Step 1: Sol Calculation of Reactions = from Symmetry of loading each support will shave Equal Reaction = total land 2 Step 3: RA = RC = 30 = 15 KN HA=0 (because there is No Horitated face instass FBD of Support E:- FFE from Ale DFE FED 15kW + cose- *15 2 V=0 SinB = 3/5 FEF Sino = 15 FEF = 25 KN C Compression) Apply SH=0 FEF COSO FED 25x15 = FED FED=20KN (TMsion) at Joint F FBD of Joint F FGF Step 2: Step 4: at Joint D FBD & Joint D +10= FDF D FCD=20kN 'DE = 20KN loka Similarly и Apply £H=0 FDE FCD 20kN (TOM) Apply EV=O FDF = 10kN (Ten) from symmetry" lebt side of force in members are also same FAH = 25KNC Compression) FAB =20kN (Tension) FGH FGF FBH=10KNC Tension) FBC=20kN (Tension) = 16-68 kN ( Compression) FCF 16-25 30-40° Apply Ev=o 'Loked FCF CS (16-26) = 10 Cos (36-63°) FCF = 8.35HN (Compression) Apply EH-O 25KN Fax + FCF sin (1626) + 10 sin (36-67") =25 F4F = 16-68 kN Compresse) from symmetry of truss FCF = FHC = 8:35 KWC Compression) at Joint C FBD of Joint 'C' = FGC 8.35IN Εντο lokN 4 F 8.35KN 3 0/5 C Rom Ale Cob&= 415 (-) Fac + 8-35x2x cose + 10 = 0 FGC = 8.35x2x+40 FGC23-36 kN (Tension) AH TRUSSES METHOD OF JANTS DETERMINE THE FORCE in EACH MEMBER OF THE ROOF TRUSS STATE IF THE memBeRS ARE in JENsion OR COMPRESSion. Assume * FOR every JoinT FH=0 Fr=0 *"compression" * REACTION Me=0 Ra 10(4)+ 10 (8) + 10 (12) 16 16 IF Reaction Ra = 15KN ↑ IS TOWARDS The JoinT * Tension IF Reaction is "Away From TheJoin T ALL mem BeRS ARE Pin-connecTED 6H Re = 30-15 Re-15KN T AT JoinTA AH H 4m B 10 KN 4m MEMBER AB BC CP DE EF 6H AH PHI CH C6 CF PF C D 4m 4m 10KN 10KN Re AXIAL FORCE C/ E 3m +- !3m ISKN 21=0 ↓二个 음 AH = 15 AB AH = 15 (3) AH= 25 KN Compression 374-0 AH AB = (25) Tension AB=20KN A 20
Step 1: Sol Calculation of Reactions = from Symmetry of loading each support will shave Equal Reaction = total land 2 Step 3: RA = RC = 30 = 15 KN HA=0 (because there is No Horitated face instass FBD of Support E:- FFE from Ale DFE FED 15kW + cose- *15 2 V=0 SinB = 3/5 FEF Sino = 15 FEF = 25 KN C Compression) Apply SH=0 FEF COSO FED 25x15 = FED FED=20KN (TMsion) at Joint F FBD of Joint F FGF Step 2: Step 4: at Joint D FBD & Joint D +10= FDF D FCD=20kN 'DE = 20KN loka Similarly и Apply £H=0 FDE FCD 20kN (TOM) Apply EV=O FDF = 10kN (Ten) from symmetry" lebt side of force in members are also same FAH = 25KNC Compression) FAB =20kN (Tension) FGH FGF FBH=10KNC Tension) FBC=20kN (Tension) = 16-68 kN ( Compression) FCF 16-25 30-40° Apply Ev=o 'Loked FCF CS (16-26) = 10 Cos (36-63°) FCF = 8.35HN (Compression) Apply EH-O 25KN Fax + FCF sin (1626) + 10 sin (36-67") =25 F4F = 16-68 kN Compresse) from symmetry of truss FCF = FHC = 8:35 KWC Compression) at Joint C FBD of Joint 'C' = FGC 8.35IN Εντο lokN 4 F 8.35KN 3 0/5 C Rom Ale Cob&= 415 (-) Fac + 8-35x2x cose + 10 = 0 FGC = 8.35x2x+40 FGC23-36 kN (Tension) AH TRUSSES METHOD OF JANTS DETERMINE THE FORCE in EACH MEMBER OF THE ROOF TRUSS STATE IF THE memBeRS ARE in JENsion OR COMPRESSion. Assume * FOR every JoinT FH=0 Fr=0 *"compression" * REACTION Me=0 Ra 10(4)+ 10 (8) + 10 (12) 16 16 IF Reaction Ra = 15KN ↑ IS TOWARDS The JoinT * Tension IF Reaction is "Away From TheJoin T ALL mem BeRS ARE Pin-connecTED 6H Re = 30-15 Re-15KN T AT JoinTA AH H 4m B 10 KN 4m MEMBER AB BC CP DE EF 6H AH PHI CH C6 CF PF C D 4m 4m 10KN 10KN Re AXIAL FORCE C/ E 3m +- !3m ISKN 21=0 ↓二个 음 AH = 15 AB AH = 15 (3) AH= 25 KN Compression 374-0 AH AB = (25) Tension AB=20KN A 20
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
At step 3, at joint F, how does the angles 16.26 and 36.67 gets from?
