Determine the slope and deflection at point B of the beam ABC shown using the AreaMoment Method. Use E = 200 GPa and I = 1665 x 106 mm4SUBJECT: THEORY OF STRUCTURESTHE SECOND PIC IS MY PROGRESS 1. Determine the slope and deflection at point B of the beam ABC shown using the Area-Moment Method. Use E = 200 GPa and I = 1665 x 106 mm².270 KN45 kN/mAB3 m3 m3 mT MAA270 KN3mm3mMEI ALLYM/E!wrt pt (Ⓡ45 kN/m pick ref. pointBM/2l wrtpt.3m(2) Solve for reactionsZMA=0;(3)14- = 270 (0) = 45 (4+ 2) = 0MAZ27 M4 - 270 (3)-45 (3) (6+2) = 0маMA == + 1822.30 KN-m, (4ZFY = 0;AY-270-45(3) = 0+ZFx = 0;✓ An=2°✓Xn=2°45 kN/m-MA = "1822 50 KM²m#1PL = 405 (3) - 1215 14-47EEl0-45 (6²) ₂-810 IN-MEI2E145 (3²)221O202-5 KN.ME1==Ay = 4015 KN, 1Ax = 0·bh(2)+1(2) +z = b+2MAL=3L=3BPL45kN/m-Ama LEGML=3mw=95 KNYM

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Answered: 1. Determine the slope and deflection…

1. Determine the slope and deflection at point B of the beam ABC shown using the Area- Moment Method. Use E = 200 GPa and I = 1665 x 106 mmª. 270 KN 45 kN/m A B 3 m -3 m -3 m T

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Beams

In structural engineering, a beam is a component made of a variety of materials (including steel alloys, and wood) that can resist loads applied laterally to the beam axis. Members, elements, rafters, shafts, and purlins are all terms used to describe beams.

Question

Determine the slope and deflection at point B of the beam ABC shown using the AreaMoment Method. Use E = 200 GPa and I = 1665 x 106 mm4

SUBJECT: THEORY OF STRUCTURES

THE SECOND PIC IS MY PROGRESS

1. Determine the slope and deflection at point B of the beam ABC shown using the Area-
Moment Method. Use E = 200 GPa and I = 1665 x 106 mm².
270 KN
45 kN/m
A
B
3 m
3 m
3 m
T

MAA
270 KN
3mm
3m
MEI ALLY
M/E!
wrt pt (Ⓡ
45 kN/m pick ref. point
B
M/2l wrt
pt.
3m
(2) Solve for reactions
ZMA=0;
(3)
14- = 270 (0) = 45 (4+ 2) = 0
MAZ
27 M4 - 270 (3)-45 (3) (6+2) = 0
ма
MA =
= + 1822.30 KN-m, (
4ZFY = 0;
AY-270-45(3) = 0
+ZFx = 0;
✓ An=2°
✓Xn=2°
45 kN/m
-MA = "1822 50 KM²m
#1
PL = 405 (3) - 1215 14-47
E
El
0
-45 (6²) ₂-810 IN-M
EI
2E1
45 (3²)
221
O
202-5 KN.M
E1
=
=
Ay = 4015 KN, 1
Ax = 0
·bh
(2)+1
(2) +z = b
+2
MA
L=3
L=3
B
PL
45kN/m
-Ama LEGM
L=3m
w=95 KNYM

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