For the beam illustrated in the figure, find the locations...

For the beam illustrated in the figure, find the locations and magnitudes of the maximum tensile bending stress due to M and the maximum shear stress due to V. Parameters are a = 290 mm, b = 135 mm, c = 19 mm, h = 43 mm, and F = 4450 N. The moment of inertia is $ oxed{ } imes 10^5 ext{ mm}^4 $ The maximum tensile bending stress due to M is $ oxed{ } ext{ MPa} $. The maximum shear stress due to V is $ oxed{ } ext{ MPa} $.

Transcript

00:01 Here, in this question, the free body diagram is given.

00:05 Here, this is the free body diagram.

00:07 This is reaction force at a point o, reaction force at a point b, and there is a force of 4450 acting downwards.

00:17 So here, we can say that here, moment about point o will be equal to 0, that is total moment about 0 will be equal to 0.

00:28 So here we are considering the clockwise direction to be positive.

00:34 So total moment about 0 .0 will be we have 4 .450 newton acting downwards into the distance from o2 that force which is 2090 mm which is in clockwise direction and the next force rb which is in anti -clockwise direction so we are considering it as negative.

00:54 Rb into the distance from o2rb is 200.

00:59 2090 millimeter plus 135 millimeter is equal to total moment is equal to 0.

01:06 Upon solving we will get the value of reaction force at b to be 3036 .47 newton.

01:16 Now we are considering total force in the vertical direction will be equal to 0.

01:24 Here we are considering the upward direction to be positive and downward direction to be positive and downward direction to be negative.

01:31 So here we have r0 acting upward, so considering it as positive.

01:36 And we have 445 newton acting downwards, so negative.

01:40 Plus we have also rb force acting upward.

01:44 So the total force in the vertical direction will be equal to 0.

01:47 Upon solving, we will get here we have already found out the value of rb.

01:52 So solving for r0, we get the value of reaction force at a point to be 1 413 point.

01:59 53 newton.

02:01 Now we are going to calculate the sheer values.

02:06 So here first we have at v0 is equal to the value of r0 itself which is equal to 1 413 .53 newton.

02:19 Then we have v of l al which is equal to again r0 which is equal to 1413 .5.

02:30 Then we have v of a r which is equal to r0 minus 4 4 450 upon solving we will get the value to be minus 30 36 point 47 newton similarly we have vb which is equal to r0 minus 4 4450 itself which is again equal to minus 30 three zero three six 047 newton.

03:02 So we have got 4 shear values.

03:05 So from the obtained shear values, the maximum shear value will be, v max will be here upon avoiding the negative sign, we will get a magnitude of minus 30 ,36 .4 .7 newton as the maximum shear value.

03:24 So here, v max will be equal to 3036 .47 newton.

03:32 Remember that here we have to avoid the negative sign.

03:36 Now we are going to calculate the moment values.

03:42 So here we have moment about different points.

03:46 First we are considering moment about point o.

03:49 Here moment about point o will be equal to 0.

03:52 Now we have moment about point a will be equal to r0 into 0...

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