WORLD FOR CIVIL

Topic 3.3a: Statically Indeterminate Structures - Example 1

The structure shown in Diagram 1 is formed by member ABCD (which is pinned to the floor at point B), Brass member AF, and Steel member CE, both of which are pinned to and support member ABCD, and both of which are pinned to the ceiling as shown. A downward external load of 20,000 lb. is applied at point D. For this structure we would like to determine the stress in the steel and brass members, and the movement of point D due to the load. As we analyze this structure, we ignore the fact that member ABCD will experience some bending (which in a sense is a deformation, and which we will deal with when the topic of beams is discussed). Ignoring the bending in this case effects the result only to a small degree.

Our first step is to apply our static equilibrium procedure to our structure

Part I - Static Equilibrium Analysis
1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions. In Diagram 2, a FBD of the structure is shown. At point B, where member ABCD is pinned to the floor, we replace the pin by support forces B_x and B_y. The brass and steel members are pinned to the ceiling. Normally we would replace the pins by horizontal and vertical support forces, however in this case we can do better (meaning less unknowns). Since both the steel and brass members have forces acting on them at only two points (each end), they are axial members and are in simple tension or compression. The 20,000 lb. load acting at end D tries to rotate the bar ABCD downward about point B. Thus putting the brass member (AF) in compression and the steel member (CE) in tension. As a result, the ceiling acts on the members as shown in the free body diagram.

2: Resolve forces into x and y components. (All forces are either in x or y direction.)
3: Apply the equilibrium conditions.
: B_x = 0
: B_y -F_Br + F_St - 20,000 lb. = 0
: + F_Br (10 ft.) + F_St (6 ft.) - 20,000 lb. (12 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 do see that B_x must be zero, from the first equation, but that is no help with the other two equations, in finding B_y, F_Br, F_St.) We might try to take the structure apart in some way, or redraw the FBD, but none of this will help. Static equilibrium conditions alone are not enough to solve this problem - it is statically indeterminate. 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
Step 1 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 20,000 lb. load will be to compress the brass member and to elongate the steel member which will cause member ABCD to rotate downward about hinged point B. We diagram this, showing and labeling the deformations involved. See Diagram 3. We now write a general relationship between the deformations of the members which we can get from the geometry of the problem. Since member ABDF is pinned at point A, as it rotates downward we see we can write (from similar triangles):

Deformation of Steel / 6 ft = Deformation of Brass / 10 ft
or we can rewrite as: Deformation of Steel = .6 * 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. At first it may not be clear how we can use this equation, since it involves deformations, not forces as in the equilibrium equations. However, if we recall our stress/strain/deformation relationships, we see we can write the deformation of a member as:

deformation = [ force in member * length of member] / [ young's modulus of member * area of member], or def = FL/EA. Substituting this relationship into our deformation equation we obtain: [FL/EA]_St = .6 [FL/EA]_Br We now have our additional relationship between the forces. After substituting in the values for the members we get:
( F_St *96")/(30 x 10⁶ lb/in² * .75 in²) = .6[( F_Br * 96")/(15 x 10⁶ lb/in² * 1.5 in² )]
If we simplify this equation we obtain: F_St = .6 F_Br We now substitute this into our torque equation from static equilibrium equations (shown on right)
: B_x = 0
: B_y -F_Br + F_St - 20,000 lb. = 0
: + F_Br (10 ft.) + F_St (6 ft.) - 20,000 lb. (12 ft.) = 0
and obtain: (F_Br)(10 ft.) + (.6 F_Br) (6 ft.) - 20,000 lb. (12 ft.) = 0; and solving we have:
F_Br = 17,650 lb., and F_St = 10,600 lb. We can now also solve for By, finding By = 27,050 lb
We find the stress from: Stress Steel = 10,600 lb/ .75 in² = 14,130 lb/in², Stress Brass = 17,650 lb/1.5 in² = 11,770 lb/in².
And finally we can find the movement of point D by first finding the elongation of member CE, the steel member, from Def. = FL/EA = (10,600 lb * 96 in.)/(30 x 10⁶ lb.in² * .75 in²) = .0452 in. Point D moves in proportion to the elongation of member CE, and we may write: .0452 in./6ft = Move. D/12 ft, solving Move. D = .0904 in.

We have now solved our statically indeterminate problem and determined the values of the external support reactions acting on the structure, the stresses, and the movement of point D.