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lecture notes for civil engineering
Typology: Lecture notes
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Associate Professor Svetlana Lilkova-Markova, PhD
Assistant Professor Angel Mladenski, PhD Assistant Professor Dimitrina Kindova-Petrova
January 2014
To examine the real object a correct corresponding computational scheme must be chosen. The
computational scheme is a real body for which the unessential attributes are eliminated. To choose the
correct computational scheme the main hypotheses of Strength of materials have to be introduced.
1.4. MAIN HYPOTHESES
A. Hypotheses about the material building the body
The material is uniformly distributed in a whole body volume.
All points of the body have the same material properties.
The material properties are the same in each direction of a body.
Contrary to the Theoretical Mechanics studying the rigid bodies, Strength of Materials studies the bodies possessing the ability to deform , i.e. the ability to change its initial shape and dimensions under the action of external forces.
The deformations at each point are assumed to be small relative to the dimensions of construction. Then, their influence onto the mutual positions of the loads can be neglected (the calculations will be made about the undeformed construction).
Elasticity is the ability of the body to restore its initial shape and dimensions when the acting forces have been removed.
B. Hypotheses about the shape of the body
Each planar cross-section normal to the axis of the beam before the deformation remains planar and normal to the same axis after deformation.
C. Hypotheses about the applied forces
If we replace a set of forces acting upon an area 1 of the deformable body with other set of forces equivalent to the first one, but acting upon the area 2 of the same body, the replacing will influence on the stresses and deformations in the area , containing 1 and 2 , where the influence’s magnitude will correspond to the size of the bigger area between 1 and 2.
If the body is in equilibrium, then, each part of the body is also in equilibrium.
The magnitude of the applied external forces increases gradually from zero to the final value.
Let the initial position of the beam to be the position of equilibrium. If the applied external
forces cause the small deformations according to the hypothesis studied earlier, the final position of the
beam is also position of equilibrium. Then, investigating the beam, the assumption that the initial
position of equilibrium coincides with the final one is made.
Rigid body – a body consisting of particles the distances between which do not change
Deformable body – a body consisting of particles the distances between which change. A
deformable body is a rigid one only to the definite loading.
CHAPTER 2
INTERNAL FORCES
A beam in equilibrium under the action of a set of forces is considered. This set of forces causes the deformation of the beam where the distances between the beam points change. Then, the forces of interaction between the points also change. The additional forces of interaction arising in the body are named internal forces. They have to be studied because they are related to the resistance of the body against the applied loads, and, consequently, to the strength of the body. The internal forces are the measure of interaction between two body parts situated on the two sides of the same section.
The internal forces can be determined by the method of section , as follow: Let the beam in fig.2. to be in equilibrium under the action of a set of forces F 1 (^) , F 2 ,... F n named external forces. They include the external loads as well as the support reactions previously obtained. A plane normal to the longitudinal axis of the beam divides the body into two parts. A border section between these two parts is called the cross-section.
Fig. 2.1: A beam acted upon by a set of external forces
Further, one of the parts is removed (usually this one upon which the bigger number of loads acts) while the other will be investigated.
Fig. 2.2: The left beam part
The hypothesis of the local equilibrium has the essential role in Strength of materials and according to it, if a body is in equilibrium, then each part of the body is in equilibrium, too. This hypothesis leads to the conclusion that the left part of the beam must be in equilibrium under the action of a set of forces applied on it. However, the external forces are not in equilibrium themselves. To be
The internal forces must be introduced always with their positive senses, which for the left and right beam part are given in fig.2.4:
Left beam part Right beam part
Fig. 2.4: Positive senses of internal forces – plane case
The axial force N is supposed to be positive when its sense is out of the section. The shearing force V is supposed to be positive when its sense coincides with the sense of the positive axial force rotated at an angle of 90^0 in clockwise direction. The bending moment is supposed to be positive when the curved arrow represented the moment begins from the downer beam end and finishes in the upper one without crossing the beam. It is important to note, that the concept of internal forces always relates to the definite beam section.
