by the Static Method Determining the Ultimate Load by the Mechanism
Method Analysis of a Fixed-End Beam under Concentrated Load
Analysis of a Two-Span Beam with Concentrated Loads Selection of
Sizes for a Continuous Beam Mechanism-Method Analysis of a
Rec
analysis of beam action, the general assumption is that the beam is in a
horizontal position and carries vertical loads lying in an axis of
symmetry of the transverse section of the beam. The vertical shear V at a
given section of the beam is the algebrai
calculation procedures in both the known and unknown units. Overseas
engineers who must work in USCS because they have a job requiring its
usage will find the dual-unit presentation of calculation procedures most
helpful. Knowing SI, they can easily conve
negative. GRAPHICAL ANALYSIS OFA FORCE SYSTEM The body
in Fig. \a is acted on by forces A, B, and C, as shown. Draw the vector
representing the equilibrant of this system. ( a) Space diagram and string
polygon FIGURE 1. Equilibrant of force system. Calcul
unit 0.0929 0.2831 6.894 4.448 1.356 1.355 0.06309 6.89 *Because of
space limitations this table is abbreviated. For a typical engineering
practice an actual table would be many times this length. TABLE 2
Typical Conversion Table* To convert from To Multi
longitudinal stress Use the relation s' =pD/(4t), where s' = longitudinal
stress, i.e., the stress parallel to the longitudinal axis of the cylinder,
lb/in2 (kPa), with other symbols as before. Substituting yields s' = 7200
lb/in2 (49,644.0 kPa). 3. Compu
rigidities. Using the subscripts s and b to T>,TTTT. * ^ j i ^ denote
steel and bronze, respectively, we FIGURE 24' ComPund shafi see that
O = T5LJ(JsG5) = W(J,G,), where the symbols are as given in the
previous calculation procedure. Solving yields Ts =
. 1.88 Most Economic
Section for a Beam with Intermittent Lateral Support under Uniform
Load . 1.89 Design of a Beam
with Reduced Allowable
Stress . 1.90 Design of a
Cover-Plated Beam . 1.92 Design of a Continuous
Beam . 1.95 Contents xi This page has bee
1860(0.75) + 1512(0.866) = 3069 in4 (12.77 dm4 ). The product of
inertia is Pxy=Px,y, cos 20+ [(Ix>-Iy)l2 sin20= 1512(0.5)+ 1(1458 1860)/2]0.866 = 582 in4 (2.42 dm4 ). Analysis of Stress and Strain The
notational system for axial stress and strain used in
the system In Fig. 2b, draw a free-body diagram of the bar. The bar is
acted on by its weight W, the force P, and the reaction R of the plane on
the bar. Show R resolved into its jc andy components, the former being
directed upward. 3. Resolve the forces
Safeguard by Alternative
Method . 9.113 Optimal Inventory
to Meet Fluctuating Demand . 9.115 xxx Contents This page has
been reformatted by Knovel to provide easier navigation. Finding
Optimal Inventory by Incremental-Profit
Method . 9.116 Simulation of
C
to 0.016 in (0.4064 mm); FIGURE 23 solve for P. Or, AL = (42,900 2.6P)IO"6 = 0.016 in (0.4064 mm); P = (42,900 - 16,000)72.6 = 10,350
Ib (46,037 N). 3. Determine the stresses and deformation Substitute the
computed value of P in the stress equation 5- = P
Frame . 1.8 Graphical Analysis of a Plane
Truss . 1.9 Truss Analysis by the Method of
Joints . 1.11 Truss Analysis by the Method of
Sections . 1.13 Reactions of a Three-Hinged
Arch . 1.14 Length of Cable Carrying Known
Loads . 1.15 Parabolic Cable Tension
+ 2EhI(LPIA)]*- 5 ), where h = vertical displacement of body, ft (m). 2.
Substitute the numerical values Thus, PIA = 18/1.2 = 15 lb/in2 (103
kPa); h = 3 ft (0.9 m); L = 5 ft (1.5 m); 2EhI(LPIA) = 2(30) x 106 )(3)/
[5(15)] = 2,400,000. Then s = 23,250 lb/i
Computing Beam Deflection .
