Rectangle A
( )( )
( )( ) 493
2'
493
2'
8010.1600100121
1005.0100600121
mm
AdII
mm
AdII
xyy
yxx
==
+=
×==
+=
Rectangle B
( )( ) ( )( )( )
( )( ) ( )( )( ) 4923
2'
4923
2'
1090.1250300100100300121
10425.1200300100300100121
mm
AdII
mm
AdII
xyy
yxx
×=+=
+=
×=+=
+=
Rectangle D
( )
( )49
9
49
9
1060.5
1090.180.190.1
1090.2
10425.105.0425.1
mmI
I
mmI
I
y
y
x
x
×=
×++=
×=
×++=
Adding each segment
Determine the moment of inertia of the area shown with respect to the x-axis centroid.
Bodies Ai yi yi*Ai Ii di=yi-ybar di2Ai
1 21600 90 1944000 58320000 0 02 -9600 90 -864000 -11520000 0 0
12000 1080000 46800000 0
ybar 90 mm I 46800000 mm4
-
mmrmm
mmAIr
x
xx
45.6212000
108.462
46
=
×==Radius of Gyration
Determine the moment of inertia of the area shown with respect to the x-axis centroid.
Moments of Inertia of Composite Areas
Moments of Inertia of Composite Areas
Determine the moment of inertia and radius of gyration with respect to an axis which is parallel to the plate and passes through the centroid of the section.
The strength of a W14x38 rolled steel beam is increased by attaching a plate to its upper flange.
Determine location of the centroid of composite section with respect to a coordinate system with origin at the centroid of the beam section.
12.5095.170011.20Section Beam
12.50425.76.75Platein ,in. ,in ,Section 32
== ∑∑ AyA
AyyA
in. 792.2in 17.95in 12.50
2
3====
∑∑∑∑ A
AyYAyAY
Apply the parallel axis theorem to determine moments of inertia of beam section and plate with respect to composite section centroidal axis.
( )( )
( )( ) ( )( )4
2343
1212
plate,
4
22sectionbeam,
in 2.145
792.2425.775.69
in3.472
792.220.11385
=
−+=+=
=
+=+=
′
′
AdII
YAII
xx
xx
2.1453.472plate,section beam, +=+= ′′′ xxx III
4in 618=′xICalculate the radius of gyration from the moment of inertia of the composite section.
inAIr x
x 87.5in 17.95in 5.617
2
4
=== ′′
Relationship between Bending Moments and
Curvatures
EIM
==ρ
κ 1
Relationship between Bending Moments and
Normal Stresses
IMy
x −=σ
2
22
1
11
SM
IMc
SM
IMc
==
−=−=
σ
σ
Maximum stresses
S1 and S2 are known as the section moduli.
Stresses in a simple beam.
( )( )( ) ( )( )( ) ( )( )
KipsRKipsR
RM
RRF
BA
AB
BAy
41.21 59.23
09221211225.1220
125.1220
@
==
=−−−==
+−−==
∑∑
CqdxVqdxdV
x =−==−= ∫ 0
( )
( )
( ) ( ) ( ) ftKipsMKipsVKipsVxxM
CMCxxM
xVRCRVCxVx
x
x
x
AAx
−=−==+−=
=⇒=⇒++−=
+−=
=⇒=⇒+−=≤≤
+− 6.151 91.1 09.1059.2375.0
0059.2375.0
59.235.15.1 90
999
2
2022
101
Rectangular Cross Section( )( )
3
43
10632
2714352
1435212
2775.8
incIS
inIz
===
==
( )
( )
( ) ( ) ftKipsMKipsVxxM
CMCxxM
xVCVCxVx
x
x
x
x
−==++−=
=⇒=⇒++−=
+−=
=⇒−=⇒+−=≤≤ +
6.151 41.2102.10859.1175.0
02.108059.1175.0
59.115.159.1191.15.1 229
922
2
42242
393
ftkipsM Max −= 6.151
psiS
M MaxMax 1710
10631210006.151
=××
==σ
A cast-iron machine part is acted upon by 3 kN-m Knowing E = 165GPa and neglecting the effects of fillets, determine (a) the maximum tensile and compressive stresses, (b) the radius of curvature.
(a) Based on the cross section geometry, calculate the location of the section centroidand moment of inertia.
(b) Apply the elastic flexural formula to find the maximum tensile and compressive stresses.
