Drawing graphic work 10. Guide for performing graphic work on descriptive geometry for university students. Practical and graphic work on drawing

T G T U

P.A. Ostrozhkov, M.A. Kuznetsov, S.I. Lazarev
for university students studying in engineering fields
and technology

Graphic work No. 1
Graphic work No. 2
Graphic work No. 3
Check
test
Application

Graphic work No. 1
The relative position of two planes.
Purpose of the work: consolidation of knowledge in
positional tasks.
Task
Task No.
№ 11
decision
Task
Task No.
№ 22

Task No. 1
1. In a plane defined by three points
A, B, C (see point coordinates in
application) to construct a triangle,
formed by horizontal, frontal and
profile straight. Draw
the resulting triangle to natural
size.
2. Construct a plane parallel
given and spaced from it by
distance 50 mm.

A3 format
(290x420 mm)
menu

Mentally mark the sheet into 2 parts

On the left side of the A3 sheet we mark the coordinate axes.
z
x
0
y

According to the coordinates of the individual task, we mark points A, B and C -
vertices ∆ ABC in coordinate planes.
B"
A"
z
C"
x
0
B'
A'
C'
y

We connect the points with segments, forming a plane ∆ ABC, respectively in
projections.
B"
A"
z
C"
x
0
B'
A'
C'
y

We carry out a projection of the horizontal D”P” in the frontal plane
(parallel to the X axis) and project it into the horizontal plane
projections.
B"
z
D"
P"
A"
C"
x
B'
0
D'
A'
P'
C'
y

We draw frontal D’E’ in the horizontal plane of projection, then
draw a profile straight line. We form DEF, in which, using
using the right triangle method to find the actual size
leg EF.
B"
z
E"
D"
A"
F"
P"
C"
x
0
B'
D'
A'
E*
E'
F' P'
C'
y

We construct the natural value of DEF formed by the straight lines of the quotient
provisions.
B"
z
E"
D"
A"
F"
P"
C"
x
0
B'
E"
D'
A'
E'
F' P'
C'
F
D
nv
E
y

Constructing a plane parallel to the given one and 50 mm distant from it.
We extend the horizontal projection of the horizontal (DF) and the frontal projection
frontal (DE), then to these straight lines we restore the perpendicular from point A and on this
perpendicular we mark in an arbitrary manner so as
B"
z
K"
Here we have applied the theorem about
projecting a right angle
E"
D"
A"
F"
P"
C"
x
0
B'
E"
D'
A'
F" P'
C'y
K'
E
nv
D
E'
F

We measure the difference in distances between points K and A (segment KL) and plot it on
perpendicular lowered to point K”, forming point K*.
B"
S"
S*
K"
K*
50
mm
D"
A"
x
By connecting t. A and t. K*, we get
z natural segment AK, extending this segment
put a segment equal to 50 mm on it and mark
E"
t. S*.
From point S* we draw a straight line parallel to the segment
P"
K*K” until it intersects with the original
F"
perpendicular (A”K”), forming t.S”.
C"
From point S we draw a plane parallel to this one.
To do this, at point S we intersect two straight lines, parallel
0 to two any straight lines of a given plane.
B'
E"
D'
A'
L"
F" P'
C'y
K'
S'
E
nv
D
E'
F

Task No. 2
According to the coordinates of the individual version of the task (see appendix), we mark
points A, B, C and D, E, F.
By connecting them with segments we obtain triangles ABC and DEF, respectively in projections.
D"
B"
Z
E"
C"
A"
F"
X
0
B'
F'
E'
C'
A'
D'
U

We mark in the horizontal plane of the projection t.1 and t.2, the intersection points of side A’B’
(ABC) respectively with sides E’F’ and D’E’ DEF.
We project t.1 and t.2 into the frontal plane of projection onto the corresponding straight lines and
We connect t.1 and t.2 to each other with a segment.
At the intersection of straight line AB” and segment 1”2” we form t.K”, then project it into
horizontal plane of projection onto the corresponding straight line.
1”
E"
D"
B"
Z
K"
C"
2”
A"
F"
X
0
B'
2’
E'
K'
F'
C'
1’
A'
D'
U

In the frontal plane of projection we mark points 3 and 4, the points of intersection of sides AB” and
"АС" (АВС) with side D"F" (DEF).
We project t.3 and t.4 into the horizontal projection plane onto the corresponding sides
triangle, connect them together with a segment.
At the intersection of the segment 3’ 4’ with side D’F’, we form point L.
We project t.L into the frontal plane of projection onto the corresponding side (D”F”).
D"
3”
1”
By connecting t.K and t.L with each other, we get
the desired line KL - the intersection line
planes defined by triangles.
E"
B"
Z
K"
L"
2”
C"
4”
A"
F"
X
0
B'
2’
E'
F'
3’
K'
L'
C'
4’
1’
A'
D'
U

Using the competing point method, we determine the visibility of the planes specified
triangles ABC and DEF.
B"
S"
z
D"
3”
1”
S*
K"
K*
50
mm
E"
E"
B"
K"
L"
2”
D"
A"
F"
Z
P"
C"
4” (6”)
5”
A"
F"
C"
x
0
X
B'
0
B'
E"
D'
A'
L"
E'
F" P'
C'y
K'
S'
2’
E'
6’
3’
K'
L'
(5’) 1’
F'
C'
4’
A'
E
U
D'
nv
D
F
menu

Graphic work No. 2

Methods for converting a drawing
Purpose of work: consolidation of knowledge and basic
techniques for solving metric problems.

The task.
Given a pyramid SABCD with a base ABCD (coordinates
points, see appendix) located in
planes of general position.
Required:
1.Use the method of rotation around the level line to determine
the actual size of the ABCD base.
2.Plane-parallel movement method
determine the distance from the vertex S to the plane
ABCD bases.
3. Using the method of changing projection planes, determine
the true value of the dihedral angle at edge BC,
formed by the base and side face
pyramids.

To perform this graphic work, a sheet is used
A3 format (290x420 mm)

It is decorated with a frame, a corner stamp and filling in the main inscription.

Z
B"
D"
L"
C"
S"
X
A"
0
S'
B'
D'
L'
C'
A'
U
According to individual assignment
mark by coordinates of the point
S, A, B, C and D“, missing
coordinate of point D’ - determine
construction.
We connect the points with segments,
form a base plane
ABCD pyramids.

Z
B"
L"
D"
C"
S"
H"
X
0
A"
S'
B'
D*
D'
R.D.
L'
C'
O'1
A'
rotation axis
.
H'
D
U
We define the axis of rotation (line
level-AH).
In the horizontal plane
projections from point D’ are omitted
perpendicular to the axis of rotation
A’H’, at their intersection we form
center of rotation (point O’1)
corresponding point D.”
Rectangular method
we get a triangle
natural radius size
rotation of point D.
Rotate t.D until it intersects with
perpendicular to them
at the intersection we form point D

Z
B"
L"
D"
C"
S"
H"
X
0
A"
S'
B'
D*
D'
R.D.
L'
B*
R.B.
C'
O'1
A'
O'2
O'3
H'
D
B
U

from point B’ we lower the perpendicular
to the axis of rotation A’H’, to their
at the intersection we form the center
rotation (point O’2) corresponding
points B.
Rectangular method
triangle we get the natural
the value of the radius of rotation of point B.
Rotate t.B until it intersects with

form t.V

Z
B"
L"
D"
C"
S"
H"
X
0
A"
S'
B'
D*
D'
R.D.
L'
B*
R.B.
C'
O'1
C*
R.C.
A'
O'2
O'3
H'
C
D
B
U
In the horizontal plane of projection
from point C’ we lower the perpendicular to
rotation axis A’H’, at their intersection
forming a center of rotation
(point O’3) of the corresponding point C.
Right triangle method
we get the actual size
radius of rotation of point C.
Rotate t.C until it intersects with
perpendicular, at their intersection
we form t.S.
We do not rotate point A, since it lies
on the axis of rotation.

