Question

Coordinates (x, y, z): A(0, +7mm, 0) B(0, 0, -7mm) C(0, -7mm, 0) O (20cm, 0, 12cm) Fixed support (y direction 35 kN (x direction) 22 kN-m (z direction) 16 kN A structure is made of a series of rigidly-connected solid cylindrical segments with a radius of 7mm. Each segment aligns with one of the given coordinate axes. For the loading shown, determine the following: 1. Which internal forces cause normal stresses, and which cause shear stresses? 2. Calculate the magnitudes of the six internal forces and moments at the cross- section where B is located. (Note -- use notation that makes sense to you. / only need magnitudes, so I don't care if your answers are positive or negative here.) 3. Calculate Ox, Oy, and oy at point B based on the loading shown. (Note -- sign matters!) 4. Calculate Txy at point B based on the loading shown. (Note -- sign matters!) 5. Calculate Txz at point B based on the loading shown. (Note -- sign matters!) 6. Calculate Tyz at point B based on the loading shown. (Note -- sign matters!)

Please show all work and steps. Any help will be appreciated and thank you for your time!

Answer 1

Problem Description:

A 35 kW. 16Ant da um С. 120m POINT-O is at is at a depth of 20 cm from À

Solution:

1) Answer for Question No 1:

a) Normal Stress: Since the moment 22 KN-m is acting in the direction of x-axis, its going to create bending tension at all the points 'A', 'B' and 'C'.

b) Forces acting towards 'Y' & 'Z' direction creates shear stress as it is acting perpendicular to the points 'A', 'B' and 'C'. Hence the force 35KN and 16KN creates shear stress at points 'A', 'B' and 'C'.

2) Answer for Question No 2:

Let us consider forces along 'X' direction be represented as 'Fx',

similarly, forces along 'Y' direction be represented as 'Fy', and

forces along 'Z' direction be represented as 'Fz'.

Let moment along 'X' direction be represented as 'Mx',

similarly, moment along 'Y' direction be represented as 'My', and

moment along 'Z' direction be represented as 'Mz'.

Hence, Fx = 0KN; Fy = 35KN; Fz = 16KN and

Mx = +22 + (-35*20) + (-16*20) = -998KN-m

My = (35*0) = 0KN-m, since there is no lever arm along Y direction

Mz = -35*(12+7) = -665KN-m

3) Answer for Question No 3:

Area of Cross-scection of a circle = 3.14 * r²  = 3.14* (7)^{2 =} 153.86 mm²

Normal stress along X direction = σ_x = (Fx / A) = 0/153.86 = 0KN/mm²

Normal stress along Y direction = σ_y = (Fy / A) = 35/153.86 = 0.227KN/mm²

Normal stress along Z direction = σ_x = (Fz / A) = 16/153.86 = 0.104KN/mm²

4) Answer for Question No 4, 5 and 6:

Bending stress along XY plabe Txy = (Mx/I)*Yy

where, I = (pi/64) * diameter^4 = (3.14/64) * (7+7)^4 = 1884.78 mm⁴

Yy = Distance of point B along Y axis = 0mm

Yx = Distance of point B along X axis = 0mm

Yz = Distance of point B along Z axis = 7mm

Hence Txy = (-998/1884.78)*0 = 0KN/mm²

Txz = (Mz/I)*Yz = (-998/1884.78)*7 = -3.70KN/mm²

Tyz = (My/I)*Yz = (-998/1884.78)*0 = 0KN/mm²