Axial force; Shear force;

Bending moment; Strain energy;

deflection

Dr. V. Srinivas
 Introduction
Structures can be seen all around us in our everyday lives.

High Rise Building

Vehicle Frame

Bridge Supporting
Roadway
Residential Dwelling
 Mechanisms, Motion and Force
Mechanism
A system of moving parts
that performs some
function.
Collins Dictionary

Motion
The process of continual
change in the position of an
object “movement”

Collins Dictionary
Force
There are five main types of force:

1. Compression

2. Tension
Shear forces

3. Shear

4. Bending

5. Torsion Columns in Compression A rope in

Tension

Torsion forces
Pipe Bending
Force Compressive and Tensile Forces
Can crushed by compression

This is where a load Spring in Tension, this is

presses or squashes where a load pulls an object
objects together apart.
Force
Torsion and Shearing forces
Shearing Force

Torsion
This is where the load
causes an object to twist

This is where loads

push at right angles
to the surface of the
object. E.g. scissors

Torsion force
Force Bending and Static Forces

This is where the load causes an object to Bend

A piece of metal

Dynamic Load – a car which is

Bending a piece of metal
not fixed to any one specific
point, e.g. a car travelling along a
road

Static Load – a load which is fixed at

one point, e.g. a building
External/Internal Forces
• External forces arise from contact or gravitational
attraction

– Point and distributed loading

– Weight
• Internal forces are forces arising to hold
bodies together
– Internal stress is a form of an internal force
 Effect of External Forces

Positive Shear

M M

Positive Moment
ME221 Lecture 14 9
Internal Force Components
• Force components
P
Vz
– Axial is along beam Vy
– Shearing forces are transverse components

• Moment components
– Torsion along beam T
Mz
– Bending for transverse
My
components
 Tie

Wall
Strut

• A strut is the part, which has compressive forces acting upon it.
• A tie is the part of a structure which has tensile forces acting upon it.
Calculating Forces
The frame opposite supports a Mass of 60Kg.
Pin A in the sketch is in equilibrium.
Calculate the size of the forces acting on members B
300N
and C in the frame.

In order to calculate the force of the

Mass in Newton’s, the Mass must be
Multiplied by 9.81N, rounded to
(10).Therefore force equals 600N.

519.6N 60Kg

Triangle of Forces diagram.

60Kg
The force at B is calculated as: 600 Cos 60° = (600)(0.5) = 300N

The force at C is calculated as: 600 Sin 60° = (600)(0.866) =519.6N

 Recall from mechanics of materials that the internal forces
Internal Forces and Moments P (generic axial), V (shear) and M (moment) represent
resultants of the stress distribution acting on the cross
section of the beam.

Internal Axial Force (P) ≡ equal in magnitude but opposite

in direction to the algebraic sum (resultant) of the
components in the direction parallel to the axis of the beam
of all external loads and support reactions acting on either
side of the section being considered.

T  Tension
C  Compression

Positive Internal Forces Acting

on a Portal Frame
 Beam Sign
Convention for
Shear and Moment

Positive Sign Conventions:

Tension axial force on the

section
Shears that produces
clockwise moments
Bending moments that
produce compression in the
top fibers and tension in the
bottom fibers of the beam

Internal Shear Force (V) ≡ equal in magnitude but opposite in direction to the algebraic sum (resultant) of the
components in the direction perpendicular to the axis of the beam of all external loads and support reactions acting
on either side of the section being considered.

Internal Bending Moment (M) ≡ equal in magnitude but opposite in direction to the algebraic sum of the moments
about (the centroid of the cross section of the beam) the section of all external loads and support reactions acting
on either side of the section being considered.
 kN
Shear and bending moment
diagrams depict the variation of
these quantities along the length kN/m
of the member.