Z
B"
L"
D"
C"
S"
H"
X
0
A"
S'
B'
D*
D'
R.D.
L'
B*
R.B.
C'
O'1
C*
R.C.
A'
O'2
NV
D
B
O'3
H'
C
U
Connecting the formed dots
segments, we get
life size
base of the pyramid ABCD.

Z
B"
L"
D"
C"
S"
A" H"
H"
X
0
A"
S'
B'
D*
D'
R1
R.B.
R2
C'
A'
O'2
R3
R.C.
O'3
R.C.
R.C.
C*
H'
B'
NV
D
B
R.C.
C'
R.B.
L'
R.B.
O'1
B*
C
R.B.
A'H'
U
We bring plane ABC into
projecting position
planes, i.e. perpendicular
projection plane. For getting
front-projecting
plane needs horizontal
AH planes together with the system
all points of the plane (ABC)
put in position
perpendicular to the frontal
projection planes.

Z
We move t.S – the top of the pyramid.
B"
L"
D"
C"
S"
A" H"
H"
X
0
A"
S'
B'
D*
D'RS
R1
O'1
B*
R.S.
L'
A'
R2
R.S.
C'
C'
R.S.
C*
R3
O'2
O'3
H'
B'
NV
D
C
R.S.
A'H'
S'
B
U

Z
B"
B"
K"
L"
D"
C"
C"
S"
H"
S"
X
A" H"
0
A"
S'
B'
D*
D'
R1
L'
B*
R2
C'
O'1
A'
C'
C*
R3
O'2
O'3
H'
B'
NV
D
C
A'H'
S'
B
S” K” = 32 mm
K'
U
Along the moved horizontal
projection A’B’C’ and its original
frontal projection we build
new frontal ABC projection
and point S. Determine the distance
from t.S to a given plane. It
equal to the perpendicular segment SK,
lowered from t.S onto the plane
degenerated on the new
front-projecting
projection planes into a straight line
line.
Having received the base of the perpendicular
SK, we build its horizontal
projection on the original drawing
tasks.

Z
B"
B"
K"
L"
D"
C"
The dihedral angle is measured
linear angle made
lines of intersection of faces
dihedral angle with a plane,
perpendicular to its edge.
C"
S"
H"
S"
X
A" H"
0
A"
S'
B'
D*
D'
R.D.
L'
B*
B"
R.B.
C'
O'1
C'
C*
R.C.
A'
O'2
O'3
H'
B'
NV
D
X
C
A'H'
S'
B
S” K” = 32 mm
C"
S"
K'
P2
P1
0
A"
S"
B'
U
C'
A'

Z
B"
B"
K"
L"
D"
C"
C"
S"
H"
S"
X
A" H"
0
A"
S'
B'
D*
D'
R.D.
L'
B*
BIV
R.B.
R.C.
A'
C'
SIV
C'
O'1
When using the plane replacement method
you need to keep in mind that the figure does not change
its position in space, the plane
projections P1 are replaced by a new plane,
respectively P4. When constructing projections
figures on the new projection plane
It must be remembered that there is a transition from
one image to another, in which
the corresponding projections of the points are also
located on communication lines. Coordinates
point on the new projection plane is equal to
coordinate of a point on the plane being replaced
AIV
projections.
O'2
O'3
CIV
C*
B"
H'
B'
NV
D
C
A'H'
S'
K'
X
B
S” K” = 32 mm
U
P4
C"
S"
P2
P2
A"
P1
0
S"
B'
C'
A'

Z
B"
B"
K"
L"
D"
C"
In order for the linear angle
projected onto the projection plane in
natural size, need a new one
place projection plane P5
perpendicular to the edge of the BC dihedral
angle.A
V
P5 P4
C"
S"
H"
S"
X
0
A"
AIV
A" H"
SV
BV C V
S'
SIV
B'
D*
D'
R.D.
L'
CIV
B*
C'
B"
R.B.
C'
O'1
C*
=40°
R.C.
A'
O'2
O'3
H'
D
X
C
A'H'
S'
B
S” K” = 32 mm
K'
P4
C"
S"
B'
NV
BIV
P2
P2
P1
0
A"
S"
B'
U
C'
A'
menu

Graphic work No. 3 sheet 1
Intersection of a surface by a plane.

skills in solving positional problems on the surface
and construction of surface developments.

The task.
1. Construct projections of a section of a regular pyramid
general position plane defined by three points
A, B, C (see the appendix for the coordinates of the points). Center
circle circumscribed around the base of the pyramid
located at point K with coordinates (70,60,0).
2. Construct a complete development of a truncated pyramid according to
conditions of the previous problem.

To perform this graphic work, a sheet of format is used
A3 (290x420 mm)

Designed with a frame, corner stamp and filling in the main inscription

Z
S"
B"
A"
C"
P1
D"
F"
E"
0
P2
A'
D'
U
B'
S'
F'
K'
E'
C'
In the left half of the A3 sheet, axes are outlined
coordinates, according to their option are taken
quantities that define the surface of the pyramid
and plane ABC (see appendix). Determined
center (point K) of a circle with radius R base
pyramids in the level plane. On the vertical axis
at a distance H from the level plane and above it,
the top of the pyramid is determined.

Z
S"
B"
A"
C"
P1
D"
F"
E"
0
P2
A'
D'
U
B'
S'
F'
K'
E'
C'
Based on the coordinates of points A, B, C, it is determined
cutting plane.

Z
S"
B"
A"
To facilitate the construction of a section line
an additional drawing of the specified
geometric images.
An additional system of P1/P4 planes is selected
projections in such a way that the cutting plane
was presented as projecting.
Additional projection plane P4
perpendicular to the given plane ABC.
AIV BIV
P1
C"
P1
D"
F"
P4
E"
0
P2
DIV
A'
SIV
D'
U
B'
KIV
FIV
S'
F'
K'
EIV
E'
C'
CIV

Z
S"
B"
A"
The section line is projected onto the plane
projection P4 in the form of a straight line segment on the trace
this plane. Having a section projection on
additional plane P4 build the main
her projections.
L"
N"
C"
P1
D"
F"
P1
M"
AIV BIV
P4
E"
0
P2
DIV
LIV
A'
D'
SIV
U
B'
L'
NIV
KIV
FIV
S'
N'
MIV
K'
E
IV
M'
F'
E'
C'
CIV

Z
S"
B"
A"
L"
L"
N"
N"
M
C"
P1
A full scan is built on the right half of the sheet
pyramids.
On the frontal projection, the natural
the size of the edge of the pyramid.
We take down the characteristic points of the pyramid's cross-section to
natural size of the rib.
”
D"
F"
M"
P1
AIV BIV
P4
E” E”
0
P2
DIV
LIV
A'
D'
SIV
U
B'
L'
NIV
KIV
FIV
S'
N'
MIV
K'
E'
E
IV
M'
F'
E'
C'
CIV

S
R1
R1
Z
S"
R1
D
D
R1
B"
A"
R
R1
R
L
”
L"
N"
M"
C"
P1
F
N"
D"
F"
M"
P1
A
BIV
P4
R
E” E”
0
P2
E
R
IV
R
DIV
LIV
A'
D'
U
B'
L'
SIV
NIV
KIV
R
FIV
M
S'K'
N'
EIV
M'
F'
E'
C.I.
V
C'
IV
D
Knowing the natural size of the rib
pyramids, build its development.

