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60 kN 100 kN 130 kN Q.1 Determine the magnitude, sense, and direction of the resultant of the concurrent force system below 5 12 3 4 60 0.

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Presentation on theme: "60 kN 100 kN 130 kN Q.1 Determine the magnitude, sense, and direction of the resultant of the concurrent force system below 5 12 3 4 60 0."— Presentation transcript:

1 60 kN 100 kN 130 kN Q.1 Determine the magnitude, sense, and direction of the resultant of the concurrent force system below 5 12 3 4 60 0

2 A 25 kN 19 kN 10 kN 2 m 3 m Q2. Calculate the net moment about point A due to the forces shown below

3 1. Types of Supports 2. Types of Beams 3. Calculate Reactions 4. Draw/sketch Shear Force Diagram(SFD) 5. Draw/sketch Bending Moment Diagram(BMD)

4  Beam is the most common structural member and carries transverse load  They are sometimes referred by other names, indicative of some specialized function such as girder (penopang),stringer, purlins or joists.  The concepts of shear and bending in statically determinate beam only.

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16 1) Load  Deflection is directly proportional to the load 2) Span  Deflection is proportional to the cube of the span

17 3) Size and shape of beam  The greater the size of beam, the less the deflection.  Deflection is inversely proportional to the moment of inertia 4) Stiffness of material  The stiffer the material, the greater its resistance to bending, the less will be the deflection.  Table 11.1 Values of coefficient c for deflection formula δ=cWl 3 /EI

18 A 406 x 178 UB 54 is simply supported at the ends of a span of 5 m carries a uniformly distributed load of 60 kN/m. Calculate the maximum deflection given E = 205 000 N/mm 2

19 From Table 11.1 δ = 5Wl 3 /384 EIwhere W = 60 x 5 = 300 kN = 3 x 10 5 N l = 5000 mm E = 205 000 N/mm 2 I = 164.6 x 10 6 mm 4 δ = (5/384) x {(300 000 x 125 x 10 9 )/ (205 000 x 164.6 x 10 6 )} = 14 mm


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