Internal forces 1 problem due determine the internal normal...

Internal Forces 1. <Problem Due> Determine the internal normal force, shear force, and bending-moment acting at point D on the member. Point D is just to left of the 8-kN load. (Fig. 1). Ans.: $N_D = -4$ kN, $V_D = -14.7$ kN, $M_D = -24.2$ kN$\cdot$m 8 kN 15 kN/m 20 kN$\cdot$m 60° A 2 m 1 m D B 2 m 3 m Figure 1

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00:01 Hello everyone in the given question it is given in the following diagram that there is a bar a and it is fixed at and the end b bar it is free now here at point b which is 4 .5 meters from the from a from a 10 kilo nitrogen forces being acted downward and a distribution is there from point a to some point that is 3 meters from the point 8.

00:33 So this is the position.

00:35 And there is another point c located at 1 .5 meters distance from a.

00:41 Now we are supposed to find shearing force, normal force, and moment across c and d points.

00:50 C and d points.

00:54 And this is the pre -body diagram.

00:57 Ax acting in x direction and ay is acting in the y direction.

01:02 The point direction in the similar manner b -y is acting only the upward direction as it is free end there is no x component and now as 10 kilograms 10 kilo newton is acting in the downward direction we can see here and this is the position 6 kilo newton per meter acting in the downward direction triangle distribution now to find first let us try to find this unknown components a x a y and b -y so to find b -y i'll be using moment across a moment about a moment about a would be sigma m across a would be zero so from that part when what can we write moment because of this fourth distribution would be half into arc multiplied by six multiplied by three multiplied by 2 by 3 2 divided by 3 multiplied by 3 plus moment because of 10 kilo root beam 10 multiplied by 10 multiplied by 4 .5 minus b minus b by 6 is equal to 0 minus is because it is in the upward direction opposite to the directions of other two forces.

02:47 So by solving this we can get that by is equal to 10 .5 kilo -meter.

02:59 Now let us use horizontal force, sigma effect, equals to zero as it is an equilibrium.

03:09 Using this we shall find x...

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