Reaction Forces in a Beam Subjected to Different
Supports and Distributed Loading
Hamaad Javed
School of Mechanical and
Manufacturing Engineering (SMME)
Natioanl University of Science and
Technology (NUST)
H-12 Islamabad, Pakistan
Humna Fatima
School of Mechanical and
Manufacturing Engineering (SMME)
Natioanl University of Science and
Technology (NUST)
H-12 Islamabad, Pakistan
hamaddj88@gmail.com
humnafatima27@gmail.com
Mowahhid bin Safdar
School of Mechanical and
Manufacturing Engineering (SMME)
Natioanl University of Science and
Technology (NUST)
H-12 Islamabad, Pakistan
Mowahhid84@gmail.com
Muhammad Abdullah Zafar Ghauri
School of Mechanical and
Manufacturing Engineering (SMME)
Natioanl University of Science and
Technology (NUST)
H-12 Islamabad, Pakistan
mazghauri2004@gmail.com
Muhammad Zain ul Abideen
School of Mechanical and
Manufacturing Engineering (SMME)
Natioanl University of Science and
Technology (NUST)
H-12 Islamabad, Pakistan
Mustafa Siddiqui
School of Mechanical and
Manufacturing Engineering (SMME)
Natioanl University of Science and
Technology (NUST)
H-12 Islamabad, Pakistan
zainhike82@gmail.com
mustafasid1912@gmail.com
Abstract—— Comprehending the behaviour of reaction
forces in beams is rudimentary to the analysis and design of
different structures in engineering. The study investigates
the varying effects of different support conditions and
distributed loading on reaction forces within a beam. The
research employs analytical methods and computational
simulations on software like ANSYS and SOLIDWORKS to
reconnoitre the complex adaptability between structural
elements and external forces. The investigation commences
by examination of common support configurations such as
simply supported, fixed, and cantilevered beams. While
each particular scenario is subjective to a range of
distributed loading patterns, representing real-world
conditions encountered in engineering applications. The
study brings into play different mathematical formulations
and numerical simulations to analyse the resulting internal
forces, focusing on reactions at the supports. The findings
shed light on the nuanced relationships between support
types, loading conditions, and reaction forces. Insights
gained from this research contribute to the development of
more accurate analytical models and aid engineers in
designing structures that can efficiently withstand a variety
of loading scenarios. The practical implications of these
findings extend to a wide array of engineering disciplines,
including civil, mechanical, and aerospace engineering,
enhancing the understanding of structural behaviour and
facilitating the optimization of design processes for
improved safety and efficiency.
Keywords— ANSYS, SOLIDWORKS, Reaction Forces, Beams,
of the two can be considered the action, while the other is
its associated reaction.
•
Distributive Loading:
Distributed loads are forces
which are spread out over a length, area, or volume. Most
real-world loads are distributed, including the weight of
building materials and the force of wind, water, or earth
pushing on a surface. Pressure, load, weight density and
stress are all names commonly used for distributed loads.
Distributed load is a force per unit length or force per unit
area depicted with a series of force vectors joined together at
the top, and will be designated as w(x) to indicate that the
distributed loading is a function of.x.
For example, although a shelf of books could be treated as a
collection of individual forces, it is more common and
convenient to represent the weight of the books as
a uniformly distributed load. A uniformly distributed load
is a load which has the same value everywhere,
i.e., w(x)=c, a constant.
The magnitude of the distributed load of the books is the
total weight of the books divided by the length of the shelf
as:
It represents the average book weight per unit length.
Similarly, the total weight of the books is equal to the value
of the distributed load times the length of the shelf or
Internal Forces, Structural Behaviour
I. INTRODUCTION
•
Reaction Forces:
In accordance to Newton's Third Law of
Motion of classical mechanics, all forces exist in pairs
such that if one object exerts a force on another object,
then the second object exerts an equal and opposite
reaction force on the first.[1][2] The third law is also more
generally stated as: "To every action there is always
opposed an equal reaction: or the mutual actions of two
bodies upon each other are always equal, and directed to
contrary parts."[3] The ascription of which of the two forces
is the action and which is the reaction is arbitrary. Either
XXX-X-XXXX-XXXX-X/XX/$XX.00 ©20XX IEEE
This total load is simply the area under the curve w(x), and
has units of force. If the loading function is not uniform,
integration may be necessary to find the area. [4]
• Fixed Support:
In the case of fixed support, the beam is
firmly attached to a fixed support from where it has no
chances of being subjective to slippage.
