2b_ex

7/30/2019 2b_ex

1/4

2b-ex. Statics Examples Ex.2b.1 Ex. 2b.2 Ex. 2b.3 Ex. 2b.4 Ex. 2b.5

Back | Index | Next

Example 2b.1 Moment Equilibrium

Given: In an attempt to pull a nail out of a piece of wood, force P =

20 lb is applied perpendicular to the handle of a hammer. A block is

placed under the hammer to provide leverage. The hammer pivots

(rocks) at Point A. Let L = 10 in., q = 15o and r = 2.0 in.

Req'd: Determine the tensile force in the nail before the nail begins

to slide (the nail remains at rest so had zero acceleration).

Sol'n: Tensile force F can be determined by taking the FBD of the

hammer and summing moments about Point A. Taking

counterclockwise moments as positive:

SMA

= PL + Fr = 0

F = PL/r = [(20 lb)(10 in.)] / (2.0 in.)

F = 100 lb

Note that angle q is not needed since P is normal to the handle.

Applied force P and force ofnail on hammer. What forcesare missing if this is to be acorrect FBD?

Example 2b.2 Moment Equilibrium

Given: A crane is used to move cargo to and from

ocean-going ships. The crane is lifting an object ofmass M = 1000 kg. Boom AD has length of LAD =

30 m, and mass of mb = 400 kg. The boom is b = 40

from the horizontal, and a = 20. Angle ABC is a

right angle.

Req'd: Determine the tension in cable CD.

Sol'n: Create a free-body diagram of the boom,

including the weight of the boom itself, which is

assumed to act at its geometric center.

To determine the tension TCD, sum moments about

Point A.

Click Here to see the FBD

TCD = 26.3 18.04 kN

x Examples http://strengthandstiffness.com/2_statics/page_2b-ex.ht

d 4 27.7.2013 22:59

7/30/2019 2b_ex

2/4

Example 2b.3 Torsion

Given: Specifications a lug-nut require that it be tightened to

a torque of 150 ft-lb.

Req'd: Using the lug-wrench shown, determine force P to

obtain the required torque. Let a = 9 in.

Sol'n: The forces P at the ends of the wrench arms cause atorque at Point C of TC = P(2a), which acts about the CD

axis.

From a FDD of shaft CD, the torque at C must equal the

torque at D:

TC = TD

P(2a) = 150 lb-ft

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

P = 100 lb

Lug Wrench

FBD of lug wrench shaft.

Example 2b.4 Beam

Given: An I-beam shown is acted upon by a 4

kN point load 4 m from the left end.

Req'd: Determine the shear force and bending

moment distributions throughout the beam. Plotthe results on Shear and Moment Diagrams.

Sol'n: From the FBD of the entire beam, the

reaction forces at the supports are found by

taking moments about each end:

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: SFy= 2.4 kN + 1.6 kN 4 kN = 0 OK

A segment of the beam of length x (0< x < 4 m)

is next isolated and treated as a FBD. The

internal shear force and bending moment acting

on the segment are shown in their positive

directions (per this text).

Applying equilibrium:

S Fy = 0 2.4 + V(x) = 0

V(x) = 2.4 kN

FBD of entire beam

FBDs of segments of the beamleft: for x < 4m; right: for x > 4m. Positive sense of

shear force and moment per convention of text.

Shear diagram


d 4 27.7.2013 22:59

7/30/2019 2b_ex

3/4

S M = 0 M(x) 2.4x = 0

M(x) = 2.4x kNm

These internal loads act on any cross-section of

the beam to the left of the 4 kN load (x4m).

Moment diagram

Shear and moment equations

V(x) and M(x) can be plotted on Shear Force and Bending MomentDiagrams.

Between R1 and the 4-kN point load, the internal shear force is constant: V = 2.4 kN (the

negative sign indicates the shear force acts downward on a positive x-face, opposite

drawn). To the right of the point load the shear force has a value of +1.6 kN.

From x = 0 to 4 m, the bending moment increases linearly from of 0 to 9.6 kNm. To the

right of the load, the bending moment decreases linearly from 9.6 kNm to 0.

The change in moment between two cross-sections is the negative in area under the shear

diagram. From 0 to 4 m: Area = (2.4 N)(4 m) = 9.6 kNm].

Beams are covered in more depth in Chapter 6.

Example 2b.5 Combined Loading

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 support mast is 2000 lb. During a

storm the wind pressure applies an equivalent force of 2200 lb

acting at the center of the sign.

Req'd: Determine the reaction forces and moments at the base

of the sign.


d 4 27.7.2013 22:59

7/30/2019 2b_ex

4/4

Sol'n:

Step 1. Create a FBD of the sign/support/mast system. 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 at Point A. The weight of the mast Wm acts along its axis

BC.

Applied forces

Step 2. The forces acting at A can be relocated to B, creating

force-couple systems at B:

RBy

= Fw

= 2200 lb

MBz = Fw[(14 ft)/2] = 15400 lb-ft

RBz = Ws + Wf= 900 lb

MBy = (Ws + Wf)[(14 ft)/2] = 6300 lb-ft

Step 3. Apply equilibrium on mast BC to determine the

reaction forces and moments. The reactions are drawn in the

positive direction of the coordinate axes.

SFx = 0 : RCx = 0

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

SFz = 0 : RCz = RBz + Wm : RCz = 2900 lb

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

SMy = 0 : MCy = MBy = 6300 lb-ft

SMz

= 0 : MCz

= MBz

= 15400 lb-ft

The moment about the z-axis Mz, is generally referred to as a

torque since it twists the mast as if it were a torsion member.

Here, torque on the mast is constant throughout BC.

Applied Forces

Top Back | Index | Next

Updated: 05/24/2009 DJD


d 4 27 7 2013 22 59

2b_ex

Documents