Question

please do 14-11

petermine the stitfness matrix k for estant rix K for the truss AE i e the force in member 6. Take erand200 GPa for each member. termine the force in member 1 if this member before it was fitted into the truss. For 1move the 10-kN load. Take A 0.0015 m2 1o mm too long it so200 GPa for each member. 2 11 3 m

Answer 1

4 5 4 FREE BODY DIAGRAM

ANSWER---

Static Indeterminacy of truss =m+r_e -2 j

Where m= no of members

r_e=no of external reaction at the support

j =no of joints

D_s=m+r_e-2j =6+3-(2*4) = 1

Hence above truss is indeterminate to 1 st degree . When member 1 is fitted then forces will be induced in the member 1 . Member 1 is longer in length so compressive forces will be introduced in the member 1 . Let that force be X and this force will act at joint 2 and 3.

Q- SYSTEM OF FORCES

Force in member 1= Q₁= X

AT JOINT 2

Q₅*(3/5)=X SUM F_Y =0

Q₅=5 X/3

Q₅*(4/5)+Q₃ =0 SUM F_X =0

Q₃= -4X/3

AT JOINT 3

Q₆*(3/5) =X SUM F_Y=0

Q₆=5 X/3

Q₆*(4/5)+Q₂ =0 SUM F_X=0

Q₂= -4X/3

AT JOINT 1

Q₄+Q₆*(3/5)=0 SUM F_Y=0

Q₄= -X

PLOTTING DATA IN A TABLE

MEMBERS Q- FORCE LENGTH Q²*L

1 -X 3 3X²

2 -4X/3 4 64X²/9

3 -4X/3 4 64X²/9

4 -X 3 3X²

5 5X/3 5 125X²/9

6 5X/3 5 125X²/9

TOTAL= 48 X²

TOTAL STRAIN ENERGY ,$U=\frac{\sum Q^{2}*L}{2AE}=\frac{48X^{2}}{2AE}$

FOR TRUE VALUE OF X

$\frac{\partial U}{\partial X}=\frac{2*48*X}{2AE}=\Delta$

$\frac{48*X}{0.0015*200*10^{6 }kpa}=10 mm$

$X=\frac{10^{-2}*0.0015*200*10^{6}}{48}=62.5 KN$

FORCE IN MEMBER 1 =X = -62.5 KN

FORCE IN MEMBER 2 = -83.333 KN

FORCE IN MEMBER 3 = -83.333 KN

FORCE IN MEMBER 4 = -62.5 KN

FORCE IN MEMBER 5 =104.167 KN

FORCE IN MEMBER 6 =104.167 KN