WORLD FOR CIVIL

Topic 3.2c: Statically Determinate - Example 3

In the structure shown in Diagram 1, member BCDFG is assumed to be a solid rigid member. It is supported by a two cables, AB & DE. Cable AB is steel and cable DE is aluminum. Both cables have a cross sectional area of .5 in². For this structure we would like to determine the axial stress in cable AB and DE. We would also like to determine the movement of point F due to the applied loads. [Young's Modulus for steel : E_st = 30 x 10⁶ psi, Young's Modulus for Aluminum: E_al = 10 x 10⁶ psi.]

Part I. To solve the problem we first need to determine the external support forces acting on the structure. We proceed using our static equilibrium procedure (from Topic 1 - Statics)
1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions.

2: Resolve forces into x and y components. All the forces are acting in the y-direction, as is shown in the FBD.
3: Apply the equilibrium conditions.
Sum F_y = A_y + E_y - 30,000 lb. - 10,000 lb. = 0
Sum T_B = (-30,000 lb.)(4 ft) + E_y(10 ft) - (10,000 lb.)(12ft) = 0
Solving for the unknowns: E_y = 24,000 lb.; A_y = 16,000 lb.

Part II. Since both the steel member AB and the aluminum member DE are single axial members connected to the supporting ceiling, the external forces exerted by the ceiling on the members is also equal to the internal forces in the members. Thus F_AB = 16,000 lb. (tension), F_DE = 24,000 lb. (tension). To find the stress in each cable is now straight forward. We apply the stress equation

(from the Stress / Strain / Hooke's Law relationships shown to the right).
So, Stress AB = F/A = 16,000 lbs/.5 in² = 32,000 psi., Stress DE = F/A = 24,000 lbs/.5 in² = 48,000 psi.


Part III To find the movement of point F requires us to use a bit of geometry. Point F moves since both member AB and ED deform and member BCDFG moves downward according to these deformations.
1: Calculate deformation of members AB and ED.
Def_AB = (FL / AE)_AB = (16,000 lbs)(120 in) / (30*10⁶ lbs/in²)(.5 in²) = .128 in
Def_ED = (FL / AE)_ED = (24,000 lbs)(120 in) / (10*10⁶ lbs/in²)(.5 in²) = .576 in

2: Movement of point F. In Diagram 3 we have shown the initial and final position (exaggerated) of member BCDFG.

Point B moves down .128 inches (the deformation of member AB), point D moves down .576 inches (the deformation of member DE). Point F moves down an intermediate amount. To determine this we have drawn a horizontal line from the final position of point B across to the right side of the beam as shown. From this we see that the distance point F moves down is .128 inches + "x" (where x is the distance below the horizontal line as shown in the diagram). We can determine the value of x from proportionality (since similar triangles are involved), and write: (x / 12 ft) = (.448 in / 10 ft). Solving for x we find: x = .5376 inches.
So the Movement of F = .128 inches + .5376 inches = .666 inches