(c) Calculate the curvature
(a) Based on the cross section geometry, calculate the location of the section centroidand moment of inertia.
(b) Apply the elastic flexural formula to find the maximum tensile and compressive stresses.
3649
3649
108.22038.010868 038.0
1045.39022.010868 022.0
mm
mcISmc
mm
mcISmc
B
zBB
A
zAA
−−
−−
×=×
===
×=×
===
MPammN
SM
MPammN
SM
BB
AA
3.131108.22
3000
0.76105.394
3000
36
36
−=⋅
⋅−==
=⋅
⋅−−=−=
−
−
σ
σ
(c) Calculate the curvature
( )( )m
mmPa
mNEIM
7.47
0209.01086810165
30001 1499
=
=⋅⋅
⋅== −
−
ρρ
Stresses in a beam with an overhang.Cross Section
Reactions ( )( )( )
( )( ) kNR
kNR
A
B
6.38.105.42.3
8.103
25.45.42.3
=−=
=−
=
( )
( )
( ) ( )
( ) ( ) 6.3 025.2 8.4 0.6
6.36.1
006.36.1
125.12.36.306.32.3
6.36.32.3 30
0.3125.1
33
2
2022
101
mkNMmkNMkNVkNV
xxM
CMCxxM
mxVxV
kNCkNVCxVx
x
x
xx
x
−−=−=
=−=+−=
=⇒=⇒++−=
==⇒=⇒+−=
=⇒=⇒+−=≤≤
+−
( )
( )
4.324.146.1
4.326.34.146.1
4.142.34.148.42.3 5.43
2
4342
333
−+−=
−=⇒−=⇒++−=
+−=
=⇒=⇒+−=≤≤
xxM
CMCxxM
xVkNCkNVCxVx
x
x
x
x
( ) ( ) 6.3 025.2 0.3125.1 mkNMmkNM −−=−=Maximum Bending Moments
34
222
34
111
4012952.61
2468761 52.61
13359148.18
2468761 48.18
mmmmmm
cISmmc
mmmmmm
cISmmc
z
z
====
====
MPam
mNSM
MPam
mNSM
mx
5.5010401.0
2025
2.1510336.1
2025125.1
332
2
331
1
=×
−==
−=×
−−=−=
=
−
−
σ
σ
MPammN
SM
MPammN
SMmx
8.8910401.0
3600
9.2610336.1
36000.3
332
2
331
1
−=×
−−==
=×
−−−=−=
=
−
−
σ
σ
Maximum Tensile Stress = 50.5MPa at x=1.125m
Maximum Compressive Stress = 89.8MPa at x=3.0m
Which type of cross section is the most efficient in resisting the bending stresses??
Consider beams made of the same material, subjected to the same moment and with similar cross section areas.
Allowed
MSσ
=
4
2dA π=Circular cross section
334
098175.0322
64
ddd
ISdI circle ====ππ
Square cross section dhdhA 886.04
22 =⇒==
π
33
44
6955.062
12 12
dhh
hShI square ====
dnhdnhbhA
24
22 ππ=⇒===Rectangular cross section
h
b=h/n
3
3
3
33
4
tan
4
367.0 10164.0 2116.0 1
116.062
12 12
dSndSndSn
dnn
hh
nh
Sn
hI glerec
==
==
==
====
Design of a post using solid wood or aluminum tube.
Solid wood: σAllowed=15MPa
Aluminum tube: σAllowed=50MPa
( )( ) mkNmkNPhM Max −=== 305.212
Solid wood:
mmd
mmMdSAllowed
MaxWood
273
10232
1
363
=
⋅===σ
π
Aluminum tube
( )[ ]
mmd
mmMd
IS
dtddI
Allowed
Max
208
106002
03356.064
2
33
22
42
42
42
=
⋅===
=−−=
σ
π
Bending Members Made of Several Materials
Consider a composite beam formed from two different materials E1 and E2.
The normal strain varies linearly. ρ
ε yx −=
Piecewise linear normal stress variation
ρεσ
ρεσ
yEE
yEE
x
x
222
111
−==
−==Elemental forces on the section areNeutral axis does not pass through
section centroid of composite section.
dAyEdAdF
dAyEdAdF
ρσ
ρσ
222
111
−==
−==
Define a transformed section such that
( )ndAyEdAynEdAyEdFρρρ112
2 −=−=−=
1
2
EEn =
A bar is made from bonded pieces of steel (ESteel=29000ksi) and brass (Ebrass=15000ksi). Determine the maximum stress in the steel and brass when a moment of 40kips-in is aplied.