S
R.L.
R.L.
L
Z
S"
L
RN
R.M.
D
B"
A"
RN
L
”
R.L.
N
R.M.
P1
M
L"
N"
F
N"
M"
C"
D"
F"
D
M"
P1
A
P4
IV
B
E
IV
E” E”
0
P2
DIV
LIV
A'
D'
U
B'
L'
SIV
NIV
KIV
FIV
MIV
S'K'
N'
E
IV
M'
F'
E'
C.I.
V
C'
D
On the edges and faces of the pyramid
(on the scan) determine
vertices of spatial
broken pyramid intersection
with a plane.

S
M
L
L
Z
S"
R1
D
B"
D
R
R1
N
A"
R
L
”
L"
N"
P1
F
N"
M"
C"
D"
F"
M
M"
P1
E
A
BIV
IV
P4
E” E”
0
P2
DIV
LIV
A'
D'
U
B'
L'
SIV
NIV
KIV
D
FIV
MIV
S'K'
N'
We get the development of the pyramid.
EIV
M'
F'
E'
C.I.
V
C'
menu

Graphic work No. 3 sheet 2

Mutual intersection of surfaces.
Cone development.
Purpose of work: consolidation of knowledge and acquisition
skills in solving positional problems on the surface and
construction of surface developments

The task.
1) construct projections of the line of intersection of two
surfaces in an auxiliary way
cutting planes.
2) construct projections of the line of intersection of two
surfaces using the concentric sphere method.
3) construct a development of the lateral surface
cone with drawing the line of intersection along
condition of task 1 or 2.

To perform this graphic work, a sheet is used
A3 format (290x420 mm)

It is decorated with a frame, a corner stamp and filling in the main inscription.

K"
S"
TO'
S'
On the left half of the sheet mark
image of three surfaces
rotation according to your option (see.
application). Choose for two
intersecting surfaces
(having parallel axes) method
auxiliary cutting planes, and
for the other two intersecting
surfaces (having
intersecting axes) method
concentric spheres.

When solving a problem using
auxiliary cutting planes
determine the points of the intersection line
surfaces.
Construction begins with characteristic
edge points of the intersection line.
K"
S"
S"
3”
2”
1”
’
1’
K'
S'
3’
S'
2’
’

K"
S"
S"
3”
1”
4”
5”
R 1'
R1
1”
2”
1’
R 1'
’
S'
K'
5’
3’
R1
S'
4’
2’
’
By drawing auxiliary secants
horizontal projection planes
1- n, we get in the section of each
surface circle. Projections of two
circles on a horizontal
projection planes intersect between
themselves at two points 4’ and 5’,
belonging to the desired line
intersections. Frontal projections
these points are constructed using lines
connections, they are located in the P2 plane
on the trace of the cutting plane.

K"
S"
S"
3”
1”
4”
5”
R1
2”
R 1'
7”
6”
R2
R 2'
1”
2”
R2'
R1'
1’
7’
’
5’
S'
K'
3’
R2
R1
S'
4’
6’
2’
’

K"
S"
S"
3”
1”
4”
5”
R1
2”
R 1'
7”
6”
R2
3”
R 2'
9”
8”
R 3'
R3
1”
2”
R3'
R2'
9’
1’
’
R1'
7’
5’
S'
K'
3’
R2
R1
S'
R3
4’
6’
8’
2’
’

An intersection line is constructed using points
surfaces of revolution and
its visibility is established in
projections.
S"
S"
3”
1”
4”
5”
R1
2”
R 1'
7”
6”
R3
3”
R 2'
9”
8”
R 3'
R4'
1”
2”
R3'
R2'
9’
1’
’
R1'
7’
5’
S'
K'
3’
R2
R1
S'
R3
4’
6’
4’1
8’
2’
’

S"
S"
3”
1”
1”
4”
5”
R1
2”
R 1'
7”
6”
R3
3”
R 2'
9”
8”
R 3'
R4'
1”
2”
2”
R3'
R2'
9’
1’
’
R1'
7’
1’
5’
S'
K'
3’
R2
R1
S'
R3
4’
6’
8’
2’
’
2’
When solving a problem using
auxiliary concentric spheres
the following must be completed
conditions:
both surfaces must be
surfaces of rotation;
their axes must intersect;
each axis must be parallel
any projection plane.
We begin the construction with the definition
characteristic edge points of lines 1 and 2
surface intersections.

S"
S"
3”
1”
1”
R1
7”
6”
R 2'
9”
8”
R 3'
R4'
1”
R1
R 1'
R3
3”
3” 3”1
4”
5”
2”
From the point of intersection of the axes as from the center
a sphere of arbitrary radius is drawn.
It intersects both surfaces along
circles.
2”
2”
3’
R3'
R2'
9’
1’
’
R1
R1'
7’
1’
5’
S'
K'
3’
R2
R1
S'
R3
4’
6’
3’1
8’
2’
’
2’

S"
S"
3”
1”
1”
R1
7”
4” 4”1
6”
R 2'
R2
9”
8”
R 3'
R4'
1”
R1
R 1'
R3
3”
3” 3”1
4”
5”
2”
Changing the radius of the auxiliary secant
spheres, you can get
sequential series of line points
intersections.
2”
2”
3’
4’
R3'
R2'
9’
1’
’
R1
R1'
7’
1’
5’
S'
K'
2’
R2
3’
R2
R1
S'
R3
4’
6’
3’1
8’
2’
’
4’1

S"
S"
3”
1”
1”
R1
2”
7”
4” 4”1
6”
R 2'
R2
9”
8”
5” 5”1
R3
R 3'
R4'
1”
R1
R 1'
R3
3”
3” 3”1
4”
5”
2”
2”
3’
4’
5’
R3'
R2'
9’
1’
’
R1
R1'
7’
1’
5’
S'
K'
3’
R2
R3
R3
R1
S'
5’1
4’
6’
3’1
8’
2’
’
2’
R2
4’1

S"
S"
3”
1”
1”
R1
7”
4” 4”1
6”
R 2'
R2
9”
8”
5” 5”1
R3
R 3'
R4'
1”
R1
R 1'
R3
3”
3” 3”1
4”
5”
2”
Having constructed a sufficient number of points for
constructing intersection lines
surfaces and determining its visibility in
projections, draw a line of intersection
surfaces.
2”
2”
3’
4’
5’
R3'
R2'
9’
1’
’
R1
R1'
7’
1’
5’
S'
K'
3’
R2
R3
R3
R1
S'
5’1
4’
6’
3’1
8’
2’
’
2’
R2
4’1

S"
S"
3”
1”
1”
R1
7”
4” 4”1
6”
R 2'
R2
9”
8”
5” 5”1
R3
R 3'
R4'
1”
R1
R 1'
R3
3”
3” 3”1
4”
5”
2”
On the right half of the sheet they build
development of the lateral surface of the cone.
Divide the circle (base of the cone) by
12 equal parts.
2”
2”
3’
4’
5’
R3'
R2'
9’
1’
’
R1
R1'
7’
1’
5’
S'
K'
3’
R2
R3
R3
R1
S'
5’1
4’
6’
3’1
8’
2’
’
2’
R2
4’1

S"
S"
3”
1”
R1
R1
R 1'
7”
R 2'
R2
9”
8”
2”
2”
3’
1’
’
X
7’
R1
2’
1’
5’
S'
5’
R2'
R1' VII
IX
9’
4’
R3'
VIII
K'
VI
3’
R2
XI
V
4’
S'
6’
3’1
8’
XII
2’
III
I
II
R2
R3
R1
R3
’
5” 5”1
R3
R 3'
R4'
1”
4” 4”1
6”
R3
3”
3” 3”1
4”
5”
2”
1”
IV

It is not so easy to get a good higher education here. To do this, you will need not only to attend lectures, seminars and workshops, but also to complete various independent tasks, such as essays or coursework. In this article I would like to talk about what calculation and graphic work is.