Transcribed Image Text:Step 1:
Sol
Calculation of Reactions =
from
Symmetry of loading each support will shave
Equal Reaction = total land
2
Step 3:
RA = RC = 30 = 15 KN
HA=0 (because there is No Horitated face instass
FBD of Support E:-
FFE
from Ale DFE
FED
15kW
+
cose-
*15
2 V=0
SinB = 3/5
FEF Sino = 15
FEF = 25 KN C Compression)
Apply SH=0
FEF COSO FED
25x15 = FED
FED=20KN (TMsion)
at Joint F
FBD of Joint F
FGF
Step 2:
Step 4:
at Joint D
FBD & Joint D
+10= FDF
D
FCD=20kN
'DE = 20KN
loka
Similarly
и
Apply £H=0
FDE FCD 20kN (TOM)
Apply EV=O
FDF = 10kN (Ten)
from symmetry" lebt side of force in members are also same
FAH = 25KNC Compression)
FAB =20kN (Tension)
FGH
FGF
FBH=10KNC Tension)
FBC=20kN (Tension)
= 16-68 kN ( Compression)
FCF
16-25
30-40°
Apply Ev=o
'Loked
FCF CS (16-26) = 10 Cos (36-63°)
FCF = 8.35HN (Compression)
Apply EH-O
25KN
Fax + FCF sin (1626) + 10 sin (36-67") =25
F4F =
16-68 kN Compresse)
from symmetry of truss
FCF = FHC = 8:35 KWC Compression)
at Joint C
FBD of Joint 'C' =
FGC
8.35IN
Εντο
lokN
4
F
8.35KN
3
0/5
C
Rom Ale
Cob&= 415
(-) Fac + 8-35x2x cose + 10 = 0
FGC =
8.35x2x+40
FGC23-36 kN (Tension)
Transcribed Image Text:AH
TRUSSES
METHOD OF JANTS
DETERMINE THE FORCE
in EACH MEMBER OF THE
ROOF TRUSS STATE IF THE
memBeRS ARE in JENsion
OR COMPRESSion. Assume
* FOR every JoinT
FH=0
Fr=0
*"compression"
* REACTION
Me=0
Ra 10(4)+ 10 (8) + 10 (12)
16
16
IF Reaction Ra = 15KN ↑
IS TOWARDS
The JoinT
* Tension IF
Reaction is
"Away From
TheJoin T
ALL mem BeRS ARE Pin-connecTED
6H
Re = 30-15
Re-15KN T
AT JoinTA
AH
H
4m
B
10 KN
4m
MEMBER
AB
BC
CP
DE
EF
6H
AH
PHI
CH
C6
CF
PF
C
D
4m
4m
10KN
10KN
Re
AXIAL
FORCE
C/
E
3m
+-
!3m
ISKN
21=0
↓二个
음 AH
= 15
AB
AH = 15 (3)
AH= 25 KN Compression
374-0
AH
AB = (25)
Tension
AB=20KN
A
20
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