WORLD FOR CIVIL

Topic 3.7: Stress, Strain & Hooke's Law - Solution Exam Question 2

In the structure shown horizontal member BCDE is supported by vertical brass member, AB, an aluminum member EF, and by a roller at point C. Both AB and EF have cross sectional areas of .5 in². The structure is initially unstressed and then a load of 24,000 lb. is applied at point D. For this structure:
A. Draw a Free Body Diagram showing all support forces and loads.
B. Determine the axial stress that develops in aluminum member EF.
C. Determine the resulting movement of point E.
Est = 30 x 10 6 psi; Ebr = 15 x 10 6 psi; Eal = 10 x 10 6 psi
Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints.

Solution:
Our first step is to apply our static equilibrium procedure to our structure
Part I - Static Equilibrium Analysis

Step 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions. The 24,000 lb. load acting at end D tries to rotate the bar BCDE downward about point C. This puts the brass member, AB, in compression and the aluminum member, EF, in tension. As a result the ceiling acts on the members as shown in the free body diagram.
Step 2: Resolve forces into x and y components. (All forces are either in x or y direction.)
Step 3: Apply the equilibrium conditions.

: C_y - F_Br + F_Al - 24,000 lb. = 0
: + F_Br (4 ft.) + F_Al (12 ft.) - 24,000 lb. (8 ft.) = 0
At this point in a statically determinate problem, we would, in most cases, be able to solve for the external support reactions. However, in this case, we observe that we have three unknowns and only two independent equations - and can not solve. We might try to take the structure apart in some way, or redraw the FBD, but none of this will help. Another way to state the problem is that we need another independent equation to solve for the unknowns. The deformations of the brass and steel members will give us this additional equation..

Part II. - Deformation Equation
Our first step is to find some general relationship between the deformations of the members of the structure. We may get this from the way the problem is stated, or often from the geometry of the structure - as in this case. The effect of the 24,000 lb. load will be to compress the brass member and to elongate the aluminum member which will cause member BCDE to rotate downward about point C. We diagram this, showing and labeling the deformations involved. We now write a general relationship between the deformations of the members which we can get from the geometry of the problem. Since member BCDE rotates downward about point C, we see we can write (from similar triangles):

Deformation of Br / 4 ft = Deformation of Aluminum / 12 ft or we can rewrite as: Deformation of Aluminum = 3* Deformation of Brass, or symbolically:

This is our additional equation which we will use in combination with the static equilibrium equations to find the external forces acting on the structure. Since deformation of a member = [ force in member * length of member] / [ young's modulus of member * area of member], or def = FL/EA, we can substitute into the deformation relationship and obtain:. [FL/EA]_Al = 3 [FL/EA]_Br
We now have our additional relationship between the forces. After substituting in the values for the members we have:
( F_Al *48")/(10 x 10⁶ lb/in² * .5 in²) = 3[( F_Br * 96")/(15 x 10⁶ lb/in² *.5 in² )]
If we simplify this equation we obtain: F_Al = 4 F_Br We now substitute this into our torque equation from static equilibrium equations
: + F_Br (4 ft.) + F_Al (12 ft.) - 24,000 lb. (8 ft.) = 0
and obtain: (F_Br)(4 ft.) + (4 F_Br) (12ft.) - 24,000 lb. (8 ft.) = 0; and solving we have: F_Br = 3,690 lb., and F_Al = 14,770 lb. We can now also solve for Cy, finding Cy = 12,920 lb We find the stress from: Stress Brass = 3,690 lb/ .5 in² = 7,380 lb/in², Stress Aluminum = 14,770 lb/.5 in² = 29,540 lb/in².

And finally we can find the movement of point E by finding the elongation of member EF, the aluminum member, from Deformation of EF = FL/EA = (14,770 lb * 48 in.)/(10 x 10⁶ lb.in² * .5 in²) = .142 in.