Proceeding from one end of the

member to the other, sections are
passed. After each successive
change in loading along the
kN kN
length of the member, a FBD
(Free Body Diagram) is drawn to V and M are in the opposite
determine the equations expressing directions of the positive
the shear and bending moment beam sign convention
in terms of the distance from
a convenient origin.
(kN)
Zero Shear
Plotting these equations produces Maximum
the shear and bending moment Positive
diagrams. Bending
Moment

(kN-m)
Shear and Moment Diagrams using Sectioning Method
Generate a shear / bending diagram as follows:
1. Find reaction forces
2. Take a section on each side of an applied force or
moment and inside a distributed load
(take a new section whenever there is a change in the load or shape of
the beam)

- draw a FBD and sum forces / moments

3. Repeat 2 along the length of the beam.
w(x) distributed load

V(x) shear force

M(x) moment
Principle of
Superposition
 20 kg/m 125 kg 125 kg

1m 2m 2m 2m

+ve x
-ve
tension up Tension down
 Sketching the Deflected
Shape of a Beam or Frame An accurate sketch must satisfy
the following rules:
Following our positive beam
sign convention, a positive • The curvature must be consistent
bending moment bends a beam with the moment curve.
concave upward (or towards the • The deflected shape must
positive y direction), whereas a satisfy the boundary
negative bending moment constraints.
bends a beam concave • The original angle at a rigid
downward (or towards the joint must be preserved.
negative y direction). • The length of the deformed
member is the same as the
original length of the unloaded
member.
• The horizontal projection of a
beam or the vertical projection of
a column is equal to the original
length of the member.
• Axial deformations, which are
trivial compared to bending
deformations, are neglected.
 DERIVATION OF PURE BENDING EQUATION
Relationship between bending stress and radius of curvature.
Consider the beam section of length “dx” subjected to
pure bending. After bending the fibre AB is shortened
in length, whereas the fibre CD is increased in length.
In b/w there is a fibre (EF) which is neither shortened
in length nor increased in length (Neutral Layer).
Let the radius of the fibre E'F′ be R . Let us select one
more fibre GH at a distance of ‘y’ from the fibre EF as
shown in the fig.
EF= E'F′ = dx = R dθ
The initial length of fibre GH equals R dθ
After bending the new length of GH equals

G'H′= (R+y) dθ
= R dθ + y dθ
Change in length of fibre
GH = (R dθ + y dθ) - R dθ = y dθ
Therefore the strain in fibre GH

Є= change in length / original length= y dθ/ R dθ

Є = y/R

If σ ь is the bending stress and E is the Young’s

modulus of the material, then strain
Є = σ ь/E

σ ь /E = y/R => σ ь = (E/R) y---------(1)

σ ь = (E/R) y => i.e. bending stress in any fibre is proportional to the
distance of the fibre (y) from the neutral axis and hence maximum
bending stress occurs at the farthest fibre from the neutral axis.
 Moment of resistance

on one side of the neutral axis there are compressive stresses and
on the other there are tensile stresses. These stresses form a
couple, whose moment must be equal to the external moment M.
The moment of this couple, which resists the external bending
moment, is known as moment of resistance.

σc

Neutral Axis
σt

Note: Neutral axis coincides with the horizontal centroidal axis of the cross section
 Moment of resistance

da
y
N A

Consider an elemental area ‘da’ at a distance ‘y’ from the neutral

axis.
The force on this elemental area = σ ь × da
= (E/R) y × da {from (1)}

The moment of this resisting force about neutral axis =

(E/R) y da × y = (E/R) y² da
Total moment of resistance offered by the beam section,
M'=  (E/R) y² da
= E/R  y² da
 y² da =second moment of the area =moment of inertia about the neutral axis.
M'= (E/R) INA
For equilibrium moment of resistance (M') should be equal to applied moment M
i.e. M' = M
Hence. We get M = (E/R) INA
(E/R) = (M/INA)--------(2)
From equation 1 & 2, (M/INA)= (E/R) = (σ ь /y) ---- BENDING EQUATION.
(Bernoulli-Euler bending equation)

Where E= Young’s modulus, R= Radius of curvature,

M= Bending moment at the section,
INA= Moment of inertia about neutral axis,
σ ь= Bending stress
y = distance of the fibre from the neutral axis
 Strain energy
Energy is the capacity of a physical system to do work.
Strain energy is as the energy which is stored within a material when work has been done on the material. Here
it is assumed that the material remains elastic whilst work is done on it so that all the energy is recoverable and
no permanent deformation occurs due to yielding of the material,
Strain energy U = work done

Thus for a gradually applied load the work done in straining the
material will be given by the shaded area under the load-
extension graph

Work done by a gradually applied load.

