Mathematical Modelling (Statics) Flashcards

Question

Varignon’s Theorem:

Answer 1

Moment of a force about any point is equal to the sum of the moments of the components of the force about the same point.

Answer 2

M = r x F = (Fr x sin(alpha))u sub(M) where r is the position vector, Fr sin(alpha) is the magnitude and u sub(M) is a unit vector in the direction of moment axis.

Answer 3

* In 3D, the orientation of the axis can be obtained by the “right-hand rule” (similar to current in physics). The moment is as well drawn as a “double arrow”. A positive moment turns always anti -clockwise. Thus in 3D, there are three principal directions for the moments each rotating about one of the coordinate axis. Mx;My;Mz

Answer 4

Two equal and opposite forces separated by a distance d produce moment M of magnitude M = F(a + d) − Fa = Fd

Answer 5

Force-Couple systems: The translation of a force perpendicular to its direction creates a moment.

Answer 6

* Definition: A free body diagram is a sketch that shows a body “free” from its surroundings with all the forces and moments that act on the body. In other words, “Isolate” the body ! * When isolated, the forces and moments at the points where the body was linked to other objects should be considered. * Referring to Newton’s 3rd principle “action = reaction”, the forces and moments at the two sides where the object was separated from the environment have the same values but opposite orientations.

Answer 7

Dot product calculates the sum of the two vectors’ multiplied elements. Dot Product returns a scalar number as a result

Answer 8

The dot product is useful in calculating the projection of vectors. Dot product in Python also determines orthogonality and vector decompositions.

Answer 9

You multiple the a1 with b1 and a2 with b2 and then get the sum

Answer 10

In mathematics, the cross product or vector product (occasionally directed area product, to emphasize its geometric significance) is a binary operation on two vectors in three dimensional space. The magnitude of the product equals the area of a parallelogram with the vectors for sides.

Answer 11

Any of the minimum number of coordinates required to specify completely the motion of a mechanical system.

Answer 12

1. Draw an outlined shape of the body Imagines the body to be isolated of “free cut” from its constrains and connections and sketch its outlined shape 2. Show all force and couple moments acting on the body Identify all the external forces and couple moments acting on the body. Usually they are: a) applied loadings, b) reactions occurring at the supports and points of contact with other bodies, c) the weight of the body. 3. Label all the loadings and specify their direction relative to the x and y axis. 4. Indicate the dimensions of the body necessary for computing the moments of forces.

Answer 13

If a number of degrees of freedom are restricted (e.g. by an external support), you need to insert the same number of reactions (forces or moments) to the free body diagram

Answer 14

Equilibrium means that nothing is moving (or is moving at constant velocity Internal forces balance out!

Answer 15

A system is in static equilibrium if the effects of all forces and moments create an overall balance; This means: the resultant force and the resultant moment are 0

Answer 16

In 2D-problems we have 3 independent equilibrium equations: Sum Fix = 0; Sum Fiy = 0; Sum Miz = 0 * Hence we can determine three unknown support reactions for a structure consisting only of a single part. If more supports are used, the structure is called statically indeterminate

Answer 17

If the number of support reactions equals the number of equilibrium equations, the structure is statically determinate.

Answer 18

In some special cases, the support reactions a) all meet in one point b) are all parallel Then the structure can move (rotate or translate) AVOID THESE DESIGNS!

Answer 19

If a number of degrees of freedom are restricted (e.g. by an external support), you need to insert the same number of reactions (forces or moments) to the free body diagram The reactions (forces or moments) are calculated by applying the equilibrium conditions

Answer 20

- It can not move vertically - It can move horizontally - It can rotate 1 degree of freedom restricted = 1 reaction

Answer 21

- It cannot move horizontally - It cannot move vertically - It can rotate 2 degrees of freedom restricted = 2 reactions

Answer 22

It cannot move horizontally - It cannot move vertically - It cannot rotate 3 degrees of freedom restricted = 3 reactions

Answer 23

Shear force is parallel to the cross sectional area, use V(x) to present it. Normal is perpendicular, denoted by N(x)

Answer 24

1. Before the body is “cut”, determine the support reactions. 2. Pass an imaginary section through the body, perpendicular to the axis, at the point where the internal forces are to be determined. 3. Draw a FBD including all distributed loadings, couple moments and external forces acting on the body, as well as the internal forces at the cut. 4. Apply the equations of equilibrium to calculate the internal forces Fn, Fs, M

Answer 25

The normal force is defined as positive if it correspond to traction; compression is hence related to a negative normal force;

Answer 26

* A positive bending moment results in sagging, a negative in hogging, hence the bending moment is related to the curvature of the beam Bending leads to tension and compression at the different sides of the beam (top and bottom part of each side), which result in crumple, snap and eventually in the development of a crack

Answer 27

You follow the axis (if provided), if not use common sense If you answer for the force turns out to be -ve then you know to alter the direction of the force.

