1. The cross-sectional area of each member of the truss is 1400 mm². Calculate the stresses in members DF, CE, and BD. В D 6 m 4 m A 4 m C 3 m Е 3m 100 kN 200 kN

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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- YOUR SUB ANSWERS MUST BE IN 6 DECIMAL PLACES.

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1. The cross-sectional area of each member of the truss is 1400 mm2.
Calculate the stresses in members DF, CE, and BD.
В
D
6 m
4 m
A
F
4 m
C
3 m
Е 3m
100 kN
200 kN
Transcribed Image Text:1. The cross-sectional area of each member of the truss is 1400 mm2. Calculate the stresses in members DF, CE, and BD. В D 6 m 4 m A F 4 m C 3 m Е 3m 100 kN 200 kN
Sample Problem 2
I kip
The bars of the truss have a cross-sectional area of 1.25 in2.
Determine the average normal stress in members AB and BE due
to the loading P = 8 kip. State whether the stress is tensile or
compressive.
B
3 ft
A
E
4 ft
4 ft
0.75 P= G kiPs
P = 8kipS
1
2. Normal Stress
3
4
5
В
Sample Problem 2
3 ft
Given:
P = 8 kip
A = 1.25 in?
Required: OAB, O BE
A
E
4 ft
4 ft-
0.75 P
Solution:
* Jont A
FAB
+ Joint E
Fre (를) 8.0
FAB - 13 3333 Kips T
BE
FAE
FED
FAE
FBE - 6:0
FBE: Gkips, I
Co kips
Gkips
FBE
FAB 13:3533kipS
BE:4.8ksi, T
A
1-25 in
1
2. Normal Stress
3
4
6.
Sample Problem 3
Three forces, each of magnitude P = 4 kN, are applied to the
mechanism shown. Determine the cross-sectional area of the
uniform portion of rod BE for which the normal stress in that
portion is +100 MPa.
0.100 m
P
P
|B
C
0.150 m
0.300 m
0.250 m
1
2. Normal Stress
3
4
6.
0.100 m
Sample Problem 3
P
Given:
P = 4 kN
B
+100 MPa
OBE
Required: ABE
Solution:
0.150 m
0.300 m
0.250 m
p.yKN
P:4KN
Cx
'cy
Dy
1
2. Normal Stress
3
4
6.
0.100 m
Transcribed Image Text:Sample Problem 2 I kip The bars of the truss have a cross-sectional area of 1.25 in2. Determine the average normal stress in members AB and BE due to the loading P = 8 kip. State whether the stress is tensile or compressive. B 3 ft A E 4 ft 4 ft 0.75 P= G kiPs P = 8kipS 1 2. Normal Stress 3 4 5 В Sample Problem 2 3 ft Given: P = 8 kip A = 1.25 in? Required: OAB, O BE A E 4 ft 4 ft- 0.75 P Solution: * Jont A FAB + Joint E Fre (를) 8.0 FAB - 13 3333 Kips T BE FAE FED FAE FBE - 6:0 FBE: Gkips, I Co kips Gkips FBE FAB 13:3533kipS BE:4.8ksi, T A 1-25 in 1 2. Normal Stress 3 4 6. Sample Problem 3 Three forces, each of magnitude P = 4 kN, are applied to the mechanism shown. Determine the cross-sectional area of the uniform portion of rod BE for which the normal stress in that portion is +100 MPa. 0.100 m P P |B C 0.150 m 0.300 m 0.250 m 1 2. Normal Stress 3 4 6. 0.100 m Sample Problem 3 P Given: P = 4 kN B +100 MPa OBE Required: ABE Solution: 0.150 m 0.300 m 0.250 m p.yKN P:4KN Cx 'cy Dy 1 2. Normal Stress 3 4 6. 0.100 m
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