For a fixed support analysis there are three unknowns. The
reactions are the couple moment and the two force
components (respectively along x-axis and y-axis), or the
couple moment and the magnitude and direction of the
resultant force.
itself and perhaps a perfectly vertical load. As soon as a
lateral load of any kind pushes on the structure it will roll
away in response to the force. The lateral load could be a
shove, a gust of wind or an earthquake. Since most
structures are subjected to lateral loads it follows that a
building must have other types of support in addition to
roller supports. There exists one unknown force in roller
support. The reaction is a force which acts perpendicular to
the slot.[5]
Fig.1 Diagrammatic Representation of Fixed Support and the
Distributive Load subjected to a Beam
•
Pinned Support:
A pinned support is a type of beam
support which can resist both vertical and horizontal forces
but not a moment. They will allow the structural member to
rotate, but not to translate in any direction. Many
connections are assumed to be pinned connections even
though they might resist a small amount of moment in
reality. It is also true that a pinned connection could allow
rotation in only one direction; providing resistance to
rotation in any other direction. The knee can be idealized as
a connection which allows rotation in only one direction and
provides resistance to lateral movement. The design of a
pinned connection is a good example of the idealization of
the reality. A single pinned connection is usually not
sufficient to make a structure stable. Another support must
be provided at some point to prevent rotation of the
structure. The representation of a pinned support includes
both horizontal and vertical forces.
Fig.3 Diagrammatic Representation of Roller Support
•
Beam:
A beam is a structural element that primarily
resists loads applied laterally to the beam's axis (an element
designed to carry primarily axial load would be a strut or
column). Its mode of deflection is primarily by bending. The
loads applied to the beam result in reaction forces at the
beam's support points. The total effect of all the forces
acting on the beam is to produce shear forces and bending
moments within the beams, that in turn induce internal
stresses, strains and deflections of the beam. Beams are
characterized by their manner of support, profile (shape of
cross-section), equilibrium conditions, length, and their
material
Fig.2 Diagrammatic Representation of Pinned Support
•
Roller Support:
Roller supports are free to rotate and
translate along the surface upon which the roller rests. The
surface can be horizontal, vertical, or sloped at any angle.
The resulting reaction force is always a single force that is
perpendicular to, and away from, the surface. Roller
supports are commonly located at one end of long bridges.
This allows the bridge structure to expand and contract with
temperature changes. The expansion forces could fracture
the supports at the banks if the bridge structure was "locked"
in place. Roller supports can also take the form of rubber
bearings, rockers, or a set of gears which are designed to
allow a limited amount of lateral movement. A roller
support cannot provide resistance to a lateral force. Imagine
a structure (perhaps a person) on roller skates. It would
