Mechanics of Materials Statically Indeterminate 20 ft 40 10

Mechanics of Materials Statically Indeterminate 2.0 ft 40 1.0 ft. LOAD THE TIE RODS ARE MADE OF STRUCTURAL STEEL ROD AC HAS A DIAMETER OF .70\" ROD BC HAS A DIAMETER OF .40” THE STAINLESS STEEL CABLE CD IS .10\" DIAMETER LOAD=1000 lb. DETERMINE THE VERTICAL DISPLACEMENT OF PIN C

Solution

Let the vertical displacement of the joint C be d

Angle of BC with horizontal = tan^-1(1/2)=26.56^o

Angle of AC with horizontal = 90-40=50^o

axial deformation in cable CD = d

axial deformation in member AC = dsin50=0.766d

axial deformation in member BC = dsin26.56=0.447d

initial length of member BC = sqrt(1²+2²)=2.236 ft=26.833 in

initial legnth of member AC = 1/cos40 = 1.305 ft=15.665 in

initial length of CD = 1 ft=12 in

strain in BC = 0.447d/26.833=0.0166d

strain in AC = 0.766d/15.665=0.0489d

strain in CD = d/12=0.0833d

modulus of elasticity of steel = 29000 ksi

area of cross section of AC = pi*0.7²/4=0.385 in²

area of cross section of BC = pi*0.4²/4=0.126 in²

area of cross section of CD = pi*0.1²/4=0.00785 in²

Force in a member =stress*area= (modulus of elasticity*strain)*area of cross section

Force in member AC =(29000*0.0489d)*0.385=545.97d kips

Force in member CD = (29000*0.0833d)*0.00785=18.96d kips

Force in member BC = (29000*0.0166d)*0.126=60.66d kips

Applied force at C=1000 lb=1 kip

From vertical force equilibrium of joint C,we get

545.97d*cos40 + 18.96d + 60.66d*cos26.56 = 1

d=0.002 in

Therefore, vertical displacement of C=0.002 in