Strain energy - tension or compression-

Neglecting the weight of the bar

Consider a small element of a bar, length ds
If a graph is drawn of load against elastic extension the shaded area under the graph gives the work done

the strain energy

Strain energy - tension or compression-

Including the weight of the bar

Consider now a bar of length L mounted
vertically.
At any section A B the total load on the
section will be the external load P together
with the weight of the bar material below
AB.

Load on section A B = P ± ρgAs

The positive sign being used when P

is tensile and the negative sign when
P is compressive.

Thus, for a tensile force P the

extension of the element ds is given
by the definition of Young's modulus
E to be
Strain energy-shear

Consider the elemental bar now subjected to a shear load Q at one end causing deformation through the
angle y (the shear strain) and a shear deflection δ,
Strain energy –bending

Let the element now be subjected to a constant bending moment M causing it

to bend into an arc of radius R and subtending an angle dϴ at the center.
The beam will also have moved through an angle dϴ
Strain energy = work done = 1/2 moment x angle turned through (in radians)
Strain energy – torsion
The element is now considered subjected to a torque T, producing an angle of twist dϴ radians
DEFLECTIONS
 Deflection diagrams &
the elastic curve
• Deflections of structures can come from loads, temperature, fabrication errors or settlement
• In designs, deflections must be limited in order to prevent cracking of attached brittle materials
• A structure must not vibrate or deflect severely for the comfort of occupants
• Deflections at specified points must be determined if one is to analyze statically indeterminate structures

Factors affecting deflection

• Load
• Span
• Size and shape  here, only linear elastic material response is considered
• Stiffness of material  This means a structure subjected to load will return to its original
undeformed position after the load is removed
 It is useful to sketch the shape of the structure when it is loaded in
order to visualize the computed results & to partially check the
results
 Deflection diagrams &
the elastic curve
• This deflection diagram
rep the elastic curve
for the points at the
centroids of the cross-
sectional areas along
each of the members
• If the elastic curve
seems difficult to
establish, it is
suggested that the
moment diagram be
drawn first
• From there, the curve
can be constructed
 Deflection diagrams &
the elastic curve
• Due to pin-and-roller support, the displacement at A & D must be zero
• Within the region of –ve moment,
the elastic curve is
concave downward
• Within the region of +ve moment,
the elastic curve is concave upward
• There must be an inflection point
where the curve changes from
concave down to concave up
Example
Draw the deflected shape of each of the beams.
 Solution
In (a), the roller at A allows free rotation with no deflection while the fixed wall at B prevents both
rotation & deflection. The deflected shape is shown by the bold line.

In (b), no rotation or deflection occur at A & B

In (c), the couple moment will rotate end A. This will cause deflections at both ends of the beam since no
deflection is possible at B & C. Notice that segment CD remains un-deformed since no internal load acts
within.
 Solution
In (d), the pin at B allows rotation, so the slope of the deflection curve will suddenly change at this point
while the beam is constrained by its support.

In (e), the compound beam deflects as shown. The slope changes abruptly on each side of B.

In (f), span BC will deflect concave upwards due to load. Since the beam is continuous, the end spans will
deflect concave downwards.
Derivation of deflection formulae
Elastic Beam theory
• To derive the relationship between bending moment and
displacement of elastic curve, we look at an initially straight beam
that is elastically deformed by loads applied perpendicular to beam’s
x-axis & lying in x-v plane of symmetry
• Due to loading, the beam deforms under shear & bending
• If beam L >> d, greatest deformation will be caused by bending
• When M deforms, the angle between the cross sections becomes d

The arc dx that rep a portion of the elastic curve intersects the neutral
axis
The radius of curvature for this arc is defined as the distance, , which
is measured from ctr of curvature O’ to dx
Any arc on the element other than dx is subjected to normal strain
The strain in arc ds located at position y from the neutral axis is

  ( ds ' ds ) / ds
 Elastic Beam theory
ds  dx  d and ds'  (   y ) d
(   y ) d  d 1 
   
d  y

• If the material is homogeneous & behaves in a linear manner, then Hooke’s law applies
• The flexure formula also applies
  /E
   My / I Combining these eqns, we have:
1 M