About the concept

First of all, you need to understand the concept itself. Often, when a student hears the abbreviation RGR for the first time, he becomes confused. But there’s nothing to worry about, that’s the abbreviated name for calculation and graphic work. This is a student designed for a more complete assimilation of the material he has covered in a particular subject. It is also worth saying that RGR can be part of the course work, that is, its practical component. The essence of this type of work is to provide not only theoretical, but also practical material. Thus, the RGR will necessarily contain certain calculations, possibly graphs, tables, diagrams.

What should it be?

What important elements does the RGR consist of?

Justification of the chosen topic. This is a theoretical component where the student must talk about the importance of the work he has done.
Characteristic
Carrying out basic calculations.
Providing the results obtained in a convenient form: tables, graphs, diagrams.
Conclusions and possibly recommendations.

Structure

Calculation and graphic work must have its own structure. It is not possible to submit material for consideration in any form. So, the RGR should consist of the following points:

Table of contents. Here the student provides information about all sections of his work.
Exercise. At this stage, it is necessary to fully “voice” the task given to the student.
Initial data. The student provides all existing source data that may be needed to carry out the calculations.
This is followed by sections that will contain practical solutions and analysis of the results obtained.
Providing calculation results in the most convenient form for understanding.
Conclusions.
Bibliography.
Applications (if any).

Basic moments

There is also a list of special requirements that the student must comply with when preparing calculation and graphic work.

Design of tables and figures

Economics, statistics, theoretical mechanics... Calculation and graphic work can be performed in almost any subject where there are calculations (regardless of the student’s specialty of study). However, it is worth remembering that it is necessary not only to correctly format the text itself, but also to provide all the tables, figures and diagrams.

Computer science

What might computational and graphical work in computer science look like? So, it’s worth saying that there are no specific frameworks here. It all depends on the level of the material taught at the university for a given specialty. So, for humanities students the RGR in computer science will be one, for programmers it will be completely different. This could be simply a demonstration of PC skills (for example, in Word or Excel), or it could be programming, using different number systems to work, performing all kinds of translations between different ones, etc.

BJD

As part of the Life Safety course, some universities also offer students to complete RGR. And again, I would like to say that work in different specialties will differ from each other. After all, each profession has its own precautions and requirements. Calculation and graphic work on heavy-duty railways - what can be studied or researched here? Thus, you can calculate the most comfortable working conditions for a group of workers, you can plan the placement of jobs in a workshop or enterprise, you can analyze, etc. In fact, there are a huge number of topics to consider.

Other items

It is worth saying that calculation and graphic work can be written on almost any subject: economics, electronics, logistics, theoretical mechanics, etc. However, the goal of this work will always remain the same: to teach the student not only to correctly carry out the necessary calculations, but also to be able to correctly present them for consideration.

Send your good work in the knowledge base is simple. Use the form below

Students, graduate students, young scientists who use the knowledge base in their studies and work will be very grateful to you.

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ST. PETERSBURG STATE POLYTECHNIC UNIVERSITY

CONSTRUCTION ENGINEERING INSTITUTE

Department of "Water and hydraulic engineering"

Discipline "Road construction"

Calculation and graphic work for Road construction

Saint Petersburg

1.3.4 Path visibility

Literature

1. Determination of the required road parameters

In accordance with SNiP 2.05.02-85*, the category of a highway depends on the intensity of traffic along it. The expected traffic intensity during the construction period of the facility depends on the amount of cargo transported, construction time, vehicle brands and is determined by the formula:

Where q is the amount of cargo transported per 1 million rubles of the estimated cost of construction and installation work, t; accepted in the range of 8000-10000t;

C - estimated cost of construction and installation work on the facility, million rubles;

T - construction period of the facility, years;

n is the number of working days in a year;

Kpr - vehicle mileage utilization rate (ratio of vehicle mileage with cargo to its total mileage); for the construction conditions of the facility Kpr = 0.5-0.6;

Kgr - coefficient of utilization of the vehicle's carrying capacity (the ratio of the weight of the cargo on the vehicle to its nameplate carrying capacity), in practical calculations it is assigned Kgr = 0.7...0.8);

G is the vehicle’s carrying capacity, t. Let’s take KAMAZ-5510 as the design vehicle.

According to traffic intensity N in accordance with that given in table. 1 SNiP 2.05.02-85* by classification of highways we determine the category of the road.

7349 cars/day

According to Table 1 of SNiP 2.05.02-85*, a road with an estimated traffic intensity of 7349 vehicles/day is a category II road of regional significance.

1.2 Establishing the design speed of the road according to SNiP 2.05.02-85*

1.3 Determination of road parameters

1.3.1 Establishing the number of traffic lanes

The number of traffic lanes is determined by comparing the expected hourly traffic intensity on the road and the capacity of one lane using the formula:

where Nch is the hourly traffic intensity, car/hour;

Nп - traffic lane capacity, car/hour

Taking into account the uneven movement during the day

car/hour

The traffic lane capacity depends on the speed of the vehicles, their make, type and condition of the surface.

In this case, the traffic lane capacity is:

Here v is the estimated speed of movement, km/h;

ts- adhesion coefficient is assumed to be 0.5, which corresponds to a dry coating;

i- longitudinal slope of the road (we determine the capacity of the lane on a horizontal section, i.e. i=0);

f- rolling resistance coefficient (Table 1);

Vehicle length, m; (design vehicle KAMAZ 5510)

Distance margin equal to 5-10m;

Ke - brake performance coefficient equal to 1.4.

Table 1 - Distribution of rolling resistance coefficients

Requires 2 lanes.

1.3.2 Determination of the width of the roadway, lane and roadbed

The width of the roadbed depends on the width of the lane, the number of lanes and the width of the shoulder.

We record the values of the width of the traffic lane, roadway, shoulder and roadbed in Table 3.

1.3.3 Determination of the smallest radii of curves in plan

The smallest radius of the curve in plan, at which it is possible to use a gable profile at a given design speed, is determined by the formula:

When assigning turning radii smaller than Rн, it is necessary to provide for a superelevation device. This smallest value of the turning radius of a curved road is calculated by the formula:

The coefficient of adhesion between the wheel and the road in the transverse direction is 0.1 - 0.15;

Cross slope of the roadway (Table 2);

Superelevation slope (SNiP 2.05.02-85*, clause 4.17).

Table 2. - Cross slope values depending on the type of road surface

When making a turn, the length of the run L is determined by the expression:

where b is the width of the roadway, m;

Additional longitudinal slope for superelevation (5‰)

1.3.4 Path visibility

To ensure safe driving at the designed speed, the driver must see the road at a certain distance, which is equal to

where, m is the distance covered by the car during the driver’s reaction time, taken equal to 1 second; - braking distance length

= 5- 10 m-- distance reserve.

