Pls continue solving in finding the internal reactions. Calculate the reactions at each support. Draw the shear, moment, elastic curve and axial force diagram. ps. Pls write it on a paper and use pythagorean theorem 3.) Fixed at A, Hinge at E, Internal Hinge at B.D.F.2 GOIN4.0m<- Ax8.0mSubst. →Σ ΜΕ ΤΟ GAy (10) + MA-25 (5) + 15(s) = 0Ay (10) + MA = 50MA= 50-10 Ay →RIOM- 4 Ay +50=0Ay = 12.5 kN@ba- 0MA= -15MA- 75 KN.M@ 0MA = 50-10AYMA= 50-10(12.5)25 kN↓c3 Ay+ 4Ax + (50-10 Ay)-D3 Ay - 10 Ay +4 Ax +50=0- TAY + 4 Ax +50=0 (20)Subst →7 Ay + 4 Ax + (50-10Ay) - 507Ay-10Ay +4Ax +50=50- 3 Ay + 4 Ax = 0→Ba@2a-3a- TAY SAX50=0- 3 Ay+9Ax=0-3 Ay+ 4AX=0-3(12.6) + 4 Ax=0Ax=9,315 kN2.0mD₂3.0mt10kN/mTEXFy3.0m@ ABAyMA@ A B C D-AxAAXAy15 KNJa2.0mB, 4 BxToy25kNCΣMB=0 C²³ +Ay (3) + Ax(4) +MA-O3Ay +4 Ax +MA-0-6-DXP↑DyZMp=0 C² +Ay(1) + Ax(4) + MA-25 (2)=07 Ay + 4AX + MA= 50 -13 25 KNAM2.0mAY3.0mB2.0m3.● Fixed at A, Hinge at E, Internal Hinge at B, D and F10 KN/mE15 KN3.0m 3.0m 2.0mG

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Answered: Calculate the reactions at each…

Engineering

Civil Engineering

Calculate the reactions at each support. Draw the shear, moment, elastic curve and axial force diagram.

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

Equilibrium and Support Reactions

Civil engineers deal with structures that are composed of several members connected or with individual members such as beams and columns. To analyze such structures, the condition of statics must be applied. Under the static condition, the members are considered to be in equilibrium under the action of forces, that is, the members are assumed to be having negligible velocity.

Question

Pls continue solving in finding the internal reactions. Calculate the reactions at each support. Draw the shear, moment, elastic curve and axial force diagram. ps. Pls write it on a paper and use pythagorean theorem

3.) Fixed at A, Hinge at E, Internal Hinge at B.D.F.
2 GOIN
4.0m
<- Ax
8.0m
Subst. →
Σ ΜΕ ΤΟ G
Ay (10) + MA-25 (5) + 15(s) = 0
Ay (10) + MA = 50
MA= 50-10 Ay →
RIOM
- 4 Ay +50=0
Ay = 12.5 kN
@ba
- 0
MA= -15
MA- 75 KN.M
@ 0
MA = 50-10AY
MA= 50-10(12.5)
25 kN
↓c
3 Ay+ 4Ax + (50-10 Ay)-D
3 Ay - 10 Ay +4 Ax +50=0
- TAY + 4 Ax +50=0 (20)
Subst →
7 Ay + 4 Ax + (50-10Ay) - 50
7Ay-10Ay +4Ax +50=50
- 3 Ay + 4 Ax = 0→Ba
@2a-3a
- TAY SAX50=0
- 3 Ay+9Ax=0
-3 Ay+ 4AX=0
-3(12.6) + 4 Ax=0
Ax=9,315 kN
2.0m
D₂
3.0m
t
10kN/m
TEX
Fy
3.0m
@ AB
Ay
MA
@ A B C D
-Ax
AAX
Ay
15 KN
Ja
2.0m
B, 4 Bx
Toy
25kN
C
ΣMB=0 C²³ +
Ay (3) + Ax(4) +MA-O
3Ay +4 Ax +MA-0-6
-DX
P↑
Dy
ZMp=0 C² +
Ay(1) + Ax(4) + MA-25 (2)=0
7 Ay + 4AX + MA= 50 -13

25 KN
AM
2.0m
A
Y
3.0m
B
2.0m
3.
● Fixed at A, Hinge at E, Internal Hinge at B, D and F
10 KN/m
E
15 KN
3.0m 3.0m 2.0m
G

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Step 2: Calculate support reactions at E

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Step 3: Calculate support reactions at A

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Step 4: Calculation of Internal forces for part AB

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Step 5: Calculation of Internal forces for part BCD

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Step 6: Calculation of Internal forces for part FG

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Step 7: Calculation of Internal forces for part DF

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Step 8: Calculation of Axial forces for member DE, DF & EF

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Step 9: Calculation of Internal forces for part DE & EF

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