Part A=? Also, Make sure that you get the correct answer for part B as it shows

Question

Part A=?

Also, Make sure that you get the correct answer for part B as it shows on the pic.

Consider the truss shown in (Figure 1). Suppose that F1-55 kN and F 30 kN Part A Determine the force in member DC of the truss, and state if the member is in tension or compression. Express your answer to three significant figures and include the appropriate units. Enter negative value in the case of compression and positive value in the case of tension. HA Foc= Value Units X Incorrect; Try Again; 2 attempts remaining Figure 1 of 1 Part B 50 kN Determine the force in member HC of the truss, and state if the member is in tension or compression. 40 kN Express your answer to three significant figures and include the appropriate units. Enter negative value in the case of compression and positive value in the case of tension. HC115 kN 1.5 m Correct Part C

Explanation / Answer

let reaction forces at F be Fy in vertical direction.

let reaction forces at A be :

Ax in horizontal direction to the right

Ay in vertical direction, upwards

balancing moments about A:

Fy*4=40*2+F2*3+F1*1.5

==>Fy=63.125 kN…(1)

balancing forces in horizontal direction:

F2+F1=Ax

==>Ax=85 kN…(2)

balancing forces in vertical direction:

Fy+Ay=40+50

==>Ay=26.875 kN…(3)

consider the joint A:

angle IAB=arctan(2/1.5)=53.13 degrees

let force in member AI be Tai and force in member AB be Tab, both in tension.

balancing forces at A along horizontal direction:

Fab*sin(theta)+Ax=0

==>Fab=-106.25 kN

balancing forces at A along vertical direction:

Fai+Ay+Fab*cos(theta)=0

==>Fai=36.875 kN

at junction B:

angle IBA=90-angle IAB=36.87 degrees

balancing forces along horizontal direction:

F1+Fbi+Fba*cos(36.87)=0

==>Fbi=30 kN

balancing forces along vertical:

Fbc=Fba*sin(36.87)=-63.75 kN

at junction I:

angle CIB=theta=arctan(1.5/2)=36.87 degrees

balancing forces in horizontal direction:

Fic*cos(theta)+Fib=0

==>Fic=-37.5 kN

balancing forces in vertical direction:

Fic*sin(36.87)+Fih=Fia

=>Fih=59.375 kN

at junction C:

angle ICH=arctan(1.5/2)=36.87 degrees

angle DCH=arctan(1.5/2)=36.87 degrees

balancing forces in vertical direction:

Fcd*sin(36.87)=Fci*sin(36.87)+Fcb

==>Fcd=-143.75 kN

balancing forces in horizontal direction:

Fcd*cos(36.87)+Fch+F2+Fci*cos(36.87)=0

==>Fch=115 kN

hence answers are :

part A:

-143.76 kN

part B:

115 kN

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