 EI
  the radius of curvature at a specific point on the elastic curve

M  internal moment in the beam at the point where  is to be determined

E  the material' s modulus of elasticity

I  the beam' s moment of inertia computed about the neutral axis

 Elastic Beam theory
EI  flexural rigidity; dx  ρdθ
M
dθ  dx
EI
1 d 2 v / dx 2
v  axis as  ve , 
 [1  ( dv / dx ) 2 ]3 / 2
M d 2 v / dx 2
Therefore, 
EI [1  ( dv / dx ) 2 ]3 / 2
• This eqn rep a non-linear second DE
• V=f(x) gives the exact shape of the elastic curve d 2v M

• The slope of the elastic curve for most structures is very small dx 2
EI
• Using small deflection theory, we assume dv/dx ~ 0

By assuming dv/dx ~ 0  ds, it will approximately equal to dx

This implies that points on the elastic curve will only be displaced vertically & not horizontally

ds  dx 2  dv2  1  (dv 2  dx) 2 dx 2  dx

 The double integration method
• M = f(x), successive integration of eqn 8.4 will yield the beam’s slope
•   tan  = dv/dx =  M/EI dx
• Eqn of elastic curve
• V = f(x) =  M/EI dx
• The internal moment in regions AB, BC & CD must be written in terms of x1, x2
and x3
 The double integration method
• Once these functions are integrated & the constants determined, the functions will give
the slope & deflection for each region of the beam
• It is important to use the proper sign for M as established by the sign convention used
in derivation
• +ve v is upward, hence, the +ve slope angle,  will be measured counterclockwise from
the x-axis
 The double integration method
• The constants of integration are determined by evaluating the functions for slope
or displacement at a particular point on the beam where the value of the
function is known
• These values are called boundary conditions
• Here x1 and x2 coordinates are valid within the regions AB & BC
Once the functions for the slope & deflections are obtained, they must give the
same values for slope & deflection at point B
This is so as for the beam to be physically continuous
Example
The cantilevered beam is subjected to a couple moment Mo at its end. Determine the eqn of the
elastic curve. EI is constant.

By inspection, the internal moment can be represented throughout the beam using a single x
coordinate. From the free-body diagram, with M acting in +ve direction, we have:
Integrating twice yields: Using boundary conditions, dv/dx = 0 at x = 0 & v = 0 at x = 0 then
M  Mo C1 = C2 =0.
Substituting these values into earlier eqns, we get:
d 2v
EI  Mo
dx 2
with   dv / dx
Mox M o x2
dv   ; v
EI  M o x  C1 EI 2 EI
dx Max slope & disp occur at A (x = L) for which
M o x2 MoL M o L2
EI v   C1 x  C2 A  ; vA 
2 EI 2 EI
 Solution
The +ve result for A indicates counterclockwise rotation & the +ve result for vA indicates that it is
upwards.

In order to obtain some idea to the actual magnitude of the slope.

Consider the beam to:

Have a length of 3.6m
Support a couple moment of 20kNm
Be made of steel having Est = 200GPa
 Solution
If this beam were designed w/o a fos by assuming the allowable normal stress = yield stress =
250kNm/2.

A W6 x 9 would be found to be adequate

20kNm(3.6m)
A  6 6 12 4
 0.0529rad
[200(10 )kN / m2 ][6.8(10 )(10 )m

20kNm(3.6m) 2
vA  6 6 12 4
 95.3mm
2[200(10 )kN / m2 ][6.8(10 )(10 )m

Since 2A = 0.00280(10-6)<<1, we have obtained larger values for max  and v than would have been
obtained if the beam were supported using pins, rollers or other supports.
 Conjugate-Beam method
Conjugate beam is defined as the imaginary beam with the same dimensions (length)
as that of the original beam but load at any point on the conjugate beam is equal to the
bending moment at that point divided by EI.
The conjugate-beam method is an engineering method to derive the slope and
displacement of a beam.

• The basis for the method comes from similarity equations

• To show this similarity, we can write these eqn as shown

dV d 2M
w w
dx dx2
d M 2
d v M
 
dx EI dx2 EI
 Conjugate-Beam method
• Or integrating,

V   wdx M  wdxdx

M   M  
   dx v    dx dx
 EI   EI  
 Conjugate-Beam method
• Here the shear V compares with the slope θ , the moment M compares with the
disp v & the external load w compares with the M/EI diagram
• To make use of this comparison we will now consider a beam having the same
length as the real beam but referred to as the “conjugate beam”,
 Conjugate-Beam method
• The conjugate beam is loaded with the M/EI diagram derived from the load w on
the real beam
• From the above comparisons, we can state 2 theorems related to the conjugate
beam
• Theorem 1
• The slope at a point in the real beam is numerically equal to the shear at the corresponding
point in the conjugate beam
 Conjugate-Beam method
• Theorem 2
• The disp. of a point in the real beam is numerically equal to the moment at the
corresponding point in the conjugate beam
• When drawing the conjugate beam, it is important that the shear & moment
developed at the supports of the conjugate beam account for the corresponding
slope & disp of the real beam at its supports