1.58 Deflection of a Cantilever Frame . 1.59
Statically Indeterminate Structures . 1.61 Shear and
Bending Moment of a Beam on a Yielding
Support . 1.61 Maximum Bending Stress
in Beams Jointly Supporting a Load . 1.62
Theorem of
Design STEEL BEAMS AND PLATE GIRDERS Most Economic
Section for a Beam with a Continuous Lateral Support under a Uniform
Load Most Economic Section for a Beam with Intermittent Lateral
Support under Uniform Load Design of a Beam with Reduced
Allowable Stre
Moments Theorem of Three Moments: Beam with Overhang and Fixed
End Bending-Moment Determination by Moment Distribution Analysis
of a Statically Indeterminate Truss MOVING LOADS AND
INFLUENCE LINES Analysis of Beam Carrying Moving Concentrated
Loads Influe
longitudinal axis, in4 (cm4 ); D external diameter of shaft, in (mm); d
- internal diameter of shaft, in (mm). Substituting gives J= (7T/32)(54 34 ) = 53.4 in4 (2222.6 cm4 ). 2. Compute the shearing stress in the shaft
Use the relation ss = TRIJ, where ss
. 1.42 Beam Bending Stresses .
1.43 Analysis of a Beam on Movable Supports . 1.44 Flexural
Capacity of a Compound Beam . 1.45 Analysis of a
Composite Beam . 1.46 Beam Shear Flow and
Shearing Stress . 1.48 Locating the Shear Center of a
Section . 1.49 Cont
analysis of beam action, the general assumption is that the beam is in a
horizontal position and carries vertical loads lying in an axis of
symmetry of the transverse section of the beam. The vertical shear V at a
given section of the beam is the algebrai
Heating a Member Thermal Effects in Composite Member Having
Elements in Parallel Thermal Effects in Composite Member Having
Elements in Series Shrink-Fit Stress and Radial Pressure Torsion of a
Cylindrical Shaft Analysis of a Compound Shaft STRESSES IN
FL
Distribution of the Mean . 9.96 Estimation of
Population Mean on Basis of Sample
Mean . 9.97 Decision Making
on Statistical Basis . 9.98 Probability of Accepting a
False Null Hypothesis . 9.100 Decision Based on Proportion of
Sample . 9.101 Probability of
action line of R. 4. Draw the vector for the resultant and equilibrant In
Fig. Ia, draw the vector representing R. Establish the magnitude and
direction of this vector from the force polygon. The action line of R
passes through Q. Last, draw a vector equa
8.9966 in (228.51 mm). Likewise, AZ)5 = 21,980(9)7(30 x io6 ) =
0.0066 in (0.1676 mm), D5 = 8.99 + 0.0066 = 8.9966 in (228.51 mm).
Since the computed diameters are equal, the results are valid. HOOP
STRESS IN THICK-WALLED CYLINDER A cylinder having an
int
and equations to apply. Modern computer equipment provides greater
speed and accuracy for almost all complex design calculations. The
editor hopes that engineers throughout the world will make greater use
of available computing equipment in solving applie
distance, kips/lin ft (kN/m); L = span, ft (m); d = sag, ft (m). Substituting
yields H= 1.2(960)2 /[8(50)] = 2765 kips (12,229 kN). FIGURE 9 Cable
supporting load uniformly distributed along horizontal. Unit load = w 2.
Compute the tension at the supports
conversion factor cannot be found, simply use the dimensional
substitution. Thus, to convert pounds per cubic inch to kilograms per
cubic meter, find 1 Ib = 0.4535924 kg, and 1 in3 = 0.00001638706 m3 .
Then, 1 lb/in3 = 0.4535924 kg/0.00001638706 m3 27,680
load on each interval with its equivalent concentrated load Where the
load is uniformly distributed, this equivalent load acts at the center of the
interval of the beam. Thus WAB = 2(4) = 8 kips (35.6 kN); WBC = 2(6)
= 12 kips (53.3 kN); WAC = 8 + 12 = 20
[1/(0.5 xlO x 106 ) + 1/(0.25 >( 30 x 106 )]) = 741 lb/in2 (5109.2 kPa).
2. Compute the corresponding prestresses Using the subscripts 1 and 2
to denote the stresses caused by prestressing and internal pressure,
respectively, we find sal = pD/(2ta), where