Please double check your answer in allowable stress and allowable loads. Please double check your answer Since the member is in compression,The maximum slenderness ratio is given as,LDIAmaxAs, Amax >4.71₁given as,= 218.34Also, check for slenderness ratio,=4.71,are,FAll = F crFySolution== 4.71₁1= 133.756"min5x10³22.9Fcr=0.877Fe= 0.9 × 36.313= 32.682 MPa= 0.877 Xthen the critical stress, Fcr is= 0.877 X2x105248π²El= 36.313 MPaTherefore, the allowable stress and allowable load1².maxπ²×2×105218.34²Pall= Fall xAg= 32.682 × 1858= 62.72× 10³ N= 60.72 KNThe member force in DI is 7.288 kN compressiveThe allowable stress in DI is 32.682 MPaThe allowable load in DI is 60.72 kN From the figure shown, a truss is subjected to loads PI = 15 kN, P2 = 15 kN and P3 = 25 kN at joint C, D and E respectively. Use SI = 3m; S2 = 4 mP₁Properties of diagonal members (2L75 x 75 x 6):Ag = 1858 mm2Fy = 248 MParx = 22.9 mmry = 33.2 mmGa) What is the axial force in kN, in member DI?b) What is the allowable stress in MPa, in member DI?c) Calculate the allowable load in kN, in member DI.H6 @ SiIP₁EBS,CIVIL ENGINEERING-STEEL DESIGN...

Gross area, Yield strength, Radius of gyration about x-axis, Radius of gyration about y-axis, To…

From the figure shown, a truss is subjected to loads PI = 15 kN, P2 = 15 kN and P3 = 25 kN at joint C, D and E respectively. Use SI = 3 m; S2 = 4 m P₁ Properties of diagonal members (2L75 x 75 x 6): Ag = 1858 mm2 Fy = 248 MPa rx = 22.9 mm ry = 33.2 mm G a) What is the axial force in kN, in member DI? b) What is the allowable stress in MPa, in member DI? c) Calculate the allowable load in kN, in member DI. D H 6 @ Si P₁ CIVIL ENGINEERING- STEEL DESIGN ...

From the figure shown, a truss is subjected to loads PI = 15 kN, P2 = 15 kN and P3 = 25 kN at joint C, D and E respectively. Use SI = 3 m; S2 = 4 m P₁ Properties of diagonal members (2L75 x 75 x 6): Ag = 1858 mm2 Fy = 248 MPa rx = 22.9 mm ry = 33.2 mm G a) What is the axial force in kN, in member DI? b) What is the allowable stress in MPa, in member DI? c) Calculate the allowable load in kN, in member DI. D H 6 @ Si P₁ CIVIL ENGINEERING- STEEL DESIGN ...

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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Question

Please double check your answer in allowable stress and allowable loads. Please double check your answer

Since the member is in compression,
The maximum slenderness ratio is given as,
LDI
Amax
As, Amax >4.71₁
given as,
= 218.34
Also, check for slenderness ratio,
=4.71,
are,
FAll = F cr
Fy
Solution
=
= 4.71₁1
= 133.756
"
min
5x10³
22.9
Fcr=0.877Fe
= 0.9 × 36.313
= 32.682 MPa
= 0.877 X
then the critical stress, Fcr is
= 0.877 X
2x105
248
π²El
= 36.313 MPa
Therefore, the allowable stress and allowable load
1².
max
π²×2×105
218.34²
Pall= Fall xAg
= 32.682 × 1858
= 62.72× 10³ N
= 60.72 KN
The member force in DI is 7.288 kN compressive
The allowable stress in DI is 32.682 MPa
The allowable load in DI is 60.72 kN

From the figure shown, a truss is subjected to loads PI = 15 kN, P2 = 15 kN and P3 = 25 kN at joint C, D and E respectively. Use SI = 3
m; S2 = 4 m
P₁
Properties of diagonal members (2L75 x 75 x 6):
Ag = 1858 mm2
Fy = 248 MPa
rx = 22.9 mm
ry = 33.2 mm
G
a) What is the axial force in kN, in member DI?
b) What is the allowable stress in MPa, in member DI?
c) Calculate the allowable load in kN, in member DI.
H
6 @ Si
I
P₁
E
B
S,
CIVIL ENGINEERING-
STEEL DESIGN
...

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Step 1: Determine the given data.

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Step 2: Determine the support reactions.

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Step 3: (a) Determine the force in member DI.

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Step 4: (b) Determie the allowable stress.

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Step 5: (c) Determine the allowable load.

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