(a) Transform the bar to an equivalent cross section made entirely of brass
933.11500029000
===ksiksi
EEn
Brass
Steel
( )( ) ininininbT 25.24.075.0933.14.0 =++=
Evaluated the transformed cross sectional properties
( )( ) 433
063.512
325.212
inhbI T ===
Calculated the maximum stresses
( ) ( )( ) ksiin
ininkipsI
McMaxBrass 85.11
063.55.140
4 =⋅
=−=σ
( ) ( ) ksin MaxBrassMaxSteel 9.2285.11933.1 =×== σσ
Shear Stresses in Beams of Rectangular Cross Section
Note that at the top or at the bottom of the beam the horizontal stresses must vanish.
xx V
dxdM
= The change of moment with the distance xgenerates a shear force.
Glued beam: When it is loaded horizontal shear stresses must develop along the glued surface in order to prevent the sliding.
( )I
ydMMI
My
+−=
−=
2
1
σ
σ
Isolate a subelement mm1p1p and find the forces acting (assuming equilibrium).
( )
( )
∫∫
∫∫
∫ ∫
∫ ∫
==
−+
=−=
+==
==
ydAI
dMdAI
dMyF
dAI
MydAI
ydMMFFF
dAI
ydMMdAF
dAI
MydAF
3
123
22
11
σ
σ
bdxF τ=3
∫∫
∫==
=
ydAIbVydA
dxdM
Ib
ydAI
dMbdx
1τ
τ IbVQ
=τ
This equation is known as the shear formula.V, I (for the entire section) and b (width at y1) are constants with y, while Q varies with y.
Distribution of Shear Stresses acting on a BeamThe first moment Q of the shaded part of the cross-sectional area is obtained by multiplying the area by the distance from its centroid to the neutral axis.
⎟⎟⎠
⎞⎜⎜⎝
⎛−=⎥
⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ −+⎥
⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ −== 2
2
1111 142221
2yhbyhyyhbyAQ Shaded
For y1=h/2 then Q=0For y1=0 then Q is maximum
⎟⎟⎠
⎞⎜⎜⎝
⎛−=
⎟⎟⎠
⎞⎜⎜⎝
⎛−==
22
22
1
1
42
42
yhI
V
yhbIbV
IbVQ
τ
τParabolic distribution
( )( )
0 0 2
23
12
8 8
0
0 0 2
2
3
22
2
=⇒=⇒−=
==⇒=⇒=
=⇒=⇒=
⎟⎠⎞
⎜⎝⎛ −
⎟⎠⎞
⎜⎝⎛
h
MaxMax
h
Qhy
AV
bbh
bhVbhQy
Qhy
τ
τ
τ
AV
Max 23
=τ
Shear FlowThe shear flow represents the force over a unit length of the beam that would be required to hold the beam together. If the beam started out as two pieces separated along a horizontal line, and the two pieces were welded together, the strength of the weld would have to be at least equal to the shear flow. It wouldrepresent the required strength of a unit length of the weld. It can also be used to determine how many nails are required to hold together two pieces of a fabricated wooden beam.
bdxF τ=3
xVb
IVQqFlowShear
IbVQydA
IbVydA
dxdM
Ib
ydAI
dMbdx
Horizontal
Δ====
===
=
∫∫
∫
τ
τ
τ
_
1
From the equation:
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 V = 500 N, determine the shear force in each nail.
Calculating Q:
( )[ ]( ) 361012006.01.002.0 mmmmyAQz−×===
To determine Q, cut the cross section horizontally at the NA and choose either the top or the bottom portion, it doesn't matter which because the same answer will result using either portion.
Calculation of the moment of inertia about the centroid( )( ) ( )( )( ) ( )( )
46
32
3
1020.16
1202.01.0206.01.002.0
121.002.0
mI
I
z
z
−×=
⎥⎦
⎤⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡+=
mN3704
m1016.20)m10120)(N500(
46-
36
=
××
==−
qI
VQq
Calculating the shear flow (q): Calculating the shear force per nail for a nail spacing of 25 mm.
mNqF 3704)(m025.0()m025.0( ==
N6.92=F
Determination of the Shearing Stress in a Beam
On the upper and lower surfaces of the beam, τyx= 0. It follows that τxy= 0 on the upper and lower edges of the transverse sections.
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.
