Analyse the three span illustrated below using the manual flexibility method, in
ID: 1845253 • Letter: A
Question
Analyse the three span illustrated below using the manual flexibility method, introducing rotational releases over supports B and D. Calculate the rotations in the releases, then compute the redundant moments required to reduce these rotations to zero. Find the final internal bending moment at each mid-span and over the interior supports (points A, B, C, D and E), and the displacement at each midspan (points A, C and E). The cross section of the beam is constant with EI= 1200 kN.m2. a = 5.0 m, b = 3.0 m, and c = 4.0 m. w = 4 kN/m, Pl = 5 kN and P2 = 6 kN. The point loads are located at the middle of the spans to which they are applied. Magnitude of rotation at first release (point B, radians, absolute value): Magnitude of rotation at second release (point D, radians, absolute value): Moment at first midspan (point A, kN.m, internal convention): Moment over second support (point B, kN.m, internal convention): Moment at second midspan (point C, kN.m, internal convention): Moment over third support (point D, kN.m, internal convention): Moment at third midspan (point E, kN.m, internal convention): Displacement at middle of first span (point A, mm, positive up): Displacement at middle of second span (point C, mm, positive up): Displacement at middle of third span (point E, mm, positive up):Explanation / Answer
cross section of the beam = 800Kn/m2
span of a = 4.5m , b = 3.5m , c = 3.5m
uniform load ,w = 2kn/m
point load,P1 = 13Kn
point load ,P2 = 9Kn
1.magnitude of rotation at first release at point B
reaction at point B = Pl2/8
= 2x(4.5)2/8 = 2x20.25/8 = 40.5 = 5
2.reaction at pont B = 5Kn
magnitude of rotation at second release =
reaction= wl2/8
2x(3.5)2/8 = 2x12.25/8
24.5/8 = 30.62Kn
reaction at pont D is 30.62Kn
3.moment at first midspan(point A) = Wl2/16El
= 13(4.5)2/16x800
13x20.25 / 12800 = 263.25/12800
moment at point A= 0.02Kn/m
4.moment at point B = wl2/16El
= 2(3.5)2/ 16x800
= 2x12.25/12800
= 24.5/12800 = 1.91x10-3
moment at point B = 1.91x10-3
5.moment of second midspan point C
= wl2/16El (it is uniform load therfore 2x3.5= 7)
= 7x(3.5)2/16x800
= 7x12.25 /12800 = 85.75/12800
moment at the point C = 6.6x10-3
6.moment over the third support(point D)
= wl2/16El
= 2x(3.5)2/16x800
2x12.25/12800
= 24.5/12800 = 1.91x10-3
moment over the third support(point D) = 1.91x10-3
7.moment at third midspan (point E)
= Pl2/16El
=9x(3.5)2/16x800
= 9x12.25/12800 = 110.25/12800
= 8.61x10-3
moment at the third midspan (point E) = 8.61x10-3
8.displacement at the middle of fisrt span(point A)
displacement ,d = vt
d= Pl/8 = (13x4.5)/8
= 58.5/8 = 7.31mm
displacement at the point A = 7.31mm
9.displacement at point C
=wl2/8
= 2x(3.5)2/8
= 2x12.25/8
= 24.5/8 = 3.0mm
displacement at point C = 3.0mm
10.displacement at point E
= Pl/8
= 9x3.5/8 = 31.5/89
= 3.93mm
displacement at point E = 3.93mm
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.