2.2. INTERNAL FORCES FUNCTIONS AND DIAGRAMS The conditions of equilibrium are written about the beam part considered. These equations are:
Fix 2 ) (^) 0 ; i
Fiy 3 ) (^) 0 ; i
Fiz (2.1)
4 ) (^) 0 ; i
M (^) ix 5 ) (^) 0 ; i
M (^) iy 6 ) (^) 0 ; i
M (^) iz (2.2)
Fix 2 ) (^) 0 ; i
Fiz 3 ) (^) , 0. i
M (^) iC (2.3)
It is obvious, that each internal force can be determined by one equation. However, in a real problem, it is not enough to find the magnitude of the internal forces in the definite beam section. It is necessary to obtain the change of the internal forces in the whole beam. To perform that, the beam must be separated into the segments. The boundary point (section) of the segment is a beam point at which the concentrated force or moment is applied. If the distributed load acts upon a beam, then, both the beginning and the end of the load are the boundary points. Besides, the points at which the change of distributed load intensity exists are also boundary points. Finally, if the beam axis bends, then the bending point is a boundary point. After that, an arbitrary chosen beam section of distance x for each segment must be considered. The distance x can be measured from the beginning of the beam, but in the most of the cases x is measured from the left or the right end of the segment. Further, the imaginary cut through the section chosen has to be made to divide the beam into two parts. Then, the one beam part has to be investigated and the equilibrium conditions must be written. In this manner, the internal forces will be obtained as functions of x. The graphs of these functions are named the internal forces diagrams. To build the diagrams, first the zero line representing the beam axis must be drawn in scale. The typical values of every
function have to be drawn perpendicular to the zero line in a definite scale and the typical points have to be obtained. Finally, the points must be connected consequently. The rules about the diagrams building are:
Problem 2.1. Build the internal forces diagrams of the planar straight beam given. The support reactions are obtained by the equations:
^0 ; i
Fix AH 0 ; 0 ; i
Fiz
R (^) q 20.3 - 60 kN ; A (^) V - 60 10 0 ; AV 50 kN ;
M kNm
A
A A
Check:
MD MA AV
N
V
M
10 kN
20 kN/m 100 kNm
Rq 60
2 , 5 m
x (^) B x x
3 m 1 2
The straight beam loaded by concentrated force, concentrated moment and uniformly distributed transverse and axial loads is considered (fig.2.5).
Fig. 2.5: Straight beam under loading
F=20 kN
3m 3
M=100 kNm
To derive the differential equations of the internal forces of the segment in which the distributed loads act the infinitesimal beam part is examined.
Fig.2.6: Infinitesimal beam part
Further, the equilibrium equations of the infinitesimal beam part are written:
Fix N dN N tdx 0 ; dx t ; dN (2.4)
Fiz V dV - V qdx 0 ; dx - q ; dV (2.5)
mom (^) right F - 2 - Vdx 0. dx M dM M qdx (2.6)
The term 2
dx qdx is very small and it can be neglected. In this manner, the relation
dx
dM is obtained. (2.7)
It can be proved, if the distributed loads functions t x and q x are continuous functions ,
then the differential equations of the internal forces are:
dM x q x dx
dV x t x dx
d N x (2.8)
The distributed loads t x and q x are supposed to be positive when their senses coincide
with the positive senses of the internal forces N x and V ( )x , respectively, for the left beam part.
A small beam part of length x is considered. The distributed loads q and t
represented as continuous functions act upon this small part of the beam.
Fig. 2.7: A small beam part of length x
; - - ; V x m x. dx
dM x V x m x dx
dM x m x dx
dT x y z z z y
y (^) x (2.18)
Problem 2.3 Determine the internal forces functions of the beam shown and apply the differential equations to check the result.
First, the support reactions must be determined: ^0 ; i
Fix 15. 3 A (^) H 0 ; AH 45 kN ;
M^ A ^0 ;^7 B ^80.^2 ^50 ^0 ;^7 B ^210 ; B ^30 kN^ ; M^ B ^0 ;^ ^7 AV ^80.^5 ^50 ^0 ;^7 AV ^350 ; AV ^50 kN ; Check:
^0 ; i
Fiz A (^) V B - 80 0 ; 50 30 - 80 0 ; 80 - 80 0.
segment AC : 0 x 4 m
^0 ; i
Fix N - 45 0 ; N 45 kN ;
^0 ; i
Fiz 50 - 20 x - V 0 ; V - 20 x 50 ;
sec 0 ;^20. x M x^2 x
x M^ tion M x
Differential check (check by the differential equations of the internal forces): t 0 ; q 20 kN/m;
- t x ; 0 0 ; dx
d N x ^ - q x ; - 20 - 20 ; dx
dV x
V x ; - 20 x 50 - 20 x 50. dx
dM x
segment CB : 0 x 3 m
^0 ; i
Fix - N 15 3 - x 0 ; N - 15 x 45 ;
^0 ; i
Fiz V 30 0 ; V - 30 kN ;
30 40.
