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Question 5A fixed-end beam AB of length L supports a uniform load of intensity q as shown in FigureQ5 Beginning with the second-order differental equation of the deflection curve (the bendingmoment equation)(5 1) determine the reactions, shear forces, bending moments, slopes, and deflections of thebeamconstruct the shear-force and bending-moment diagrams, labelling all critical ordinates(5 2)MgMvRsLRAFigure Q5 Fixed-end beam

Question
Question 5
A fixed-end beam AB of length L supports a uniform load of intensity q as shown in Figure
Q5 Beginning with the second-order differental equation of the deflection curve (the bending
moment equation)
(5 1) determine the reactions, shear forces, bending moments, slopes, and deflections of the
beam
construct the shear-force and bending-moment diagrams, labelling all critical ordinates
(5 2)
Mg
Mv
Rs
L
RA
Figure Q5 Fixed-end beam
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Question 5 A fixed-end beam AB of length L supports a uniform load of intensity q as shown in Figure Q5 Beginning with the second-order differental equation of the deflection curve (the bending moment equation) (5 1) determine the reactions, shear forces, bending moments, slopes, and deflections of the beam construct the shear-force and bending-moment diagrams, labelling all critical ordinates (5 2) Mg Mv Rs L RA Figure Q5 Fixed-end beam

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check_circleAnswer
Step 1

Refer the given figure, the given figure is symmetry and loaded as equilibrium.

So the reaction force at point A and B are equal
gL
R R3
2
М, %3 М,
Thus, the reaction force are R
R3
Calculate the bending moment from the right end of the beam
M Rx- M
2
-- М, +
(Lx
.(1)
2
(ck-x7)7
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So the reaction force at point A and B are equal gL R R3 2 М, %3 М, Thus, the reaction force are R R3 Calculate the bending moment from the right end of the beam M Rx- M 2 -- М, + (Lx .(1) 2 (ck-x7)7

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Step 2

Write the differential equation for the fixed end beam.

Lx-x)
Elv" = M -M +
Integrate both side with respect to x
Elv'=-Mx+
2
C
3
(2)
2
Again integrate and apply the boundary condition.
Ev'=-Mx* gLr
2
C
12
(3)
+
2
(0) 0 C 0
..
v(L) 0. M
12
Thus, the bending moments are M, = Mg
12
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Lx-x) Elv" = M -M + Integrate both side with respect to x Elv'=-Mx+ 2 C 3 (2) 2 Again integrate and apply the boundary condition. Ev'=-Mx* gLr 2 C 12 (3) + 2 (0) 0 C 0 .. v(L) 0. M 12 Thus, the bending moments are M, = Mg 12

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Step 3

Refer the figure, write the ...

V R4- qx
2
(L-2:
Refer the figure and equation (1)
(L
12
6Lr 6x
M
Refer the figure and equation (2)
1 -3Lx + 2x)
12EI
Refer the equation (3)
qx2
(L-x2)
24EI
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V R4- qx 2 (L-2: Refer the figure and equation (1) (L 12 6Lr 6x M Refer the figure and equation (2) 1 -3Lx + 2x) 12EI Refer the equation (3) qx2 (L-x2) 24EI

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