ItVQ
xtx
IVQ
Axq
AH
ave
ave
=
ΔΔ
=ΔΔ
=ΔΔ
=
τ
τ
For a narrow rectangular beam,
AV
cy
AV
IbVQ
xy
23
123
max
2
2
=
⎟⎟⎠
⎞⎜⎜⎝
⎛−==
τ
τ
For American Standard (S-beam) and wide-flange (W-beam) beams
web
ave
AV
ItVQ
=
=
maxτ
τ
Shearing Stresses in Common Beam Types
The diagram shows a simply supported 20 ft. beam with a load of 10,000 lb. acting downward at the center of the beam. The beam used is a rectangular 2" by 4" steel beam. We would like to determine the maximum bending (axial) stress and the maximum shear stress
Reactions:
lbC
lbA
y
y
5000
5000
=
=
( )( )psi
inI
inyftlbM
Max
Max
112500
67.1012
422
50000
43
=
==
=−=
σ
( )( ) psiAV
inAlbV
Max 5.93723
8425000
2 ====
=τ
Determine the normal and shear stresses at point C.
Rectangular section
( )( )
( )( ) ( )( ) ( ) inlbM
lbVlbRlbR
C
C
BA
⋅=−
−−−=
−=−−===
179202
8368361608362880
160083616028802880 2880
( )( ) 433
333.512
0.40.112
inbhI ===
( )( ) psiin
ininlbI
yMCC 3360
333.50.117920
4 −=⋅
−=−=σ
( )( )( )( ) psi
inininlb
IbQV CC
C 4500.1333.5
5.116004 ===τ
( )( )( ) 35.15.10.10.1 ininininyAQ CCC ===
Determine the maximum permissible value PMax if the allowable stresses for bending and shear are 11MPa and 1.2MPa respectively. Data : h=150mm ; b=100mm and a=0.5m
bhAbhh
bh
cIS
PaMPV MaxMax
====
==
62
12
23
32
23
23
66 2
2
bhPbhP
AV
abhP
bhPa
SM
AllowedShearMax
MaxMax
AllowedBendingMax
MaxMax
ττ
σσ
=⇒==
=⇒==
kNPMPa
kNPMPaShear
MaxAllowed
BendingMaxAllowed
0.122.1
25.811
=⇒=
=⇒=
τ
σ
The bending stress governs the design.
A timber beam is to support the three concentrated loads shown. Knowing that for the grade of timber used,
psi120psi1800 == allall τσ
determine the minimum required depth d of the beam.
inkip90ftkip5.7kips3
max
max⋅=⋅=
=MV
Determine the maximum shear force and moment.
( )
( ) 2
2612
61
3121
in.5833.0
in.5.3
dS
ddbcIS
dbI
=
===
=Determine the section moduli
Determine the beam depth based on allowable normal stress.
( )in.26.9
in.5833.0in.lb1090psi 1800 2
3max
=
⋅×===
ddS
Mallσ
Determine the beam depth based on allowable shear stress.
( )in.71.10
in.3.5lb3000
23psi120
23 max
=
=
=
dd
AV
allτ
Required beam depth is equal to the larger of the two. in.71.10=d
Shear Stresses in Beam of Circular Sectionφφ CosrxSinry ⋅=⋅=
yCosrdA δφ ⋅⋅⋅= 2
δφφδ ⋅⋅= Cosry
2
4
42
44
2
2
4224
rIIJrI
rCosSinrI
yxzy
x
⋅=+=
⋅=
⋅=⋅⋅⋅⋅= ∫−
ππ
πδφφφπ
π
( )( )
( )( )
( ) ( )3
23
22
2
.2
32
0
20
33
23
2
0
23
2
0
rurduurQ
dSinduCosudCosSinrQ
dCosrCosrSinrydAQ
Max
Max
Max
∫
∫
∫∫
=−=−=
−=⇒=⇒=
⋅⋅⋅⋅==
ππ
π
π
φφφφφφ
φφφφ
( )( )( ) A
VrV
rr
rV
IbVQ
Max 34
34
24
32
24 ====ππ
τ
Hollow Circular Cross Section( ) ( )12
31
32 23
2 rrbrrQMax −=⇒−
=
( )
( ) ( )( )⎟⎟⎠
⎞⎜⎜⎝
⎛+++
=−⎟
⎠⎞
⎜⎝⎛ −
⎟⎟⎠
⎞⎜⎜⎝
⎛ −
= 21
22
2112
22
12
41
42
31
32
34
24
32
rrrrrr
AV
rrrr
rrV
Max πτ
( )21
22 rrA −= π
Design of a post using solid wood or aluminum tube.