On single-lane roads, car drivers must see the road at an even greater distance. It is called the visibility distance of an oncoming car and is calculated by the formula

These calculations do not meet the requirements of SNiP 2.05.02-85*, therefore, when designing the road, we will be guided by the values of the shortest visibility distance from the table. 10 SNiP 2.05.02-85*, which are equal to 250m and 450m for a stopping and oncoming car, respectively.

1.3.5 Determination of the smallest radii of vertical curves

The smallest radius of a convex curve is set from the road visibility condition:

Where d= 1,2 m-- the height of the driver's line of sight above the road surface.

The smallest radius of a concave curve is determined from the condition of limiting the magnitude of the centrifugal force:

Where v-- design speed, km/hour

These calculations do not meet the requirements of SNiP 2.05.02-85*, therefore, when designing the road, we will be guided by the values of the smallest radii of curves in the longitudinal profile from Table. 10 SNiP 2.05.02-85*, which are equal to 15,000m and 5000m for convex and concave curves, respectively.

1.3.6 Determination of roadway widening on curves

The amount of widening is set for the turning radii adopted in the project.

When driving along a curve, the width of the roadway occupied by the car increases (Fig. 4). For geometric reasons, widening one lane

Where L-- the distance between the rear axle and the front bumper of the design vehicle (see P-1 method. instructions); R-- the radius of the curve adopted in the project is 800m (according to Table 10 of SNiP 2.05.02-85*)

The speed-dependent deviations of the vehicle's average trajectory are taken into account using the empirical formula

Total broadening value

For two-lane traffic, the value e P is twice as large according to clause 4.19 of SNiP 2.05.02-85*, and in this case is equal to 0.5 m

1.3.7 Determination of the maximum longitudinal slope of the road

Maximum longitudinal slope i max is set according to the conditions of adhesion of the driving wheels of the car with the coating when starting off and according to the engine power according to formulas derived from the equation of motion of the car and road train.

According to the grip conditions when starting off:

-- for single machines

f- rolling resistance coefficient, accepted for roads of I and II categories 0.01 - 0.02, III and IV categories - 0.015 - 0.025;

g - coefficient of adhesion weight - the ratio of the weight falling on the drive axles to the entire weight of the vehicle (for trucks g = 0.65-0.75);

c - coefficient of adhesion of the wheel to the coating (c = 0.5);

j-- inertia resistance coefficient,

Where A-- acceleration taken in calculations to be 0.3--0.5 m/sec2;

g-- acceleration of gravity;

A coefficient that takes into account the inertia of the rotating parts of the car. automobile road construction cargo

For trucks

1,0+0,06TO = 4,67,

Where TO- gear ratio in the gearbox of the design vehicle = 7.82 (Table P-1 instructions).

When designing a highway, the longitudinal slope should not exceed the smallest one determined by the formulas. We compare the resulting slope with the slope from clause 4.20 of SNiP 2.05.02-85* and enter the data in the table. 3 explanatory notes.

Table 3. - Technical parameters of the highway

Name of parameters	Parameter meaning
		by calculation	Accepted in the project
Basic design speed, km/h		Not defined
Number of traffic lanes, pcs.
Lane width, m		Not defined
Width of the roadway, m		Not defined
Curb width, m		Not defined
Width of the subgrade, m		Not defined
Minimum radii of curves in plan, m: Without superelevation device With superelevation device	Not defined
Visibility distances, m: Road surface Oncoming car
Smallest radii of vertical curves, m Convex Concave
Amount of roadway widening, m	Not standardized
Maximum longitudinal slope, ‰
	Not standardized	asphalt concrete	asphalt concrete

2. Design of the longitudinal profile of the subgrade, drainage

2.1 Design of the longitudinal profile

The longitudinal profile contains the ground line (black profile), the terrain along the road axis, the ground line and the design line (red profile). In general, the longitudinal profile characterizes the geological conditions and the altitude position of the edge of the roadbed. The altitude position of the edge relative to the ground surface line, assessed by working marks, decisively determines the operational, strength and economic indicators of the road, as well as its durability. To obtain optimal results when designing a longitudinal profile, the following must be ensured:

Necessary conditions for vehicle movement and cost-effective operation of vehicles;

Smooth and safe movement of vehicles reaching the design speed;

Stability, reliability and durability of the road;

Uninterrupted functioning of the road;

Cost-effectiveness of road construction.

The necessary operating conditions are ensured by laying the design line with gentle longitudinal slopes.

SNiP 2-05.02-85* recommends using slopes of up to 30%. If it is economically infeasible to implement this recommendation due to the terrain, it is allowed to use longitudinal slopes not exceeding the following maximum values: for a road category of category II - 40%.

The smooth movement of cars is achieved by fitting circular vertical curves into the fractures of the design line, and safety is achieved by assigning such radii of vertical curves that provide the calculated visibility distances (at convex fractures) and limit the centrifugal force to within 5% of the weight of the car (at concave fractures). Vertical curves must be inscribed at fractures, where the algebraic difference of adjacent slopes D i equal to or exceeds on roads I-II categories - 5%. Ascents are considered positive slopes, descents are considered negative. Value D i at turns of associated slopes (two ascents or descents) is defined as the difference between the conjugate slopes, and at turns of opposite slopes (descent and ascent, ascent and descent) - as their sum.

The lowest values of the longitudinal profile parameters, which still ensure smooth and safe movement of vehicles, are given in Table 10 of SNiP. In projects, one should strive to use the largest possible parameter values - this increases the convenience and safety of movement.

2.2 Requirements for cuvette design

On vertical curves, the ditches repeat the actual circular outline of the edge of the roadbed. The cuvettes are designed in the following sequence:

1. Based on the values of the working marks, the places where it is necessary to install ditches are established.

2. the slope of the ditch bottom and the type of reinforcement are set;

3. The line of the bottom of the cuvette is drawn roughly onto the drawing;

4. analytically determine the distance from the nearest picket to the points with zero working marks and to the points of intersection of the bottom of the ditch with the black profile (to do this, it is necessary to consider the geometric figure obtained in the drawing: a triangle or a trapezoid, as well as draw up and solve the corresponding proportion);

5. design marks of the ditch bottom are indicated at all its fractures, on pickets and in places where it comes to the surface;

6. the design slopes of the ditches are recorded;

7. the distances between the fractures are indicated and the points of the beginning and end of the ditch, as well as points with zero marks, are linked to the picketage;

8. the calculations are checked (the elevations of the bottom of the ditch at the points of exit to the surface must correspond to the elevations of the ground; the difference between the design elevations of the edge of the roadbed and the design elevations of the bottom of the ditch must be equal to the accepted depth of the ditch; in addition, the specified distances, slopes and marks);

9. The drawing and the corresponding columns are finalized. Design data related to cuvettes is indicated in red.

2.3 Pavement design

Road pavement is the most critical element, therefore, both strength and durability, as well as the overall cost of the road, depend on its correct design. Non-rigid clothes are those whose layers either do not have resistance to bending or have it to a small extent. These include asphalt concrete, crushed stone (with or without treatment), gravel, cement-soil, soil-gravel and similar clothes. Design and calculation of non-rigid clothing is carried out in accordance with Instructions for the design of non-rigid road pavements VSN 46-83.

When designing non-rigid clothing it is necessary:

Take into account the purpose of the road, its category, composition and intensity of traffic, specific pressure on the pavement and the size of the imprints of car tires, climatic and soil-hydrogeological conditions of construction, the availability of road building materials and their design parameters;

Determine the base material, as well as the need to introduce frost protection and drainage layers into the structure;

Accept the minimum thickness of structural layers according to technological requirements.

