Graphics Statics UNIT. Learning Objectives. Space Diagram, Bow s Notation and Vector Diagram

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Colin Stephens
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1 UNIT 7 Learning Objectives Graphics Statics The graphical statics presents a less tediuos and practical solutions of a problem in statics by graphical method. The accuracy of the graphical solution may not match with that of the analytical one but is generally sufficient for all practical purposes. Space Diagram, Bow s Notation and Vector Diagram The relative positions of the various vectors acting on a system are represented, in a figure called the Space Diagram. It is drawn to a linear scale to show the points of application and the directions of all the vectors. In naming the vectors, a standard practice or notation is used. Bow s notation is generally followed. In Bow s notation, each space on either side of the line of action of each vector is given a name. The vector Diagram represents the magnitudes and directions of all the vectors acting on the system. It is drawn to the scale of vectors. Equilibrant and Resultant of Two Concurrent Forces These are determined the help of the law of triangle of forces. Example 7.1 : Determine the equilibrant and hence the resultant of two forces of 150 N and 250 N acting at a point O if the angle between them is 60 0.

2 440 Construction Technology Fig 7.1 Fig 7.2 Force Magnitude Inclination with OX Equilibrant ca 350 N Resultant ac 320 N Draw the space diagram to show the two given forces making 60 0 with each other. 2. Name the two forces as per Bow s notation, the 150 N by AB and the 250 N by BC as shown in Fig 7.1. The equilibrate will then be represented by CA. 3. Select a convenient point (in Fig 7.2) to present the space A. Draw through a, ab parallel to the direction of force AB (150 N). Mark the point b on ab represents 150 N to the selected scale say 1 mm for 5 N. 4. From b, draw bc parallel to the 250 N force i.e., BC. The length bc is selected such that the magnitude of BC is represented by it to the same scale of 1 mm for 5 N. 5. Join ca to get abc, the triangle of forces for the point O. Fig. 7.2 is known as the Vector Diagram. 6. ca represents the magnitude and direction of the equilibrant of the given forces. Measure its magnitude, Draw a parallel to this direction in the space diagram, tabulate results and measure the angle made by it with OX. 7. ac represents the magnitude and direction of the resultant of the two given forces. Measures its magnitude and inclination with OX and tabulate result.

3 Paper - III Engineering Mechanics 441 Note 1. To every space in the space diagram, there will be a corresponding point in the vector diagram. 2. To every vector in the space diagram, there will be a straight line in the vecot diagram. 3. The equilibrant and the resultant will be collinear, equal and opposite. 4. The vector diagram is a closed figure for a system of forces in equilibrium Equilibrant and Resultant of more than two Concurent Forces These are determined by the law of polygon of forces. This is only an extension of the method of triangle of forces. Example 7.2 Determine the equilibrant and resultant of 4 pulls of 300 N, 600N, 400 N and 200 N making angles of 30 0, 120 0, and respectively with a fixed direction OX. Procedure 1. In the space diagram (Fig ) draw the direction OX and the direction of all the given forces making the stated angles with OX. 2. Name the given forces as AB, BC, CD and DE by using Bow s notation starting with 200 N and going clock wise about O. (See Fig ) Let EA be the equilibrant to he system.

4 442 Construction Technology Results Force Magnitude Inclination with OX Equilibrant ea 320 N Resultant ae 320 N Draw ab (Fig ) parallel to the force AB=200 N and to represent its magnitude to a scale of 1 mn for 10 N. 4. From b draw be parallel to the next force in the order i.e., BC and mark as such that be represents the forces of 300 N to the scale selected. 5. From c draw cd parallel and proportional to force CD=600N and form an draw be parallel and proportional to DE =400 N. 6. abcde is the vector diagram. Hence by law of polygon of forces ea represents the equilibrate of the given system of cocurrent forces. Measure its magnitude to scale (=320 N) and draw a parallel to its direction through O in the space diagram. Measure inclination of this line with OX (=298 0 ) and tabulate results. 7. ae represents magnitude and direction of the resultant. Measure its magnitude and inclination and tabulate results. Note : The equilibrnat of a system of copanar concurrent forces is also a coplanar force and is concurrent with the system. Hence the resultant passes through O, the point of concurrency. Its direction will be parallel to the closing sidce eea of the polygon of forces for the point O. Reactions of Simply Supported Beams To find the reactions at the supports of simply supported beams, proceed as follows: 1. Draw space diagram, vector diagram and funicular polygon for all the forces on the beam excepting the reactions. 2. Produce the first ray a/o and the last ray say d/o of the funicular polygon to cut the lines of action of the reactions at the respective supports at p, q respectively. The line pq will be the closing line ofthe funicular polygon. 3. Draw a ray parallel to the closing line pq through the pole O of the vector diagram to meet the load line at say e. 4. ea will represent the reaction EA and de will represent the reaction DE.

5 Paper - III Engineering Mechanics 443 Graphical Method of Determing Centroid As in the analytical method, the composite area is divided into elementary figures whose area and centroid can be determined easily. The areas of all such elements are considered as parallel forces acting in a convenient direction through the respective centroid. The line of action of the resultants force is determined graphically as per the method. The centoid of the given composite area lies on the line of action of the resultant force. If there is no symmetry about any axis, the centroid is then located at the intersection of the resultant forces in the two assumed directions. Example 4.9 : Determine graphically the centroid of a Tee section 180 mm/120mm / 20 mm. The Tee section is symmetric about the axis of the web. Hence its centroid lies on this axis. The areas of the flange and web will be treated as horizontal forces through their centroids located by intersecting the diagnonals. Example: 4.10 : Determine graphically the centroid of an unequal angle 100 mm x 80 mm x 10 mm. Diagram is displayed in the next page

6 444 Construction Technology

7 Paper - III Engineering Mechanics 445

8 446 Construction Technology Long Answer Type Questions 1. Determine graphically the euilibrant of the forces shown in Fig. 2. Two forces 200 N and 300 N act at an aggle of Find the magnitude and direction of the resultnat by graphical method. The 200 N Force is horizontal. 3. Determine the distance of the centroid of the sections shown in Fig. from the bottom most edge and the central vertical axis.

If three points A (h, 0), P (a, b) and B (0, k) lie on a line, show that: a b 1.

If three points A (h, 0), P (a, b) and B (0, k) lie on a line, show that: a b 1. ASSIGNMENT ON STRAIGHT LINES LEVEL 1 (CBSE/NCERT/STATE BOARDS) 1 Find the angle between the lines joining the points (0, 0), (2, 3) and the points (2, 2), (3, 5). 2 What is the value of y so that the line