Solid wood: τAllowed=7.5MPa
Aluminum tube: τAllowed=25MPa
( )( )
mmd
mMPakNVd
dV
AV
Allowed
AllowedMax
1.52
1072.25.73
12163
16
43
434
1
2321
21
=
⋅===
=⎟⎠⎞
⎜⎝⎛
==
−
ππτ
τπ
τ
Solid wood:
Aluminum tube
( )( )
( )
mmd
mMPa
kNVr
r
r
r
Vrr
rrrrAV
Allowed
Max
5.52
10689.025
12436.1436.1
1625
1637
167
13
434
2
322
22
22
22
21
22
2112
22
=
⋅==
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
=⎟⎟⎠
⎞⎜⎜⎝
⎛+++
=
−
τ
πτ
( ) 22
21
22 16
7 rrrA ππ =−=
Shearing Stresses in Thin-Walled Members
Consider a segment of a wide-flange beam subjected to the vertical shear V.
The longitudinal shear force on the element is
xI
VQH Δ=Δ
ItVQ
xtH
xzzx =ΔΔ
≈=ττ
The corresponding shear stress is
NOTE: 0≈xyτ
0≈xzτin the flanges
in the web
Previously found a similar expression for the shearing stress in the web
ItVQ
xy =τ
Shear Stresses in the Webs of Beams with Flanges
The variation of shear flow across the section depends only on the variation of the first moment. I
VQtq ==τ
For a box beam, q grows smoothly from zero at A to a maximum at C and C’ and then decreases back to zero at E.
The sense of q in the horizontal portions of the section may be deduced from the sense in the vertical portions or the sense of the shear V.
For a wide-flange beam, the shear flow increases symmetrically from zero at A and A’, reaches a maximum at C and the decreases to zero at E and E’.
Shearing Stresses in Thin-Walled Members
The continuity of the variation in q and the merging of qfrom section branches suggests an analogy to fluid flow.
Knowing that the vertical shear is 50 kips in a W10x68 rolled-steel beam, determine the horizontal shearing stress in the top flange at the point a.
For the shaded area, ( )( )( )3in98.15
in815.4in770.0in31.4=
=
The shear stress at a, ( )( )( )( )in770.0in394
in98.15kips504
3==
ItVQτ
ksi63.2=τ
Shear Stresses in the Webs of Beams with Flanges
( )( ) ( )[ ]
( )( )[ ] ( )( )[ ]
( )( )
( )⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛ −=
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛ +
=
+=+=
+−+=⇒=
fw
Web
fw
Web
fw
fw
Web
ww
Web
Ahh
ItV
hhAItV
hhbtItVthhb
ItV
thhhhbItVhyFor
222
222
8
28
28
082
_
1
1
11
111
1
τ
τ
τ
τ
( )( ) ( )[ ]
( )( ) ( )( )[ ]
( ) ( )
( ) ( ) ⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛ +⎟⎠⎞
⎜⎝⎛=
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛ +=
++=
+−+=⇒=
221
221
222
2222
8
28
80_
11
11
111
21111
hAhhAItV
hAhhAItV
hhtthhbItV
hthhhhbItVyFor
wfw
Web
wfw
wfw
Web
ww
Web
τ
τ
τ
A cantilever beam with T cross section is loaded at the tip by a vertical force. Determine at the section n-n: (a) Maximum compressive stress. (b) Maximum tensile stress. (c) Maximum shear stress.
Calculation of the centroid using bb as reference
( )( )( ) ( )( )( )( )( ) ( )( ) in
AAy
y 1667.05.025.04
15.0225.05.04=
++−
==∑∑
Calculation of the moment of inertia about the centroid
( )( ) ( )( )( ) ( )( ) ( )( )( )
4
23
23
417.1
1667.0125.012
25.01667.025.05.0412
5.04
inI
I
z
z
=
⎥⎦
⎤⎢⎣
⎡−++⎥
⎦
⎤⎢⎣
⎡++=
Calculating the bending stresses ( ) inlbPxM z −=== 18000121500
Compressive Stress
Tensile Stress
( )
( ) psiyI
M
psiyI
M
topz
ztopx
bottomz
zbottomx
8.8472667.0417.1
18000
23300833.1417.1
18000
,
,
=−−=−=
−=−=−=
σ
σ
Calculating Q:To determine Q, cut the cross section horizontally at the NA and choose either the top or the bottom portion, it doesn't matter which because the same answer will result using either portion.