The design of flexible clothing consists of:

1. In the choice of materials for structural layers,

2. Assigning the number of these layers,

3. Placing them in the structure,

4. Determination of the thickness of each layer based on strength calculations,

5. Calculations for frost resistance.

From the table 25 SNiP, we choose an improved capital pavement made from an asphalt concrete mixture, laid in a warm state. From the guidelines in Fig. 24, we select an asphalt concrete coating on a crushed stone base.

Pavement design

3. Hydraulic calculation of culverts

3.1 Hydraulic calculation of the pipe

Hydraulic calculation of a pipe includes determining:

Pipe diameter and type of channel strengthening;

The height of the water head and the height of the embankment above the pipe;

Pipe lengths.

Calculation of free-flow pipes is made according to table. P-15, which is made up of the condition that the pipes have slopes no less than critical i cr. In practice, pipes are laid along the slope of the terrain. Since it is less than critical by more than 2 times, it is necessary to increase the backwater N, obtained from the table, by the value:

22.3*(0.006-0)=0.13 m

Where l-- pipe length, m; i 0 -- pipe slope.

Based on a given calculated flow rate to determine the pipe diameter Qр=2.4 m3/s and head type I according to table. P-15 guidelines determine the height of water pressure in front of the pipe N, the speed of water flow in the pipe v and pipe diameter d.

H=1.27m,v=2.47m/s atd=1.5m, bell-shaped pipe head.

Based on the speed of water flow (Table P-16 guidelines), we assign strengthening of the channel type riprap made of cobblestones or broken stones.

To determine the height of the embankment above the pipe N we should be guided by the instructions of SNiP 2.05.03.84* table. 1.

In addition, the height of the embankment must ensure that pavement can be placed above the pipe.

H us = d+ h up to + 0.5=1.5+0.68+0.5=2.68 m.

The approximate length of the pipe can be determined by the expression:

l= B+ 2mH us=15+2*1.5*2.68=23.04m,

Where B- width of the roadbed, m; m- embankment slope steepness coefficient equal to 1.5.

From Table P-17 we find:

Link thickness = 0.14m,

Head length = 2.74 m.

3.2 Calculation of the small bridge opening

The small bridge hole is calculated in the following sequence:

The domestic depth of water flow in the unconstrained bed of the watercourse is determined;

A water flow pattern under the bridge is established;

The size of the bridge hole is determined;

The calculation data is being clarified in relation to typical sizes of small bridges.

3.2.1 Determination of domestic depth

The following data is taken into account: estimated flow rate Q R= 15.0 m 3 /s; i 1 = 0,100; i 2 = 0.060; bed slope i R = 0.007; we ask h b =0.95 m. We determine the open section area, wetted perimeter p and hydraulic radius R:

where is the slope of the channel.

where is the channel coefficient established according to the table; y=0.25 - exponent. Knowing the cross-sectional area and speed in everyday conditions, we find the flow rate:

The resulting flow rate Q is compared with the calculated Q p . If the difference between Q and Q p is less than 10%, we accept the assigned domestic depth and speed as real:

The resulting flow rate differs from the calculated one by 3.6%.

3.2.2 Establishing the water flow pattern under the bridge

To establish the pattern of water flow under the bridge, it is necessary to know the critical flow depth:

where is the flow speed at which the soil or strengthening of the riverbed is not eroded - riprap made of cobblestones;

g=9.8 - acceleration due to gravity.

Since the outflow is free and the spillway is not flooded.

3.2.3 Determining the size of the bridge opening

With free outflow, the bridge opening at the level of the free surface is determined by the formula:

where =0.9 is the flow compression coefficient, depending on the shape of the abutments.

The resulting value is rounded to a standard size.

3.2.4 Clarification of calculated data

Let's determine the actual speed under the bridge:

Let's determine the depth of the flow under the bridge:

Flow depth in front of the structure:

where is the speed coefficient depending on the shape of the supports.

3.2.5 Determination of bridge height and length

The minimum height of the bridge is found by the expression:

where Z=0.75 is the smallest elevation of the bottom of the span above the main water supply;

K=0.96 - structural height of the bridge.

We find the length of the bridge using the formula:

where B = 7.5 - bridge hole; m = 1.5 - embankment slope steepness coefficient; = 3.0 - bridge height; d = 0 - width of the intermediate support; p = 0.1 - distance from the front edge of the abutment to the base of the embankment; q = 0.3 - distance from the rear edge of the abutment to the top of the embankment slope.

Literature

SNiP 2.05.02-85* Highways.

1. Guidelines for performing calculation and graphic work "Road Construction" (for students of the Civil Engineering Institute of Correspondence Studies).

2. V.G. Popov, Construction of highways. A manual for foremen and workers of road organizations, Moscow 2001.

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Starting control of drawing training in 8th grade (for 7th grade)

Last name, class:______________________________________________________________

What is the subject of study in the subject "Drawing"? _____________________________________________________________________________________________________________________________________________________________________________
What does it stand for: E S K D?

________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Name the main lines of the drawing: ________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Write down the known dimensions of the drawing font and the angle of inclination of the letters for italic font type B: __________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Name the types of triangles: ________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Name the types of quadrilaterals: ________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
List the names of geometric bodies: ________________________________________________________________________________ ________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
Types of angles and their designation: ________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________
What is pairing? ______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

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7th grade

On HORIZONTAL

Build FOUR circle radius 30 mm each
into 3, 4, 5 and 6 equal parts

7th grade

GRAPHIC DRAWING WORK

“Dividing a circle into equal parts” Task:

On HORIZONTAL On a sheet of notebook located in the center of the working field, draw an axial (center) line for further construction of circles on it.

Line type: dash-dotted thin line.

Start constructing circles from the middle of the drawn center line.
Build FOUR circle radius 30 mm each using a compass. Please note that the construction of a circle begins with the construction of a perpendicular second center line passing through the center of the circle.
Divide the constructed circles from left to right using previously learned methods.into 3, 4, 5 and 6 equal parts. Save auxiliary for constructing the line.
As a result of the work done, you should get four regular polygons inscribed in circles.

7th grade

GRAPHIC DRAWING WORK

“Dividing a circle into equal parts” Task:

On HORIZONTAL On a sheet of notebook located in the center of the working field, draw an axial (center) line for further construction of circles on it.

Line type: dash-dotted thin line.

Start constructing circles from the middle of the drawn center line.
Build FOUR circle radius 30 mm each using a compass. Please note that the construction of a circle begins with the construction of a perpendicular second center line passing through the center of the circle.
Divide the constructed circles from left to right using previously learned methods.into 3, 4, 5 and 6 equal parts. Save auxiliary for constructing the line.
As a result of the work done, you should get four regular polygons inscribed in circles.

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DRAWING - 7TH GRADE

_________________________________________

Graphic work on the topic “Drawing lines.

Working with drawing tools."

Draw a square with a side of 15 cm in your workbook on a separate sheet of paper.
Divide the square with a diagonal drawn from the lower left corner.
In the resulting division areas, perform the following constructions:

A) in one area, draw horizontal lines at intervals of 1 cm.

B) in another area, draw vertical lines at intervals of 0.5 cm.

_______________________________________________________________

Test work on the topic “Subject “Drawing”. Blueprints"

What does the subject "Drawing" study?
What is a drawing called?
List the drawing tools used in drawing lessons at school.
List the areas of industry where drawings are used.
What can be determined from a product drawing by “reading” it?

When answering questions, you should not rewrite the question itself.