sec^0 ;^303500 ;
M x
M (^) tion M x
Differential check: t 15 kN / m ; q 0 ; t x ; 15 15 ; dx
d N x q x ; 0 0 ; dx
dV x
V x .; - 30 - 30. dx
dM x
segment BD : 0 x 2 m
^0 ; i
Fix N 0 ; ∑ 0 ; i
Fiz V 0 ;
M^ sec tion ^0 ; M ^50 ^0 ; M ^50. Differential check: t 0 ; q 0 ; t x ; 0 0 ; dx
d N x - q x ; 0 0 ; dx
dV x
V x ; 0 0. dx
dM x
This approach is applicable when a complicated distributed loads act upon a straight beam as well as a curved one. The essence of the method is the integration of the internal forces differential equations (2.8) in every beam segment. To determine the integration constants the boundary conditions of equilibrium of typical beam sections must be written. These beam sections are separated by cuts at infinitesimal distance from the section. It is important the unknown support reactions must not take part in the boundary conditions. After the internal forces functions have been determined, the internal forces diagrams can be drawn.
Problem 2.4 Apply the integration method to find the internal forces functions of the beam shown.
The beam contains two segments and the differential equations (2.8) are written and integrated for each of them, as follow: segment AC : 0 x 4 m
t x - 5 ;
dN x t x dx
dN x
The function q ( )x is a parabola of a type q x ax^2 bx c. To find constants a , b and c
the conditions q 0 0 ; q 5 30 и q ' 5 0 will be used. It is obtained: a - 1 , 2 ; b 12 ; c 0.
Then, q x -1,2 x 2 12 x.
The differential equation is written: x x x x dx
dV x q x dx
dV x - ; --1,2^2 12 1,2^2 - 12 ,
and after integration it is carried out: V x 0 , 4 x^3 - 6 x^2 C 2.
Boundary section В: H^ ^0 ;^2 )^ N ^5 ^ ^5 ; M^ B ^0 ;^3 )^ M ^5 ^ ^0.
The integration constants are: C 1 (^) - 20 ; C 2 (^) 37 , 5 ; C 3 (^) 0. The full expressions of the segment AB’s internal forces functions have been already found and the diagrams can be built. N x 5 x - 20 ; V x 0 , 4 x^3 - 6 x^2 37 , 5 ; M x 0 , 1 x^4 - 2 x^3 37 , 5 x. To find the internal forces functions in segment ВС the method of section will be applied and the right segment part will be considered. H^ ^0 ; N ^5 ; V^ ^0 ; V ^0 ; M^ ^0 ; M ^0.
The differential equations of the internal forces have to be considered:
2.5.2. CHECK OF THE INTERNAL FORCES DIAGRAMS a) Check about the type of the diagrams The last two differential equations in (2.8) are rearranged in a form q x dx
dV x dx
d М x 2 -
2 . (2.19)
It is obvious, if in some beam segment q x const , then the shearing force function V x must be linear, while the bending moment function M x must be square. If q x 0 , then V x must be constant, while M x must be linear function. Furthermore, if in some beam segment the distributed load q x points down , then the function V x must decrease , and the convexity of M diagram must direct down. However, if q x points
up , then the function V x must increase , and the convexity of M diagram must direct up.
b) Check about the steps and kinks in the diagrams
If a concentrated transverse force F is applied at some beam section, then the step in V diagram must exist at the same section where the magnitude and the sense of the step coincide with these ones of the force. Besides, the kink in M diagram must exist at the same beam section where the sense of the kink is in the sense of the force.
If a concentrated axial force F is applied at some beam section, then the step in N diagram
must exist at the same section where the magnitude of the step is equal to this one of the force while the sense of the step is the force’s sense rotated at an angle of 90 0 clockwise. If a concentrated moment is applied at some beam section, then the step in M diagram must exist at the same section where the magnitude and the sense of the step coincide with these ones of the moment.
c) Area check The differential equations (2.8) are considered for any segment and the rearrangements are made, as follow:
d M x V xdx ; (^)
l l dM x V xdx 0 0
where l is the length of the segment. The integral in the right side of the equation represents the area
AV (^) , l of V diagram. Then, using (2.20), it is obtained: M l - M 0 AV , l. (2.21) The other two equations in (2.8) are integrated in the same manner:
d N x - t xdx ; (^)
l l dN x t xdx 0 0
d V x - q xdx ; (^)
l l dV x q xdx 0 0
Introducing
l Rt t xdx 0
( ) and
l Rq qxdx 0
which are the resultant forces of the distributed loads t x and q x , respectively, the relations (2.22)
and (2.23) become:
