Dissolution - Stephanie Mohr vs Joseph Mohr - Order of Disposition- Contempt Dismissed
Mohr´s method – for determining a deflection or an angle ...
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1 Mohr´s method – for determining a deflection or an angle of rotation in given place Analogy of • differential condition: M´´= -q • Differential equation of elastic curve: EIw´´ = -M Procedure: 1. Moment diagram of the given real beam 2. Fictive (dual) beam to given beam 3. Fictive beam loaded by the moment area of the real beam 4. Deflection: 5. Angle of rotation: EI M w _ = EI V w - = = ϕ ´ - - V a M - Bending moment and shear force of the fictive beam from the load of the moment diagram of the real beam Fictive beam: has to fulfil the same deformation conditions against M a V as the real beam fullfills against deflection and angle of rotation. Real beam Fictive beam advantage – M function doesn ´t has to be continuous disadvantage – complicated for continuous load, suitable for force and moment load in points Example 1 Determine by Mohr´s method: - deflection w in the middle of the beam; - angle of rotation ϕ in the left support. w=0 w=0 a b c F l/2 l/2 Reactions: 2 b a F R R = = 4 c Fl M = R a R b 4 2 2 2 a c Fl l F l R M = ⋅ = ⋅ = R a R b ~ ~ R a ~ 4 Fl ] kNm [ 16 2 1 2 4 ~ ~ 2 2 b a Fl l Fl R R = ⋅ ⋅ = = ] kNm [ 48 6 2 ~ 6 1 ~ 2 ~ ~ 3 3 1 1 a c Fl l l Q Q l R M = - = - ⋅ = 1 ~ Q l/6 ] kNm [ ~ q ] kNm [ M M=0 M=0 16 2 4 2 1 ~ ~ 2 2 1 Fl l Fl Q Q = ⋅ ⋅ = = a c b a c Fictive beam EI V EI M w ~ ~ = = ϕ Mohr´s method: EI V EI M w ~ ~ = = ϕ Mohr´s method: R a R b ~ ~ R a ~ 4 Fl 1 ~ Q l/6 EI Fl EI V 2 a 16 1 ~ = = ϕ 0 c = ϕ EI Fl EI V 2 b 16 1 ~ - = = ϕ EI Fl EI M w 3 c 48 1 ~ = = 0 b a = = w w R a ~ –R b ~ ] kNm [ ~ 2 V ] kNm [ ~ 3 M + – + b a c a ϕ b ϕ c w Elastic curve R a ~ EI Fl EI V 2 a 16 1 ~ = = ϕ 0 c = ϕ EI Fl EI V 2 b 16 1 ~ - = = ϕ EI Fl EI V 2 a 16 1 ~ = = ϕ 0 c = ϕ 0 b a = = w w EI Fl EI V 2 b 16 1 ~ - = = ϕ EI Fl EI V 2 a 16 1 ~ = = ϕ 0 c = ϕ EI Fl EI M w 3 c 48 1 ~ = = 0 b a = = w w EI Fl EI V 2 b 16 1 ~ - = = ϕ EI Fl EI V 2 a 16 1 ~ = = ϕ 0 c = ϕ 2° 3° Make the assessment after serviceable limit state: , F=10kN l=1m E=210GPa Square section a=100mm w all = l/1000 . ... 1 119 , 0 48 3 satisf mm mm EI Fl w c p = = 4 6 10 33 8 m . , I y - - - = = =
Transcript of Mohr´s method – for determining a deflection or an angle ...
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Mohr´s method – for determining a deflection or an angle of rotation in given place
Analogy of
• differential condition: M´´= -q
• Differential equation of elastic curve: EIw´´ = -M
Procedure:1. Moment diagram of the given real beam
2. Fictive (dual) beam to given beam
3. Fictive beam loaded by the moment area of the real beam
4. Deflection:
5. Angle of rotation:
EI
Mw
_
=
EI
Vw
−
== ϕ´−−
VaM- Bending moment and shear force of
the fictive beam from the load of themoment diagram of the real beam
Fictive beam: has to fulfil the same
deformation conditions against M a V as the real beam fullfills against deflection
and angle of rotation.
Real beam Fictive beam
advantage – M function doesn ´t has to be continuousdisadvantage – complicated for continuous load, suitable for force and moment load in points
Example 1Determine by Mohr´s method:
- deflection w in the middle of the
beam;
- angle of rotation ϕ in the leftsupport.
w=0 w=0
a b
c
F
l/2 l/2
Reactions:2
ba
FRR ==
4c
FlM =
Ra Rb
4222ac
FllFlRM =⋅=⋅=
Ra Rb
~~Ra
~
4Fl
]kNm[162
1
24
~~ 2
2
ba
FllFlRR =⋅⋅==
]kNm[4862
~
6
1~
2
~~ 3
3
11ac
FlllQQ
lRM =
−=−⋅=
1
~Q
l/6
]kNm[~q
]kNm[M
M=0 M=0
16242
1~~2
21
FllFlQQ =⋅⋅==
a
c
ba
c
Fictive beam
EI
V
EI
Mw
~~
== ϕMohr´s method:
EI
V
EI
Mw
~~
== ϕ
Mohr´s method:
Ra Rb
~~Ra
~
4Fl1
~Q
l/6
EI
Fl
EI
V2
a16
1~
==ϕ
0c
=ϕ
EI
Fl
EI
V2
b16
1~
−==ϕ
EI
Fl
EI
Mw
3
c48
1~
==
0ba
== ww
Ra
~
– Rb
~
]kNm[~ 2V
]kNm[~ 3
M
+
–
+
ba c
aϕ
bϕ
cw
Elastic curve
Ra
~
EI
Fl
EI
V2
a16
1~
==ϕ
0c
=ϕ
EI
Fl
EI
V2
b16
1~
−==ϕ
EI
Fl
EI
V2
a16
1~
==ϕ
0c
=ϕ
0ba
== ww
EI
Fl
EI
V2
b16
1~
−==ϕ
EI
Fl
EI
V2
a16
1~
==ϕ
0c
=ϕ
EI
Fl
EI
Mw
3
c48
1~
==
0ba
== ww
EI
Fl
EI
V2
b16
1~
−==ϕ
EI
Fl
EI
V2
a16
1~
==ϕ
0c
=ϕ2°
3°
Make the assessment after serviceable limit state: ,
F=10kN
l=1m
E=210GPa
Square section a=100mm
wall= l/1000
....1119,048
3
satisfmmmmEI
Flwc p==
4610338 m.,I y
−−−−====
2
ExamplesEI
V
EI
Mw b
bb
b
~~
== ϕBy Mohr´s method wb, φb :
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