You need to write its serial number and answer

Above the square, draw an isosceles trapezoid with bases of 120 mm (lower) and 70 mm (upper). The height of the trapezoid is 50 mm.

Under the square, place a rectangle with sides 140 mm and 50 mm.

Divide the rectangle into four parts by diagonals.

_________________________________________________________

The work must be done clearly and accurately,

trying to draw all lines of the same thickness.

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Workbook

Introduction to the Subject of Drawing

The history of the emergence of graphic methods of images and drawings

Drawings in Rus' were made by “draftsmen”, a mention of which can be found in the “Pushkar Order” of Ivan IV.

Other images - drawings, were a bird's eye view of the structure.

At the end of the 12th century. In Russia, large-scale images are introduced and dimensions are indicated. In the 18th century, Russian draftsmen and Tsar Peter I himself made drawings using the method of rectangular projections (the founder of the method is the French mathematician and engineer Gaspard Monge). By order of Peter I, the teaching of drawing was introduced in all technical educational institutions.

The entire history of the development of the drawing is inextricably linked with technical progress. Currently, the drawing has become the main document of business communication in science, technology, production, design, and construction.

It is impossible to create and check a machine drawing without knowing the basics of the graphic language. Which you will meet while studying the subject "Drawing"

Types of graphic images

Exercise: label the names of the images.

The concept of GOST standards. Formats. Frame. Drawing lines.

Exercise 1

Graphic work No. 1

"Formats. Frame. Drawing lines"

Examples of work performed

Test tasks for graphic work No. 1

Option #1.

1. What designation according to GOST has a format of size 210x297:

a) A1; b) A2; c) A4?

2. What is the thickness of the dash-dot line if in the drawing the solid main thick line is 0.8 mm:

a) 1mm: b) 0.8 mm: c) 0.3 mm?

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Option #2.

Select and underline the correct answers to the questions.

1. Where in the drawing is the main inscription located:

a) in the lower left corner; b) in the lower right corner; c) in the upper right corner?

2. How much should the axial and center lines extend beyond the contour of the image:

a) 3...5 mm; b) 5…10 mm4 c) 10…15 mm?

Option #3.

Select and underline the correct answers to the questions.

1. What arrangement of A4 format is allowed by GOST:

A) vertical; b) horizontal; c) vertical and horizontal?

2. . What is the thickness of a solid thin line if in the drawing the solid main thick line is 1 mm:

a) 0.3 mm: b) 0.8 mm: c) 0.5 mm?

Option number 4.

Select and underline the correct answers to the questions.

1. At what distance from the edges of the sheet is the drawing frame drawn:

a) left, top, right and bottom – 5 mm each; b) left, top and bottom – 10 mm, right – 25 mm; c) left – 20 mm, top, right and bottom – 5 mm each?

2. What type of line are the axial and center lines made in the drawings:

a) a solid thin line; b) dash-dotted line; c) dashed line?

Option #5.

Select and underline the correct answers to the questions.

1. What are the dimensions of the A4 format according to GOST:

a) 297x210 mm; b) 297x420 mm; c) 594x841 mm?

2. Depending on which line the thickness of the drawing lines is selected:

a) dash-dotted line; b) a solid thin line; c) a solid main thick line?

Fonts (GOST 2304-81)

Font types:

Font sizes:

Practical tasks:

Calculations of drawing font parameters

Test tasks

Option #1.

Select and underline the correct answers to the questions.

What value is taken as the font size:

a) the height of a lowercase letter; b) height of capital letter; c) the height of the spaces between the lines?

Option #2.

Select and underline the correct answers to the questions.

What is the height of the capital letter of rift No. 5:

a) 10 mm; b) 7 mm; c) 5 mm; d) 3.5 mm?

Option #3.

Select and underline the correct answers to the questions.

What is the height of lowercase letters that have protruding elements? c, d, b, r, f:

a) the height of the capital letter; b) the height of a lowercase letter; c) greater than the height of the capital letter?

Option number 4.

Select and underline the correct answers to the questions.

Are uppercase and lowercase letters different in writing? A, E, T, G, I:

a) differ; b) do not differ; c) do they differ in the spelling of individual elements?

Option #5.

Select and underline the correct answers to the questions.

What does the height of the numbers of a drawing font correspond to:

a) the height of a lowercase letter; b) the height of the capital letter; c) half the height of a capital letter?

Graphic work No. 2

"Drawing of a flat part"

Cards - tasks

1 option

Option 2

Option 3

Option 4

Geometric constructions

Dividing a circle into 5 and 10 parts

Dividing a circle into 4 and 8 parts

Dividing a circle into 3, 6 and 12 parts

Dividing a segment into 9 parts

Fixing the material

Practical work:

Based on these types, build a third one. Scale 1:1

Option #1

Option No. 2

Option #3

Option No. 4

Fixing the material

Write your answers in your workbook:

Option #1

Option No. 2

Practical work No. 3

"Modeling from a drawing."

Directions for use

To make a cardboard model, first cut out its blank. Determine the dimensions of the workpiece from the image of the part (Fig. 58). Mark (outline) the cutouts. Cut them along the outlined contour. Remove the cut out parts and bend the model according to the drawing. To prevent the cardboard from straightening after bending, draw lines on the outside of the bend with some sharp object.

The wire for modeling must be soft and of arbitrary length (10 – 20 mm).

Fixing the material

Option No. 1 Option No. 2

Fixing the material

In your workbook, draw a drawing of the part in 3 views. Apply dimensions.

Option No. 3 Option No. 4

Fixing the material

Working with cards

Fixing the material

Using colored pencils, complete the task on the card.

Amount (increase)

Clipping

Reinforcement task

Oval -

Algorithm for constructing an oval

1. Construct an isometric projection of a square - rhombus ABCD

2. Let us denote the points of intersection of the circle and the square 1 2 3 4

3. From the top of the rhombus (D) draw a straight line to point 4 (3). We obtain segment D4, which will be equal to the radius of the arc R.

4. Let's draw an arc that will connect points 3 and 4.

5. At the intersection of segment B2 and AC, we obtain point O1.

When the segment D4 and AC intersect, we obtain point O2.

6. From the resulting centers O1 and O2 we will draw arcs R1 that will connect points 2 and 3, 4 and 1.

Fixing the material

Complete a technical drawing of the part, two views of which are shown in Fig. 62

Graphic work No. 9

Part sketch and technical drawing

1. What is called sketch?

Fixing the material

Exercise tasks

Practical work No. 7

"Reading Blueprints"

Graphic dictation

“Drawing and technical drawing of a part based on a verbal description”

Option #1

Frame is a combination of two parallelepipeds, of which the smaller one is placed with a larger base in the center of the upper base of the other parallelepiped. A through stepped hole runs vertically through the centers of the parallelepipeds.

The total height of the part is 30 mm.

The height of the lower parallelepiped is 10 mm, length 70 mm, width 50 mm.

The second parallelepiped has a length of 50 mm and a width of 40 mm.

The diameter of the bottom step of the hole is 35 mm, height 10 mm; diameter of the second stage is 20 mm.

Note:

Option No. 2

Support is a rectangular parallelepiped, to the left (smallest) face of which is attached a half-cylinder, which has a common lower base with the parallelepiped. In the center of the upper (largest) face of the parallelepiped, along its long side, there is a prismatic groove. At the base of the part there is a through hole of a prismatic shape. Its axis coincides in the top view with the axis of the groove.

The height of the parallelepiped is 30 mm, length 65 mm, width 40 mm.

Half-cylinder height 15 mm, base R 20 mm.

The width of the prismatic groove is 20 mm, the depth is 15 mm.