( )[ ] 384.02833.15.0833.1 inyAQz =⎟
⎠⎞
⎜⎝⎛==
( )( )( )( ) psi
tIQV
z
zyMaxxy 1778
5.0417.184.01500
, ===τ
Calculating the maximum shear stress
A beam shown is loaded by a 3 kN force. For points A and B located at section n-n, (0.5m from the support) shown in the figure below, determine the average shear stress.
Calculation of the centroid using the bottom as reference
( )( )( ) ( )( )( ) ( )( )( )( )( ) ( )( ) ( )( )
mmAAy
y
AAy
y
33.684800
328000
206020802010010206060208011020100
===
++++
==
∑∑∑∑
Calculation of the moment of inertia about the centroid( )( ) ( )( )( ) ( )( ) ( )( )( ) ( )( ) ( )( )( )
46
23
23
23
1063.8
33.58206012
206033.8802012
8020667.412010012
20100
mI
I
z
z
−×=
⎥⎦
⎤⎢⎣
⎡−++⎥
⎦
⎤⎢⎣
⎡−++⎥
⎦
⎤⎢⎣
⎡+=
Calculating the bending stresses ( ) mNPxM z −=== 7505.01500
( )
( ) MPayI
M
MPayI
M
topz
ztopx
bottomz
zbottomx
49.40517.01063.8
750
93.50683.01063.8
750
6,
6,
−=−×
−=−=
=−×
−=−=
−
−
σ
σ
Calculating Q for point A and B:
( )( )[ ]( ) 351067.4005.00517.001.01.0 myAQz−×=−==
( )( )[ ]( ) 37100.7202.00683.002.006.0 myAQz
−×=−==
( )( )( )( ) MPaAverage 608.0
02.01063.81071500
6
7
=×
×= −
−
τ( )( )( )( ) MPa
ItVQ
Average 081.01.01063.8
1067.415006
5
=×
×== −
−
τ
A beam 12in long is to support a load of 488lb as shown. Basing the design only on a bending stress, the designer has selected a 3in column channel as shown. The direct shear has been neglected.
( )( )
psiI
Mc
Max
Max
99266.1
5.11098
±=
±=±=
σ
σ
However, due to the combined bending and direct shear the stress should be maximum just before 3in and at the point where the web joints the flange
y1=1.5-0.273=1.227
( )( ) psiI
McMax 812
66.1227.11098
−=−=−=σ
( )( )( ) 3525.0273.0410.1227.15.121'' inAyQ =+== ( )( )
( )( ) psiIt
VQxy 681
170.066.1525.0366
−=−=τ
y(in) Q(in3) σ(psi) τ(psi)1.5 0 992 0
1.227 0.525 812 6811.0 0.568 661 7370.75 0.605 496 7850.5 0.631 331 8180.25 0.648 165 840
0 0.653 0 847
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−−⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛=
⎥⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−−⎟
⎠⎞
⎜⎝⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛−−+⎟
⎠⎞
⎜⎝⎛=
wf
f
fw
ff
tythhbtQ
ythty
thhbtQ
2
2221
2
2221
222
For the simply supported W 10x45 beam (ABCD) determine the maximum bending stress at 6 ft from the left of the beam and the maximum horizontal shear stress at a point 4 in above the bottom of the beam.
Reactions( )( )
( )( ) ( )( )( ) ( )lbDlbB
BM
lbDBF
ftftlblbDBF
yy
yD
yy
yy
35009500
82420001250000
130000
4/200050000
@
=⇒=
+−−==
=+⇒=
−−+==
∑∑∑
psiinS
ftlbMMax 2688
1.4911000
3 −==
−−=σ
At 6 ft from the left:
D=10.10F=8.02Tf=0.62Tw=0.35
( )( )lbftM
lblbftMlbV
−−=+−−=
=
110002450020000
4500
6
6
6
- - - Flange Flange Web Cross Section Info. Cross Section Info.
