Contact- 9740501604 Indeterminate Beams Statically determinate beams o If the reaction forces can be calculated using equilibrium equations alone. o Number of unknown doesn’t exceed the no. of equations. Examples- (1) Simply supported beam (2) Cantilever beam
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Contact- 9740501604
Indeterminate Beams
Statically determinate beams
o If the reaction forces can be calculated using equilibrium equations alone. o Number of unknown doesn’t exceed the no. of equations.
Examples-
(1) Simply supported beam
(2) Cantilever beam
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Statically indeterminate beams
o If the reaction forces can’t determine using equilibrium equations only, other methods have to be used.
o No. of unknowns exceeds the no. of equations.
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Examples – (1) Propped cantilever beam
(2) Fixed-Fixed beam
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(2) Continuous beam
Degree of static indeterminacy
The degree of static indeterminacy (DSI) represents the difference between the number of static unknowns (reactions and internal forces) and the number of static equations (equilibrium equations).
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Example- 1
∑퐹 = 0 ............... (1)
∑퐹 = 0 ............... (2)
푀 + 푀 = 0 ............ (3)
There are 6 unknown in this figure.
Degree of static indeterminacy = 6 – 3 = 3
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Method of superposition
Example - 1
Sol. 3.75 N
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→ Due to UDL, 푌 =
→ Due to point load, 푌 = =
So, 푌 - 푌 = - = 0
→ R = ퟑ.풘푳ퟖ
= . . = 3.75 N
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Example – 2
Find maximum bending moment?
Sol.
Due to symmetric beam,
푅 = 푅 =
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Separating main figure into propped cantilever beam and cantilever beam with moment at free end.
Deflection due to moment, 훿 = (hogging) = ↓
Deflection due to UDL, 훿 = ↓
Deflection due to point load, 훿 = ↑
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So net deflection at point B is zero due to fixed support,
훿 = 훿 + 훿 + 훿 = 0
+ - = 0
푀 =
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→From above figure, maximum bending moment occurs at mid point.
푀 = × - 푀 - ×
= - -
=
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Example – 3
Find the deflection at mid point?
Sol.
Due to symmetric beam,
푅 = 푅 =
Separating main figure into propped cantilever beam and cantilever beam with moment at free end.
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Now separating figure (2) into two cantilever beams,
→From figure (4), we know
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푦 = and 휃 =
푦 = 푦 + 훿 = 푦 + × 휃
= + ×
= ↓
→From figure (5), we know
Deflection due to reaction force, 푦 = = .
= ↑
→From figure (3), we know
Deflection due to moment at free end, 푦 = = ↓
→ So net deflection is zero at point B due to fixed support.
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푦 + 푦 + 푦 = 0
- + = 0
푀 =
→For finding deflection by double integration method,
퐸퐼 = 푀 = 푅 . 푥 -푀 (taking moment at x distance from point B)
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= 푥 -
퐸퐼 = . - 푥 + 퐶
퐸퐼푦 = . - . + 퐶 . 푥 + 퐶
Boundary conditions
At x = 0,
y = 0 and = 0
From above equations, we get
퐶 = 퐶 = 0
So,
퐸퐼푦 = . - .
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So deflection at mid point C, x =
퐸퐼푦 = . ( / ) - . ( / ) = -
풚풄 = 휹풄 = 푾푳ퟑ
ퟏퟗퟐ푬푰 (downward)
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Example – 4
Find bending moment and deflection at mid-point C?
Sol. - H.M.
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Example – 5
Find bending moment at point C?
Sol. - H.M.
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