Bending moment = shear area

quartzaardvarkUrban and Civil

Nov 29, 2013 (3 years and 6 months ago)

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Area method Copyright Prof Schierle 2011 1
Bending moment=
shear area
+
-
Area method Copyright Prof Schierle 2011 2
Area Method
1Load diagram
2Beam diagram
3Shear diagram
4Bending diagram
Consider the beam part between m and n
Equilibrium:
Vn
= 0
Vm
-V
n
-w x = 0 V
n
=Vm
-wx
where
wx= load between m & n
Mn
= 0
Mm-Mn+Vmx-wx2/2=0 M
n= Mm+Vmx-wx2/2
where
Vmx-wx2/2= shear between m & n
Area Method:
Vn
= Vm
+ load between m & n(down -)
Mn
= Mm
+ shear area between m & n
Area method Copyright Prof Schierle 2011 3
Area Method implications
•Positive shear = increasing bending
•Negative shear = decreasing bending
•Zero shear = maximum bending
Uniform load yields:
•Linear shear
•Parabolic bending moment
•Moving from left to mid-span:
•Decreasin
g
shear reduces amount
of bending increase
Area method Copyright Prof Schierle 2011 4
Area Method implications
Point load yields:
•Constant shear between point loads
•Linear bending moment
Area method Copyright Prof Schierle 2011 5
Examples
Design
Defines beam size for actual loads and allowable stress of selected material
Analysis
Checks if a given beam satisfies allowable stress of the actual material
Assume:
Wood
Allowable bending stressF
b
= 1200 psi
Allowable shear stress (parallel to fiber)F
v = 95 psi
Steel
Yield strengthF
y = 50 ksi
Allowable bending stress (0.6 F
y)F
b
= 30 ksi
Allowable shear stress (0.4 F
y)F
v = 20 ksi
Note:
F = allowable stress
f = actual stress
Area method Copyright Prof Schierle 2011 6
Beam analysis
Reactions
Mc=0 =16 Rb-1000 (20)-300(4)18-200(16) 8
Rb=(20000+21600+25600)/16 R
b = 4200 lb
Mb=0 =-16 Rc-1000(4)-300(4)2+200(16)8
Rc=(-4000-2400+25600)/16 Rc = 1200 lb
Area method Copyright Prof Schierle 2011 7
Inflection point :
0 bending
change deflection
curve
Beam analysis
ReactionsRb = 4200 lb
Rc = 1200 lb
Shear
V
ar
= 0 -1000V
ar= -1000 lb
Vbl
= -1000 -300 (4)Vbl
=-2200 lb
Vbr
= -2200 + 4200 V
br = +2000 lb
Vcl = +2000 -200(16) = -R
c
Vcl = -1200 lb
Find X (V = 0 → Mmax)
Vbr-wX=0
X=Vbr/w = 2000/200X = 10 ft
Bending moment
Mb=4(-1000-2200)/2Mb= -6400 lb’
Mx=-6400+10(2000)/2M
x= +3600 lb’
Section modulus
S=bd2/6=(3.5)11.25
2/6 S = 74 in
3
Bending stress
fb=M/S= 6400(12)/74f
b=1038 psi
1038<1200
Shear stress
fv
= 1.5V/(bd) = 1.5(2200)/[3.5(11.25)]f
v
= 84 psi<95
Area method Copyright Prof Schierle 2011 8
1
2
Steel beam design
1 Actual beam
2Beam diagram -ignore loads at supports
(has no effect on shear & bending, but on columns)
Assume:
L = 30’, P = 36 k, F
b = 30 ksi, Fv = 20 ksi
Reactions
R = 2P/2 = 2 (36)/2R = 36 k
Shear
V
ar = Vbl =RV
ar
= Vbl = 36 k
Vbr
= V
cl = 36 -36V
br = Vcl = 0
Vcr
= Vdl = 0 -36 V
cr
= Vdl = -36 k
Vdr
= -36 + 36V
dr
= 0
Bending moment
Mb
= Mc
= 36 (10)M
b
= 360 k’
Section modulus required
S = M/Fb
= 360 k’(12”)/ 30 ksiS = 144 in
3
Use W18x76S = 146 > 144
Shear stress
fv = V/(d tw) = 36k/(18.21x0.425) f
v
= 4.65 ksi<20
Note: shear stress is rarely critical in steel beams
Area method Copyright Prof Schierle 2011 9
Area method Copyright Prof Schierle 2011 10
Concrete beam analysis
1Actual beam
2Beam diagram –ignore load at supports
(has no effect on beam shear & bending, but column)
Assume:
Span L = 30’, point load P = 20 k
Beam DL = 1.33’x2’x150 pcf /1000w = 0.4 klf
Reaction
R = (2 P+w L)/2 = (2x20+0.4x30)/2R = 26 k
Shear
V
ar
= RV
ar
= 26 k
Vbl
= 26 -0.4 (10)V
bl
= 22 k
Vbr
= 22 -20V
br= 2 k
Vcl
= 2 -0.4 (10) V
cl
= -2 k
Vcr
= -2 -20 V
cr
= -22 k
Vdl = -22 -0.4 (10)V
dl = -26 k
Bending moment
Mb
= 10 (26+22)/2M
b
= 240 k’
Mmax
= Mb + 2(5)/2Mmax
= 245 k’
Note: Concrete stress will be covered in Arch 313
w
1
2
0 shear =
max bending
Area method Copyright Prof Schierle 2011 11
Extra Credit 
To optimize beams
find the relationship
L/C
such that
+M = Abs (-M)
Hint: Area Method
C
C
L
+M
-M
-M
Area method Copyright Prof Schierle 2011 12
The end