Designation Area Depth Width thick thick x-x axis x-x axis x-x axis y-y axis y-y axis y-y axis
- A d bf tf tw I S r I S r
- in2 in in in in in4 in3 in in4 in3 in
W 10x45 13.20 10.12 8.022 0.618 0.350 249.0 49.1 4.33 53.20 13.30 2.00
( )( )( ) psilbAV
webMax 9.1447
35.062.0212.104500
=−
==τ
At 4 in above the bottom of the beam
( ) ( )[ ]( )
( )( ) ( ) ( )( ) ( ) ( )( )[ ]
psiy
psihy
yin
lb
yhthhbIt
VIt
VQ
web
web
web
web
3.13940
8.12162
486.835.086.81.1002.835.02498
4500
48
max,1
min,1
1
21
2224
21
21
21
2
=⇒=
=⇒=
−+−=
−+−==
τ
τ
τ
τ
( )( )
( )( )( )( ) psi
ItVQ
ininQinb
inIlbV
Hor 138835.0249889.264500
37.4153.635.0
2494500
2
46
===⇒
⎥⎥⎥⎥
⎦
⎤
====
τ
At the centroid of the beam
( )( )( )( ) psi
ItVQ
inQinb
inIlbV
Max 1.139435.0249003.274500
003.2735.0
2494500
3
46
===⇒
⎥⎥⎥⎥
⎦
⎤
====
τ
D=10.10F=8.02Tf=0.62Tw=0.35
For a WT 8 x 25 T-beam determine the maximum bending and shear stress in the beam. We will also determine the bending stress at 4 ft from the left end of the beam.
Reactions( )( ) ( )( )
( )( )( ) ( )( )( ) ( )lbClbB
BM
lbCBF
ftftlbftftlbCBF
yy
yC
yy
yy
66703330
62415008410000
100000
4/15004/10000
@
=⇒=
++−==
=+⇒=
−−+==
∑∑∑
psiinS
ftlbMMax 21270
77.612000
3 ==
−=σ
( ) psiinI
inyftlbM
iny 94502.42
24.4289.113.88000
24.64
==
=−−=−=
=σ
At 4ft and 2in above the bottom of the beam
Designation Area of T Width thick thick - x-x axis x-x axis x-x axis x-x axis
- A d bf tf tw d/tw I S r y
- in2 in in in in - in4 in3 in in
WT8x25 7.36 8.13 7.073 0.628 0.380 21.40 42.20 6.770 2.400 1.890
D=8.13F=7.073Tf=0.628Tw=0.38
At the centroid of the beam
( )( )( )( ) psi
inlb
ItVQ
web 275638.02.42
367.760004 ===τ
Calculation of the centroid using bb as reference( )( )( ) ( )( )( )
( )( ) ( )( ) inA
Ayy 275.1
38.0502.7628.0073.7751.338.0502.7314.0628.0073.7
=++−
==∑∑
Calculating Q: To determine Q, cut the cross section horizontally at the NA and choose either the top or the bottom portion, it doesn't matter which because the same answer will result using either portion.
( )[ ] 3367.72227.638.0227.6 inyAQz =⎟
⎠⎞
⎜⎝⎛==
We can obtain the normal and shear stresses from flexure and shear formulas:The normal stresses obtained from the flexure formula have their maximum values at the farthest distance from the neutral axis. The normal stresses are calculated at the cross section of maximum bending moment.
IMy
−=σ
Maximum Stresses in Beams
IbVQ
=τ
Normal stresses in a beam of linearly elastic material: (a) side view of beam showing distribution of normal stresses, and (b) cross section of beam showing the z axis as the neutral axis of the cross section.
The shear stress obtained from the shear formula usually have their highest value at the neutral axis. The shear stresses are calculated at the cross section of maximum shear force. In most circumstances, these are the only stresses that are needed for design purposes. However to obtain a more complete picture of the stresses, we will need to determine the principal stresses and maximum shear stresses at various points in the beam.
Points A and E are at the top and bottom of the beam. Point C is in the midheight of the beam and points B and D are in between.If Hooke’s law applies, the normal and shear stresses at each of these five points can be readily calculated from the flexure and shear formulas.All the elements of vertical and horizontal faces, are in plane stress, because there is no stresses acting perpendicular to the plane of the figure.
Beams of Rectangular Cross Section
Points A and E elements are in uniaxialcompressive and tensile stresses respectively.Point C (neutral axis) element is in pure shear.Points B and D elements have both normal and shear stresses.
Stresses in a beam of rectangular cross section: (a)simple beam with points A, B, C, D, and E on the side of the beam; (b)normal and shear stresses acting on stress elements at points A, B, C, D, and E; (c)principal stress; and (d)maximum shear stresses.