Consequently from Theorem 1 & 2, the conjugate beam must be supported by a

pin or roller since this support has zero moment but has a shear or end reaction
When the real beam is fixed supported, both beam has a free end since at this
end there is zero shear & moment
Conjugate-Beam method
 Example

Solution
The conjugate beam loaded with the M/EI diagram is shown. Since M/EI diagram is +ve, the
distributed load acts upward.
The external reactions on the conjugate beam are determined first and are indicated on the free-body
diagram.
Max deflection of the real beam occurs at the point where the slope of the beam is zero.
Assuming this point acts within the region 0x9m from A’ we can isolate the section.
 Solution
Note that the peak of the distributed loading was determined from proportional triangles,

w / x  (18 / EI ) / 9
V ' 0

   Fy  0
45 1  2x 
   x  0
EI 2  EI 
x  6.71m (0  x  9m) OK
 Solution
Using this value for x, the max deflection in the real beam corresponds to the moment M’.
Hence,

With anticlockw ise moments as  ve,  M  0

45  1  2(6.71)  1
(6.71)    6.71 (6.71)  M '  0
EI 2  EI  3
 Solution
The –ve sign indicates the deflection is downward .

201 .2kNm3
 max  M ' 
EI
 201 .2kNm3

[ 200 (10 6 )k N / m 2 ][ 60 (10 6 ) mm 4 (1m 4 /(10 3 ) 4 mm 4 )]
 0.0168 m  16 .8mm
Relationships of Forces and Deformations
There are a series of relationships among forces and deformations along a beam, which can be useful in analysis.
Using either the deflection or load as a starting point, the following characteristics can be discovered by taking
successive derivatives or integrals of the beam equations.

Source: University of Michigan, Department of Architecture

Symmetrically Loaded Beams

• Maximum slope occurs at the ends of the beam

• A point of zero slope occurs at the center line.

This is the point of maximum deflection.

• Moment is positive for gravity loads.

• Shear and slope have balanced + and - areas.

• Deflection is negative for gravity loads.

Source: University of Michigan, Department of Architecture

Cantilever Beams

• One end fixed. One end free

• Fixed end has maximum moment, but zero

slope and deflection.

• Free end has maximum slope and

deflection, but zero moment.

• Slope is either downward (-) or upward (+)

depending on which end is fixed.

• Shear sign also depends of which end is

fixed.

• Moment is always negative for gravity loads.

Asymmetrically Loaded Beams
Diagram Method

• The value of the slope at each of the endpoints is different.

• The exact location of zero slope (and maximum deflection) is

unknown.

• Start out by assuming a location of zero slope (Choose a

location with a known dimension from the loading diagram)

• With the arbitrary location of zero slope, the areas above

and below the baseline (“A” and “B”) are unequal

• Adjust the baseline up or down by D distance in order to

equate areas “A” and “B”. Shifting the baseline will remove
an area “a” from “A” and add an area “b” to “B”

Source: University of Michigan, Department of Architecture

Asymmetrically Loaded Beams (continued)
• Compute distance D with the equation:

A B
D
L
• With the vertical shift of the baseline, a horizontal shift
occurs in the position of zero slope.

• The new position of zero slope will be the actual location of

maximum deflection.

• Compute the area under the slope diagram between the

endpoint and the new position of zero slope in order to
compute the magnitude of the deflection.

Source: University of Michigan, Department of Architecture

Deflection: by Superposition of Equations
• Deflection can be determined by the use of
equations for specific loading conditions

• By “superposition” equations can be

added for combination load cases. Care
should be taken that added equations all
give deflection at the same point, e.g. the
center line.

• Note that if length and load (w) is entered

in feet, a conversion factor of 1728 in3/ft3
must be applied in order to compute
deflection in inches.
Source: University of Michigan, Department of Architecture
 Example: Equations Method
• To determine the total deflection of the beam for
the given loading condition, begin by breaking up
the loading diagram into two parts.