Answer 28

-They can determine the properties of structures -They reveal the internal loading conditions of structures and can predict the failure of structure -They can calculate the support reaction force and moments

Answer 29

Positive bending moments will be those that put the lower section of the beam into tension * Sagging moment is positive * Hogging moment is negative

Answer 30

A torque (torsional moment) is a moment rotating around the axis of the beam; The torque is independent from normal, shear force or bending moments

Answer 31

Stress is a quantity that describes the distribution of internal forces within a body units Pa

Answer 32

Sigma = F / A F- internal force A- cross-sectional area

Answer 33

Tensile stress (+ve) Compressive stress (-ve)

Answer 34

The engineering stress is defined as force divided by the initial (perpendicular) area. * It normally generates a change in volume of the material which is measured as strain. * The extension in length is normally related to reduction in width

Answer 35

* The engineering strain is the ratio between change in dimension (delta l = l-l sub0) to initial dimension. * Strain is always dimensionless.

Answer 36

Stiffness is the extent to which an object resists deformation in response to an applied force.

Answer 37

Young’s modulus (Modulus of Elasticity) is a mechanical property that measures the tensile or compressive stiffness of a solid material when the force is applied lengthwise. It quantifies the relationship between tensile/compressive 𝜎 (force per unit area) and axial strain 𝜀 (proportional deformation) in the linear elastic region of a material.

Answer 38

x10^6 (x10^3)^2

Answer 39

Stress acts parallel to the cross-section * The shear stress is defined as shear force divided by the initial (parallel) area. * It normally generates a change in shape (square to parallelogram) which is measured as shear strain.

Answer 40

tau = Fs [N/m²] / A0 Shear Stress = shear force / initial (parallel) area

Answer 41

* Shear strain corresponds to the change of angle. * For small strains we have tan(theta) = approximately theta

Answer 42

(Delta x) / y = tan(theta)

Answer 43

G = tau / gamma

Answer 44

The second moment of area is THE characteristic quantity for bending of beams. Also named moment of inertia.

Answer 45

Consider a plate submerged in a liquid. The pressure of a liquid at a distance z below the surface is given by p = gamma x z, where gamma is the specific weight of the liquid. The force on the area dA at that point is: dF = p x dA. The moment about the x-axis due to this force is z (dF). The total moment is: Intergral sub(A) z dF = Integral Sub(A) gamma x z^2 dA = gamma x Integral sub(A) (z^2 x dA). This sort of integral term also appears in solid mechanics when determining stresses and deflection. This integral term is referred to as the moment of inertia of the area of the plate about an axis.

Answer 46

* A geometric property of a section;

Answer 47

* It tells us how the area of the section is distributed about a particular axis

Answer 48

The bending of a beam is mainly influenced by: 1. Young’s modulus, 2. Moment of inertia;

Answer 49

The 2nd moment of area is normally calculated with respect to the centre of gravity or the middle axis of the beam * Sometimes a mixed 2nd moment of area is used: I sub(xy) = Integral sub(A) xy dA

Answer 50

S sub(x) = Integral Sub(A) y dA S sub(y) = Integral Sub(A) x dA

Answer 51

I sub(x) = Integral Sub(A) x^2 dA I sub(y) = Integral Sub(A) y^2 dA

Answer 52

I sub(p) = Integral sub(A) r^2 dA = I sub(x) + I sub(y)

Answer 53

d A = l x d z The equation for the second moment of area becomes: I sub(y) = Integral z^2 dA = Integral z^2 l.dz And the limits of integration are from zmin to zmax. * Attention: Here y, z are used as coordinates ( the diagram on qmplus had axis z on the y, and y on the x)

Answer 54

Centre CoG

Answer 55

the second moment with respect to an axis parallel to the axis through the CoG but within a distance can be computed as follows: The second moment of area with respect to axis y through the CoG is I sub(y) = Integral z^2 dA = Integral z^2 L.dz The second moment with respect to the translated axis (bar) y is: I sub (bar y) = I sub(y) + (d sub(z) ^2) x A where d sub(z) is the perpendicular distance between the two parallel axis

Answer 56

First Moment of Area is area times distance (to some reference line) S sub(z) = A.d = A.(bar)y If it is an arbitrary shape, we summarise all the differential elements:

Answer 57

Because they are related to the second moment of area I sigma = M sub(e) x y / 𝐼 sub (z) The farther the area of the section is from the neutral axis, the greater the bending section modulus of the section 𝐼 sub(z). 𝐼𝑧 reflects how its points are distributed with regard to a given axis.

Answer 58

Second Moment of Area is area times distance squared (to some reference line).