Stress trajectory: Gives the directions of the principal stresses.
Stress Contours: Curves connecting points of equal principal stress.
Principal-stress trajectories for beams of rectangular cross section: (a) cantilever beam, and (b) simple beam. (Solid lines represent tensile principal stresses and dashed
lines represent compressive principal stresses.)
Stress contours for a cantilever beam (tensile principal stresses only).
As for a rectangular beam we identified the point A to E, where A and E are located at the top and bottom of the beam, point C at the neutral axis and points B and D are in the web where it meets the flange. The stresses at these points can be determined using the flexure and shear formulas. They have the same general appearance as in the rectangular beam section but the stresses are different.The largest principal stresses usually occur at the top or bottom of the beam (points A and E) where the stresses obtained from the flexure formula have their largest value.However, depending upon the relative magnitudes of the bending moment and shear force, the largest stresses sometimes occur in points B and D (in the web where it meets the flange).The maximum shear acting directly on a cross section of a wide flange beam stresses always occur at the neutral axis (point C). However, the maximum shear stresses acting on inclined planes usually occur either at the top and bottom of the beam (points A and E) or in the points B and D because of the presence of normal stresses.
Wide-Flange Beams
A beam AB with a span length L = 6ft supports a concentrated load P = 10800lbacting a distance c = 2ft from the right-hand support (see figure below). The beam is made of steel and has a rectangular cross section (width b=2in and height h = 6in).
Determine the principal stresses and maximum shear stresses at cross section mn located at a distance x = 9in from the end A of the beam. (Consider only the in-plane stresses)
Solution
( ) ( )lbRlbR
RM
RRF
BA
AB
BA
7200 3600
021080060
0108000
@
==
=−⇒=
=−+⇒=
∑∑
xMlbV
x
x
x
36003600
40
==
≤≤( )
( ) ( ) inlbMlbV
inx
−==
==
32400936003600
9
9
9
Where y has units in inches and σx has units in psi. The stresses calculated are positive when in tension. Note that a positive value of y (upper half of the beam) gives a negative stress, as expected.
( )( )( )( )
yinin
yinlbbh
MyI
MyX 900
623240012
1233 −=
−−=−=−=σBending stresses:
Shear stresses:Ib
VQ=τ
4222
22
2
⎟⎟⎠
⎞⎜⎜⎝
⎛−=⎟
⎟
⎠
⎞
⎜⎜
⎝
⎛ −+⎟
⎠⎞
⎜⎝⎛ −= yhbyh
yyhbQ
The figure shows a stress element cut from the side of the beam at cross section mn. The normal stress σx and the shear stress τxy are shown acting in their positive directions.
( )( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−⎟
⎠⎞
⎜⎝⎛== 2
2
32
2
3 46
4212 yh
bhVyhb
bbhV
IbVQτ
The shear stresses τxy acting on the x face of the stress element are positive upwards, whereas the actual shear stresses τ act downward. Therefore
In which y has units of inches and τxy has units of psi
( )( )( )
( ) ( )222
3 9504
662
36006 yyinininlbτ XY −−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
Point y (in) σx (psi) τxy (psi)A 3 -2700 0B 2 -1800 -250C 1 -900 -400D 0 0 -450E -1 900 -400F -2 1800 -250G -3 2700 0
The normal stresses vary linearly from a compressive stress of -2700psi at the top of the beam (point A) to a tensile stress of 2700psi at the bottom of the beam (point G). The shear stresses have a parabolic distribution with a maximum stress at the neutral axis (point D).
Calculation of stresses on cross section mnWe divide the height of the beam into six equal intervals and label the corresponding points from A to G.
Principal Stresses and Maximum Shear Stresses
The principal stresses and maximum shear stresses at each of the seven points A through G may be determined from the following equations:
( )22
2,1 22 xyyxyx τ
σσσσσ +⎟⎟
⎠
⎞⎜⎜⎝
⎛ −±⎟⎟
⎠
⎞⎜⎜⎝
⎛ +=
( )22
2 xyyx
MAX τσσ
τ +⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
Point y (in) σx (psi) τxy (psi) σ1 (psi) σ2 (psi) τmax (psi)-2700 1350
934602450602934
1350
-1834-1052-450-152-340
034
152450
105218342700
A 3 -2700 0B 2 -1800 -250C 1 -900 -400D 0 0 -450E -1 900 -400F -2 1800 -250G -3 2700 0
