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MD Statics 2022 solutions

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‘Mechanical dynamics’
Part 1: Introduction / Statics
FH-Kärnten - Villach / Oct/Nov 2022
Dipl.-Ing. Georg Schmeja
References
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Hibbeler, Statics and Mechanics of Materials (SI-Edition) / Pearson Education
Horst Herr, Technische Mechanik
(Alfred Böge, Technische Mechanik)
(ÖH-TU-Graz, Prüfungsbeispielsammlung Statik)
All pages marked with "© Pearson Education" contain copies out of:
"Hibbeler: Statics and Mechanics of Materials (SI-Edition)"
Schmeja / 2022/2023
Mechanical Dynamics - Statics
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Technical Mechanics
• Technical Mechanics
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–
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Statics
Kinematic (geometric relations of motion)
= Dynamics
Kinetics (motion by forces)
Dynamics (acceleration and forces)
Material-/Structural Strength (stresses, deflection and stability)
Hydromechanics (Hydrostatics / Hydrodynamics)
• Knowledge about statics has to be considered as a precondition to
understand dynamics.
• This dynamic-course will also teach some basics of statics to
safeguard the principle knowledge of mechanics among all
students.
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Statics means ...
• ΣF = 0
• ΣM = 0
= Equilibrium
• F = m*a
• M = F*x
• 3 degrees of freedom in 2 dimensional systems
• 6 degrees of freedom in 3 dimensional systems
• Moment [DE/EN) = torque, bending-moment, torsional moment
• Momentum [EN] = Impuls [DE]
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Some essential (basic) units
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Mass m in [kg]
Time t in [s]
Length/Distance s in [m]
Force F in [N] before in [kp]
Torque (moment) M in [Nm]
Velocity v in [m/s]
Acceleration a in [m/s²]
Work / Energy W or E in [J] = [Ws]
Power P in [W]
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F = m*a
M = F*r
W = F*s
P = W/t
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Symbols
Symbol
Reaction-Forces
Degrees of Freedom
 supported
 hinged
 clamped
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Basic Triangular Formulas
• General
C
γ
b
– Sine law:
a / sin  = b / sin  = c / sin γ


– Cosine law:
A
c² = (a² +b² - 2ab cos γ)
a
c
B
• Right angle triangle γ = 90°
– c = Hypotenuse
– a, b = Cathetus
sin  = oposit leg / hypotenuse = a / c
cos  = adjacent leg / hypotenuse = b / c
tan  = oposit leg / adjacent leg
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Vector Addition and Forces
© Pearson Education
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Vector Addition and Forces
© Pearson Education
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Vector Addition and Forces
© Pearson Education
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Vector Addition and Forces
© Pearson Education
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Vector Addition and Forces
© Pearson Education
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Vector Addition and Forces
© Pearson Education
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Graphic Solution
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Vector Addition and Forces
© Pearson Education
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Vector Addition and Forces
© Pearson Education
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Vector Addition and Forces
© Pearson Education
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Moment
© Pearson Education
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Moment
© Pearson Education
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Moment
d) explain and indicate the distribution of shear-force and momentum
The sketch shows a beam with 1500 mm length. A mass of m=25 kg is fixed to the free end
a) Calculate the weight in N
b) Calculate the max. moment
c) How could the moment be called
d) explain and indicate the distribution of shear-force and moment
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Moment
© Pearson Education
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Moment
© Pearson Education
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Moment and Force
© Pearson Education
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Moment and Force
© Pearson Education
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Graphic indications
• http://www.statik-lernen.de/grundl_zustandslinien_2.html
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Free body diagram
© Pearson Education
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Free Body Diagrams
© Pearson Education
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Free Body Diagrams
© Pearson Education
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Free Body Diagrams
© Pearson Education
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Replacing supports by forces
and moments
© Pearson Education
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Free body diagram for hinges,
clamps, bearings ...
© Pearson Education
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Principles
free body
NOT possible
© Pearson Education
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Draft the free body diagram
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Draft the free body diagram
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Draft the free body diagram
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Draft the free body diagram
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Draft the free body diagram
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Draft the free body
diagram of the piston
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Draft the free body
diagram of the crane
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Free body diagram
© Pearson Education
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Free body diagram
© Pearson Education
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Free body diagram
Solution to 5-32
© Pearson Education
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© Pearson Education
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FA
Determine the size and direction of the reaction force in A
ΣFx = 0 = F2*cos45 – F3*cos60 +F5*cos60 +FA = 0
Fx = - 294,9 N
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Determine the forces in beams 1,2,3,4 by drafting. F = 40 kN
Tip: Start with F1 and F2; then determine F3 and F4 from F2
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Graphic Solutions
Graph of geometry
Graphical solution with
multiple forces not in one
point:
Create the graph of forces and the
resultant
Graph of forces
Choose a pole 0 and draft the polelines, starting with No. 0
Shift the pole-lines into the graph of
geometry
Shift the resultant to the intersection
of first and last pole-line
Graph of geometry
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© Pearson Education
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Determin the size of F_res using the
“Seileckverfahren”
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The trick of the “final line”:
Start with pole-line 0/1 (a) in
one of the supports.
Draw a line between this point
and intersection of 5 and the
actionline of F_By (b). Connect
a and b (final line”s”).
Shift s back into the plan of
geometry and measure the
distribution between F_Ay and
F_By
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Determin:
F_Ax; F_Ay; F_By
Determine F_A and F_B analytically.
FAy = F1+F2y+F3y+F4+F5y-FAy
FAy = 11,8 kN  FAres = wurzel(FAx² + FAy²)
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If the lines of action are known, the resultant of four
forces can be determined by the “Cullmann-Line”.
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Cullmann Solution
Determin the forces in A and B:
• Draft the geometry graph and the force graph
• Indicate the resultant force in the geometry graph and find the intersection
with the line of action of one support
• Draw the Cullmann-Line to the 2nd support
• Shift the Cullmann-Line into the plan of support-forces, thus defining the
support reactions
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a)
Determine F3 with alpha = 32°
b)
Determine the reaction forces in D
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Homework
A pantograph is designed out of a bent lever.
Determin Fz to gain equilibrium.
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Center of Gravity
© Pearson Education
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Center of Gravity
© Pearson Education
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Center of Gravity
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Truss evaluation
• Staticly defined if
s=2*k–3
s = number of beams
k = number nodes
no bending of beams
no stiffness of joints
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Method of joints
© Pearson Education
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Zero-force
members
© Pearson Education
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Method of
joints
© Pearson Education
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Method of joints
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Resolve by using the
cullmann principle
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Friction (static/dynamic)
• FR_0 = µ0 * FN
• FR = µ * FN
Static friction
Dynamic friction
• Fricion is independent from the size of the contact area
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Friction
• M56
© Pearson Education
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Friction
© Pearson Education
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Loco towing force
• What is the maximum theoretic force a locomotive with 4 axles and
an axle-load of 20 t can develop?
• Estimation µ = 0,12 – 0,35.
• What is the relevance of the given range for µ ?
• 80t * 9,81 * 0,12 = 94 kN  for operation
• 80t * 9,81 * 0,35 = 274 kN  for design (gearbox, axle, ..)
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Friction
© Pearson Education
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Seilreibung (Belt friction)
F1 = F2 * e µ
 in rad
A rope is slung around
a wheel n=2,25
times. µ = 0,35
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Which force is
transmitted into A
when the cylinder
is turning and F_G
= 2kN
•
What is the
braking torque if
d=500 mm
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References
•
•
•
•
Hibbeler, Statics and Mechanics of Materials (SI-Edition) / Pearson Education
Horst Herr, Technische Mechanik
(Alfred Böge, Technische Mechanik)
(ÖH-TU-Graz, Prüfungsbeispielsammlung Statik)
All pages marked with "© Pearson Education" contain copies out of:
"Hibbeler: Statics and Mechanics of Materials (SI-Edition)"
Schmeja / 2022/2023
Mechanical Dynamics - Statics
71
Helpful links
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http://www.bau.hs-wismar.de/boddenberg/tm1/
http://www.statik-hilfen.de/Mechanik/Mechanik_01.html
http://www.tm-aktuell.de/TM1/
http://www.tm-aktuell.de/TM2/
http://www.tm-aktuell.de/TM3/
http://www.tm-aktuell.de/TM5/Animationen/animationen.html
http://www.statik-lernen.de/grundl_kraefte_1.html
http://www.engr.uky.edu/statics/Content/ContentHome.html
• http://www.strandbeest.com
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