Hole width 10 mm, length 60 mm. The hole is located at a distance of 15 mm from the right edge of the support.

Note: When drawing dimensions, consider the part as a whole.

Option No. 3

Frame is a combination of a square prism and a truncated cone, which stands with its large base in the center of the upper base of the prism. A through stepped hole runs along the axis of the cone.

The total height of the part is 65 mm.

The height of the prism is 15 mm, the size of the sides of the base is 70x70 mm.

The height of the cone is 50 mm, the lower base is Ǿ 50 mm, the upper base is Ǿ 30 mm.

The diameter of the lower part of the hole is 25 mm, height 40 mm.

The diameter of the upper part of the hole is 15 mm.

Note: When drawing dimensions, consider the part as a whole.

Option No. 4

Sleeve is a combination of two cylinders with a stepped through hole that runs along the axis of the part.

The total height of the part is 60 mm.

The height of the lower cylinder is 15 mm, the base is Ǿ 70 mm.

The base of the second cylinder is Ǿ 45 mm.

Bottom hole Ǿ 50 mm, height 8 mm.

The upper part of the hole is Ǿ 30 mm.

Note: When drawing dimensions, consider the part as a whole.

Option No. 5

Base is a parallelepiped. In the center of the upper (largest) face of the parallelepiped, along its long side, there is a prismatic groove. There are two through cylindrical holes in the groove. The centers of the holes are spaced from the ends of the part at a distance of 25 mm.

The height of the parallelepiped is 30 mm, length 100 mm, width 50 mm.

Groove depth 15 mm, width 30 mm.

Hole diameters are 20 mm.

Note: When drawing dimensions, consider the part as a whole.

Option No. 6

Frame It is a cube, along the vertical axis of which there is a through hole: semi-conical at the top, and then turning into a stepped cylindrical one.

Cube edge 60 mm.

The depth of the semi-conical hole is 35 mm, the upper base is 40 mm, the bottom is 20 mm.

The height of the bottom step of the hole is 20 mm, the base is 50 mm. The diameter of the middle part of the hole is 20 mm.

Note: When drawing dimensions, consider the part as a whole.

Option No. 7

Support is a combination of a parallelepiped and a truncated cone. The cone with its large base is placed in the center of the upper base of the parallelepiped. In the center of the smaller side faces of the parallelepiped there are two prismatic cutouts. A through hole of cylindrical shape Ǿ 15 mm is drilled along the axis of the cone.

The total height of the part is 60 mm.

The height of the parallelepiped is 15 mm, length 90 mm, width 55 mm.

The diameters of the cone bases are 40 mm (lower) and 30 mm (upper).

The length of the prismatic cutout is 20 mm, width 10 mm.

Note: When drawing dimensions, consider the part as a whole.

Option No. 8

Frame is a hollow rectangular parallelepiped. In the center of the upper and lower base of the body there are two conical tides. A through hole of cylindrical shape Ǿ 10 mm passes through the centers of the tides.

The total height of the part is 59 mm.

The height of the parallelepiped is 45 mm, length 90 mm, width 40 mm. The thickness of the walls of the parallelepiped is 10 mm.

The height of the cones is 7 mm, the base is Ǿ 30 mm and Ǿ 20 mm.

Note: When drawing dimensions, consider the part as a whole.

Option No. 9

Support is a combination of two cylinders with one common axis. A through hole runs along the axis: at the top it is prismatic in shape with a square base, and then cylindrical in shape.

The total height of the part is 50 mm.

The height of the lower cylinder is 10 mm, the base is Ǿ 70 mm. The diameter of the base of the second cylinder is 30 mm.

The height of the cylindrical hole is 25 mm, the base is Ǿ 24 mm.

The base side of the prismatic hole is 10 mm.

Note: When drawing dimensions, consider the part as a whole.

Test

Graphic work No. 11

“Drawing and visual representation of the part”

Using the axonometric projection, construct a drawing of the part in the required number of views on a scale of 1:1. Add dimensions.

Graphic work No. 10

“Sketch of a part with design elements”

Draw a drawing of a part from which parts have been removed according to the markings applied. The projection direction for constructing the main view is indicated by an arrow.

Graphic work No. 8

“Drawing of a part with transformation of its shape”

General concept of shape transformation. Relationship between drawing and markings

Graphic work

Making a drawing of an object in three views with transforming its shape (by removing part of the object)

Complete the technical drawing of the part, making, instead of the protrusions marked with arrows, notches of the same shape and size in the same place.

Logical thinking task

Topic “Design of drawings”

Crossword "Projection"

1.The point from which the projecting rays emanate during central projection.

2. What is obtained as a result of modeling.

3. Cube face.

4. The image obtained during projection.

5. In this axonometric projection, the axes are located at an angle of 120° to each other.

6. In Greek, this word means “double dimension.”

7. Side view of a person or object.

8. Curve, isometric projection of a circle.

9. The image on the profile projection plane is a view...

Rebus on the topic “View”

Rebus

Crossword "Axonometry"

Vertically:

1. Translated from French as “front view”.

2. The concept in drawing of what the projection of a point or object is obtained on.

3. The boundary between the halves of a symmetrical part in the drawing.

4. Geometric body.

5. Drawing tool.

6. Translated from Latin, “throw, throw forward.”

7. Geometric body.

8. The science of graphic images.

9. Unit of measurement.

10. Translated from Greek as “double dimension”.

11. Translated from French as “side view”.

12. In the drawing, “she” can be thick, thin, wavy, etc.

Technical Dictionary of Drawing

Term	Definition of a term or concept
Axonometry
Algorithm
Analysis of the geometric shape of an object
Boss
Shoulder

Shaft
Vertex
View
Main view
Additional view
Local view
Screw
Sleeve

Dimensions
screw
Fillet
Geometric body
Horizontal
Ready room
Edge

Dividing a circle
Division of a segment
Diameter
ESKD

Drawing tools
Tracing paper
Pencil
Drawing Layout
Construction
Circuit
Cone
Pattern curves
Circular curves

Pattern
Rulers
Line - leader
Extension line
Transition line
Dimensional line
Solid line
Dashed line
Dashed line
Lyska

Scale
Monge method
Polyhedron
Polygon
Modeling
Main inscription
Applying dimensions
Drawing outline
Break
Oval
Ovoid
Circle
Circle in axonometric projection
Ornament
Axonometric axes
Axis of rotation
Projection axis
Axis of symmetry
Hole

Groove
Keyway
Parallelepiped
Pyramid
Projection plane
Prism
Axonometric projections
Projection
Isometric rectangular projection
Frontal dimetric oblique projection
Projection
Groove

Scan
Size
Overall dimensions
Structural dimensions
Coordinating sizes
Part element dimensions
Gap
Drawing frame
Edge
Technical drawing
Symmetry
Pairing
Standard
Standardization
Arrows
Scheme

Thor
Mating point
Protractor
Squares
Simplifications and conventions

Chamfer
Drawing formats
Frontal
Projection center
Pairing Center
Cylinder
Compass

Drawing
Working drawing
Drawing
Dimensional number
Reading the drawing

Washer
Ball
Slot
Engraving
Font
Hatching Hatching in axonometry
Ellipse
Sketch

Workbook

Practical and graphic work on drawing

The notebook was developed by Anna Aleksandrovna Nesterova, teacher of the highest category of drawing and fine art, teacher of the Municipal Budget Educational Institution “Secondary School No. 1 of Lensk”

Introduction to the Subject of Drawing
Materials, accessories, drawing tools.