remain in place as long as the structure must only support
Fig.4 A statically determinate beam, bending (sagging)
under a uniformly distributed load
II. PROBLEM ANALYSIS (A)
Q. The truss shown consists of nine members and is
supported by a ball and socket at B, a short link at C, and
two short links at D. (a) Check that this truss is a simple
truss, that it is completely constrained, and that the reactions
at its supports are statically determinate. (b) Determine the
force in each member for P=(-1200N)j and Q= 0
Equations of Analysis
ΣFx=ΣFy=ΣFz=0 (1st Condition of Equilibrium should be
satisfied)
Στx= Στy= Στz=0 (2nd Condition of Equilibrium should be
satisfied)
• ΣFx=0
Bx=0 N
• ΣFz=0
Dz=Bz=0 N
• τCD=0
1.8By+1.2P=0
By= -800 N
• τBC=0
-3Dy+0.75P=0
Dy= 100 N
• ΣFy=0
By+Cy+Dy+P=0
Cy=100 N
Analysis at B Joint
Figure 5. Problem (A)
Numerical Solution
Figure 7. Free Body Diagram of B Joint and Reference
Axis
Joint B connects members BA, BC and BE
Involving vector approach for the sake of ease, we may state
the following results proceeding as follows: |AB|=√(0.6)2 + (0.75)2 + (3)2
Figure 6. Free Body Diagram and Reference Axis for
Problem (A)
AB = 3.15 m
BC= 0.75m i
BE= 0.6m k
𝐹
FAB= 𝐴𝐵 = 𝐹𝐴𝐵 (−0.19 𝒊 + 0.95 𝒋 + 0.24 𝒌 )
|AB|
FBC=
FBE=
𝐹𝐵𝐶
|BC|
𝐹𝐵𝐸
|BE|
= 𝐹𝐵𝐶 𝒊
= −𝐹𝐵𝐸 𝒌
Performing ΣFy=0
|CE|=3.35m
0.95FAB+800=0
FAB=-842.11 N (Compression Force)
Performing ΣFx=0
-0.19FAB+FBC=0
FBC=-160 N (Tension Force)
Performing ΣFz=0
0.24FAB-FBE=0
FBE=202.11 N (Tension Force)
FAC=
𝐹𝐴𝐶
3.317
= 𝐹𝐴𝐵 (1.2 𝒊 + 3 𝒋 + 0.75 𝒌 )
FAC = 𝑭𝑨𝑪 (0.362 i + 0.9 j +0.226 k)
𝐹
FCE= 𝐶𝐸 = 𝐹𝐶𝐸 (1.8 𝒊 + 3 𝒌 )
3.35
FCE = 𝑭𝑪𝑬 (0.514 i +0.857 k)
FCD=FCD k
FBC= -160 i
Performing ΣFy=0
0.9FAC+100=0
𝑭𝑨𝑪 = 111.11 N (Compression Force)
Performing ΣFx=0
0.362FAC+0.514FCE-160=0
FCE=-3.3 N (Compression Force)
Performing ΣFz=0
0.226FAC+0.587FCE+FCD=0
FCD=2.5 N (Tension Force)
Simulation at B Joint
Analysis at C Joint
Simulation at C Joint
Analysis at D Joint
Figure 8. Free Body Diagram of C Joint and Reference
Axis
C joint is connected to members AC, BC, CE and CD
|AC|=√(1.2)2 + (3)2 + (0.75)2
|AC|=3.317m
|CE|=√(1.8)2 + (3)2
III. PROBLEM ANALYSIS (B)
Q. The truss shown consists of nine members and is
supported by a ball and socket at B, a short link at C, and
two short links at D. (a) Check that this truss is a simple
truss, that it is completely constrained, and that the reactions
at its supports are statically determinate. (b) Determine the
force in each member for P=0N and Q=(-900N k)
Figure 9. Free Body Diagram of D Joint and Reference
Axis
Joint D connects DA, DC and DE
|AC|=√(1.2)2 + (3)2 + (2.25)2
|AC|=3.937 m
𝐹
FAD= 𝐴𝐷 = 𝐹𝐴𝐷 (−1.2 𝒊 + 3 𝒋 + 2.25 𝒌 )
3.937
FCD=225 N k
FDE= -FDE i
Performing ΣFy=0
3
F + 300N = 0
3.937 AD
𝐅𝐀𝐃 =-393.7 N (Compressive Force)
Performing ΣFx=0
−1.2
F -FDE=0
3.937 AD
FDE= 120 N (Tension Force)
= -800 N
•
τBC=0
Figure 10. Problem (B)
Numerical Solution
Simulation at D Joint
So,
FAC = (40.222 i + 99.99 j +25.111 k) N
FAB = (-154.3 i + 771.5 j +194.91 k)N
FAD = (-120 i + 300 j +225 k)
FDE= (-120 i) N
FCD = (225 k) N
FCE = (119.92 i + 199.94 k ) N
FBE = (202.11 k )N
FBC = (-160 i ) N
Figure 11. Free Body Diagram and Reference Axis for
Problem (B)
Equations of Analysis
ΣFx=ΣFy=ΣFz=0 (1st Condition of Equilibrium should be
satisfied)
Στx= Στy= Στz=0 (2nd Condition of Equilibrium should be
satisfied)
• ΣFx=0
Bx=0 N
• ΣFz=0
Dz=Bz-900=0 N
• τBEk=0
|AC|=√(1.2)2 + (3)2 + (0.75)2 =3.317
|CE|=√(1.8)2 + (3)2 =3.5
FAC=
𝐹𝐴𝐶
|AC|
= 𝐹𝐴𝐶 (1.2 𝒊 + 3 𝒋 + 0.75 𝒌 )
𝐹𝐴𝐶 (0.362𝒊 + 0.9 𝒋 + 0.226 𝒌 )
𝐹
FCE= 𝐶𝐸 = 𝐹𝐶𝐸 (−(1.8)𝒊 + 3 𝒌 )
|CE|
3Dz-(-0.6Q)=0
Dz= 300N
Bz= -600 N
• τBEj=0
1.8Dy+1.8Cy=0
Cy= -Dy= -900 N
• τBC=0
𝐹𝐶𝐸 (−0.514𝒊 + 0.857 𝒌 )
𝐹𝐶𝐷= 𝐹𝐶𝐷𝑘
Performing ΣFy=0
0.9FAC-900=0
FAC=991.1 N (Tension Force)
Performing ΣFx=0
0.362FAC+0.514FCE=0
FBC=-6.8 N (Compression Force)
Performing ΣFz=0
0.226FAC+0.857FCE+FCD=0
FCD=3.5 N (Tension Force)
-3Dy- 3Q=0
Dy= 900 N
•
ΣFy=0
By+Cy+Dy=0
By=0 N
Analysis at C Joint
Simulation at C Joint
Analysis at D Joint
Figure 12. Free Body Diagram of C Joint and Reference
Axis
Joint C connects members AC, BC, CE and CD
Involving vector approach for the sake of ease, we may state
the following results proceeding as follows: -
Figure 13. Free Body Diagram of D Joint and Reference
Axis
Joint C connects members DA, DC and DE
Involving vector approach for the sake of ease, we may state
the following results proceeding as follows: |AD|=√(1.2)2 + (3)2 + (2.25)2 =3.937
𝐹
FAD= 𝐴𝐷 = 𝐹𝐴𝐷 (1.2 𝒊 + 3 𝒋 + 2.25 𝒌 )
Figure 14. Free Body Diagram of B Joint and Reference
Axis
Node B aligned members BE
FAB=FBC=0
Analysis at E Joint
|AD|
𝐹𝐴𝐷 (0.362𝒊 + 0.9 𝒋 + 0.226 𝒌 )
FCD=375 N
FDE=-FDE 𝒊
Performing ΣFy=0
3
FAD+900=0
3.937
FAD= -1181.1 N (Compression Force)
Performing ΣFx=0
−1.2
FAD- FDE=0
3.937
FDE=360 N (Tension Force)
Simulation at D Joint
Analysis at B Joint
Figure 15. Free Body Diagram of E Joint and Reference
Axis
AE is zero force member.
So,
FCE = (359.7 i + 599.73 k) N
FAD = (-360 i + 900 j +675 k)
FDE= (-360 i) N
FCD = (375 k) N
FBE = (600 k )N
IV. DOMINANT FACTORS FOR VARIATION IN THEORETICAL
AND ANSYS SIMULATION
•
In theoretical calculations, the parameters of
Modulus of Elasticity, shear stress and material type
does not affect the results, but when simulating on
Ansys, the results vary as all parameters come to
account.
•
The truss is considered in equilibrium though it
may not remain static after applied loads and forces.
•
Truss may be deformed which can be the reason
for discrepancy in reaction forces.
REFERENCES
[1]
[2]
Taylor, John R. (2005). Classical Mechanics. University Science
Books. pp. 17–18. ISBN 9781891389221.
Shapiro, Ilya L.; de Berredo-Peixoto, Guilherme (2013). Lecture Notes
on Newtonian Mechanics: Lessons from Modern Concepts. Springer
[3]
[4]
[5]
[6]
[7]
[8]
Science & Business Media. p. 116. ISBN 978-1461478256.
Retrieved 28 September 2016.
This translation of the third law and the commentary following it can
be found in the "Principia" on page 20 of volume 1 of the 1729
translation.
https://engineeringstatics.org/distributed-loads.html
https://web.mit.edu/4.441/1_lectures/1_lecture13/1_lecture13.html
Vector Mechanics for Engineers: Statics", the 11th Edition, Ferdinand
Beer and Russel Johnson
R. Nicole, “Title of paper with only first word capitalized,” J. Name
Stand. Abbrev., in press.
[9] Y. Yorozu, M. Hirano, K. Oka, and Y. Tagawa, “Electron spectroscopy
studies on magneto-optical media and plastic substrate interface,” IEEE
Transl. J. Magn. Japan, vol. 2, pp. 740–741, August 1987 [Digests 9th
Annual Conf. Magnetics Japan, p. 301, 1982].
[10] M. Young, The Technical Writer’s Handbook. Mill Valley, CA:
University Science, 1989.