• Compute the total deflection by superimposing the

deflections from each of the individual loading
conditions. In this example, use the equation for a
midspan point load and the equation for a uniform
distributed load.
Source: University of Michigan, Department of Architecture
Example: Equations Method

• For a W18x55 with an E modulus of 30000 ksi

and a moment of inertia of 890 in4

• Using an allowable deflection limit of L / 240.

• Check deflection

Source: University of Michigan, Department of Architecture

 Example: Asymmetrical Loading –
Superposition of Equations

Standard equations provide values of

shear, moment and deflection at points
along a beam.

Cases can be “superpositioned” or overlaid

to obtain combined values.

To find the point of combined maximum

deflection, the derivative of the combined
deflection equation can be solved for 0.
This gives the point with slope = 0 which is
a max/min on the deflection curve.

Source: Steel Construction Manual AISC 1989

 Force flow
Force flow bears a natural resemblance with "water flow", transferring loads from acting points to
foundation via the shortest and smoothest geometric path.
This can easily be explained by the principle of least action.
Since the force flow produces strain in material and corresponding strain energy aswell, the structure,
when functioning in a normal, balanced and stable way, whould have a minimum amount of strain energy.
Extra length and redundant turns of the path will inevitably increase the strain energy reserved, which
render the design not the optimum one.
In light of this, in preliminary design, the task engineers who undertake should design or plan the direction
of force flow and find out the shortest and smoothest path.
This is the basic concept of elevation arrangement in structural design.

Hence, conceptual analysis regarding elevation arrangement could be conducted through the spatial path
and the intensity of the force flow. In the early stage of design, designers, with reference to the functions of
the structure, should plan for shortest path that the force flow will take, to ensure that the force will be
smoothly transferred to the foundation. At the same time, being aware of the force flow, variable cross
sections could be used to increase the load-bearing efficiency.

Several structural systems with different force flow patterns are summarised as follows.
vertical load-transferring structure

Typical vertical load-transferring structure is high-rise

structure.
Under the action of wind or seismic load, vertical
deadweight and liver loads, high-rise structures vertically
transfer these loads downwards to the foundtaions.
The intensity of the force flow increase with the
accumulation of stories.
Thus, there are different force flow intensities within such
vertical structure.
For instance, due to that the upper columns bears less force
than bottom columns, the upper columns could be designed
to be have less reinforcements than bottom columns.
Lateral load transferring structure

Typical lateral load-transferring structure comes to beams and slabs. As shown, a beam could transfer vertical
loads via the internal force flow to supports. From the figure, we know that tensile reinforcements shoudl be
allocated in major tensile direction at the lower part of the member as marked out with solid lines. While the
concrete in major compressive stress as marked out by dotted lines, it aligns with the direction of force flow.
Broken-line-typed load transferring structure

Typical broken-line-typed load-transfeering structures go to all truss structures. External loads are
transferred through members of truss structure in the manner of a broken line to the support.
Streamline load-transferring structure
The feature of a streamlined load-transferring structure is that the force flow takes its own streamlined path and directly goes to
supports. There are plane and spatial streamlined load-transferring structure. Figure shows us the force flow of a deck arch
bridge, in which the loads on the deck are transferred to the main arch via the vertical bars and then to supports. If the structure
has a curved surface, the loads will be transferred through the curved surface and then to supports.

Streamlined load-transferring structure show the convergence of the force flow in a clear way and therefore nables the design of
a variable section and reduces material in accordance with the change of intensity of a force flow. Take the arch as shown for
example. The intensity of the force flow at the top of the arch is low and correspondingly the section of the arch can be
comparatively small; the intensity of the force flow at the foot is the highest, and the cross section area should be accordingly the
largest. The design of a variable section should agree with the intensity change of the force flow.

Plane Streamline load-transferring structure Spatial Streamline load-transferring structure

Principles of Reinforcing
Strut and Tie Modeling
Strut and Tie Modeling
Strut and Tie Modeling
Strut and Tie Modeling
 kN/m

kN

Calculate and draw the shear

force and bending moment
equations
 kN/m

kN/m

kN

Calculate and draw the axial

force, shear force and bending
moment equations