Fig 6 - uploaded by Muhammed Abbas
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The diagrams of example 1. Step 1: Computing the slope of the loading diagram m by dividing the maximum intensity of the load by the loading length to get / m 20 9 = . Step 2: Developing the shear parabola by applying Eq.(2.9) to obtain
Source publication
This study presents an improvement of the graphical method for plotting the shear and moment diagrams for the structural members under linearly varying loads (triangular and trapezoidal loads). Based on the parabolic nature of the shear function, when the loading varies linearly, and on the relations among load, shear, and moment, a mathematical eq...
Contexts in source publication
Context 1
... 1: Sectioning the beam at a distance x where the internal shear force S equals zero as shown in Fig.6b (the distance x represents the location of the zero-shear point measured from the point A). ...
Context 2
... x 5 2m = by using the section technique. The equilibrium equation for the moments in the free body diagram shown in Fig.6b was applied to get the maximum moment of . . ...
Context 3
... parabolic function in step 2 intersects the X-axis when V 30kN = as shown in Fig.6c This example represents the case of a beam subjected to a common and simple triangular loading. ...
Context 4
... equation, ( ) / 2 V 10 9 x = , which it developed simply from knowing the slope of the loading diagram. Regarding computing the parabolic areas, the proposed geometric method is characterized by a systematic process depending on simple equations dealing with a simple geometry. It is necessary to note that the distances b and h of the parabola in Fig.6c are measured according to the vertex (point N), which has a zero-slope tangent. Even if the number of the steps for the two solutions appears equal, actually Solution I took more computation efforts and time since both the sectioning and the equilibrium equations were utilized twice for determining the zero-shear point and the maximum ...
Context 5
... 1: Sectioning the beam at a distance x where the internal shear force S equals zero as shown in Fig.6b (the distance x represents the location of the zero-shear point measured from the point A). ...
Context 6
... x 5 2m = by using the section technique. The equilibrium equation for the moments in the free body diagram shown in Fig.6b was applied to get the maximum moment of . . ...
Context 7
... parabolic function in step 2 intersects the X-axis when V 30kN = as shown in Fig.6c This example represents the case of a beam subjected to a common and simple triangular loading. ...
Context 8
... equation, ( ) / 2 V 10 9 x = , which it developed simply from knowing the slope of the loading diagram. Regarding computing the parabolic areas, the proposed geometric method is characterized by a systematic process depending on simple equations dealing with a simple geometry. It is necessary to note that the distances b and h of the parabola in Fig.6c are measured according to the vertex (point N), which has a zero-slope tangent. Even if the number of the steps for the two solutions appears equal, actually Solution I took more computation efforts and time since both the sectioning and the equilibrium equations were utilized twice for determining the zero-shear point and the maximum ...
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