2.2 Statics Review Examples
        Ex. 2.2.1 | Ex. 2.2.2 | Ex. 2.2.3 | Ex. 2.2.4 | Ex. 2.2.5

Example 2.2.1 Moment Equilibrium

Given: To pull a nail out of a piece of wood, a force of P = 20 lb is applied to the handle of a hammer. A block is placed under the hammer to make the job easier. The hammer pivots at Point A.

Req'd: Determine the maximum tension force in the nail. Let L = 10 in., q = 15o, and r = 2.0 in.

Sol'n: The tension force in the nail, F, can be determined by taking the FBD of the hammer, and summing the moments about Point A.

SMA = -P·L + F·r = 0

F = (20 lb·10 in) / (2.0 in)

F = 100 lb

Note that the angle q is not needed.

Applied force P and force of nail on hammer. What forces are missing if this is to be a correct FBD?

Example 2.2.2 Moment Equilibrium

Given: The crane at right is used to transport cargo to and from ocean-going ships. The crane is lifting an object of mass M = 1000 kg. The crane mast has a length of LAD = 30 m, and a mass of mmast = 400 kg. The mast is at an angle of b = 40° from the horizontal, and a = 20°. Note that ABC is a right angle.

Req'd: Determine the tension in the cable CD.

Sol'n: Create a free-body diagram of the crane mast. Be sure to include all forces acting on the mast, including the weight of the mast itself.

To determine the tension in the cable, TCD, all that is necessary is to sum the moments about Point A.

Click Here to see the FBD

MA = 0
    -(M·g)·[L·cos(b)] - (mmast·g)·[(L/2)·cos(b) ] + TCD·[L·sin(a)] = 0

TCD = ( 1000 kg.·9.8m/s2 + 0.5· 400 kg.·9.8m/s2) ·
cos( 40° )

sin( 20° )

TCD = kN

Example 2.2.3 Torsion

Given: Specifications for a certain lug-nut on a car require that the nut be tightened to a torque of 150 ft-lb.

Req'd: Determine the force P needed to obtain the correct torque. Let a = 9 in.

Sol'n: The forces P at the ends of the wrench arms cause a resultant torque at Point C of TC = P(2a).

From a free-body diagram of the shaft CD, the torque at C must equal the torque at D. Therefore:

TC = TD

P(2a) = 150 ft-lb

P = 150 ft-lb / (18 in) = 150 ft-lb / (1.5 ft)

P = 100 lb

Lug Wrench


FBD of lug wrench shaft.

Example 2.2.4 Beam

Given: The I-beam shown at right is acted upon by a point load 4 m from the left end.

Req'd: Determine the shear force and bending moment distributions throughout the beam. Plot the results on Shear and Moment Diagrams.

Sol'n: From the FBD of the entire beam, we can find the reaction forces at the beam supports by taking moments about each end (taking counterclockwise moments as positive):

SMx=0 = 0 -(4 kN)(4 m) + R2(10 m) = 0
SMx=10 = 0 (4 kN)(6 m) - R1(10 m) = 0

R1 = 2.4 kN    R2 = 1.6 kN

Check: 2.4 kN + 1.6 kN = 4 kN   OK

A section of the beam of length x (from the left support) is next isolated and treated as a FBD on which the internal shear force and bending moment are shown in their positive directions (we will get into positive convention stuff later). Applying the equilibrium equations to the FBD gives:

S FY = 0 2.4 + V(x) = 0
V(x) = -2.4 kN

S M = 0 M(x) - 2.4·x = 0
M(x) = 2.4·x kN-m

These results are valid for a section of the beam to the left of the 4 kN point load (x=4m).

A section of the beam to the right of the 4-kN load is next isolated in the same manner. Equilibrium gives:

S FY = 0 V(x) = 1.6 kN

S M = 0 M(x) = 1.6·(10 - x) kN-m

These results are valid for sections of the beam to the right of the point load (x=4m).

Next, the values for V(x) and M(x) can be plotted as shown at right.
Between R1 and the 4-kN point load, the internal shear force has a constant value of -2.4 kN (the shear force acts downward on a positive x-face). To the right of the point load the shear force has a value of +1.6 kN.

As we move from x = 0 to the right, the bending moment increases linearly from a value of 0 to 9.6 kN·m at x = 4 m. To the right of the point load, the bending moment decreases linearly from 9.6 kN·m to 0.

From this example it is clear that the change in moment is simply negative the accumulated area under the shear diagram [(2.4 N)(4 m) = 9.6 kN-m]. We will cover this subject in much more depth in Chapter 6.


FBD of entire beam


FBDs of sections of the beam


Shear diagram


Moment diagram


Shear and Moment Equations

Example 2.2.5

Given: A 14' by 8' road sign hangs over Hwy. 101 near UCSB. The bottom of the sign is 20 ft above the roadway. The weight of the sign is 300 lb, the weight of the supporting framework is 600 lb, and the weight of the mast is 2000 lb. During heavy storms, the road sign is subjected to wind gusts. The wind pressure results in an equivalent force of 2200 lb acting at the center of the sign.

Req'd: Determine the reaction forces and moments acting at the base of the sign.

Sol'n:
Step 1: The first step is to create a free-body diagram of the sign/support/mast system. Note that the weight of the sign, Ws, the weight of the supporting framework, Wf, and the wind force, Fw, all act through the center of the sign, A, while the weight of the mast, Wm, acts along the axis of the mast.
Applied forces

Step 2: The forces acting at A can be relocated to B, creating a force/couple systems at B where

  • RBy = Fw = 2200 lb
  • RBz = Ws + Wf = 300 lb + 600 lb = 900 lb
  • MBy = (Ws + Wf)·[(14 ft)/2] = 6300 lb
  • MBz = Ws·[(14 ft)/2] = 15400 lb

Step 3: Now we can apply the equilibrium equations to determine the reactant forces and moments.

SFx = 0 :  RCx = 0

SFy = 0 :  RCy + RBy = 0 :   RCy = -2200 lb

SFz = 0 :  RCz = RBz + Wm = 2900 lb

SMx = 0 :  MCx = RBy·(20' + 4') = 52800 ft-lb

SMy = 0 :  MCy = MBy = 6300 ft-lb

SMz = 0 :  MCz = MBz = 15400 ft-lb

Note that the moment about the z-axis, Mz, is often referred to as a torque since it is twisting the mast as if it were a torsion member. The torque on the mast is constant throughout the height in this case.



Applied Forces