For the beam shown below, i) draw a free body diagram, ii) calculate reactions, iii) calculate the internalforces at 2, 7, 10 and 13 m from the left end of the beam iv) sketch the shear axial force, shear forceand bending moment diagrams .

Figure Q9 (a) shows a simply supported beam with forces acting atpoints B and point E.Figure Q9 (a)a) For the beam to be in equilibrium, calculate the support reaction force atpoint C and at point D.

Calculate and report the reaction loads at the bearings (all directions, using the co-ordinate system defined in Figure 2).

The horizontal beam of length 𝐿𝑏𝑚 ≔ 9.397 𝑚 shown below weighs 𝑀𝑏𝑚 ≔ 230.0 𝑁, and itscenter of gravity is at its center. Given 𝐿𝑐𝑎𝑏 ≔ 11.00 𝑚 and ℎ ≔ 5.00 𝑚, Find (a) the tension inthe cable if the weight of the can is 𝑊𝑐𝑎𝑛 ≔ 350.0 𝑁, and (b) the horizontal and verticalcomponents of the force exerted on the beam at the wall. (3 sig figs)

Beam is cantiliver i.e one end fix other end free . At the free end there is a point load of 300 pound and span of beam is 6 feet.Do the following(a) Max Stresses ( bending ) and Stresses at Corners , top and bottom of X-Sections (b) Max Shear Stress within the sections and shear stress at 1/4 of x-section. Draw shear flow of X-SectionsTry to keep weight of x section same . Preferably depth also. The x-sections areRectangular W or IChannelTZCircle

To solve this problem, we need to use the concepts of shear stress in beams and the location of the shear center. ### Part (i): Shear Stress at Points A, B, and the Neutral AxisThe shear stress (\(\tau\)) in a beam is given by: \[ \tau = \frac{VQ}{It} \] where: - \(V\) is the shear force. - \(Q\) is the first moment of area about the neutral axis. - \(I\) is the second moment of area (moment of inertia). - \(t\) is the thickness of the section at the point where the shear stress is being calculated. Given: - \(V = 11 \, \text{kN} = 11000 \, \text{N}\) - \(I = 5.626 \times 10^6 \, \text{mm}^4\) - Thickness at point A (\(t_A\)) = 4 mm- Thickness at point B (\(t_B\)) = 4 mm#### Shear Stress at Point ATo find \(Q\) at point A, we need to consider the area above point A: \[ Q_A = \text{Area} \times \text{distance from centroid of area to neutral axis} \] The area above point A is a rectangle with height 150 mm and width 4 mm: \[ \text{Area} = 150 \, \text{mm} \times 4 \, \text{mm} = 600 \, \text{mm}^2\] The distance from the centroid of this area to the neutral axis is: \[ \text{Distance} = \frac{150 \, \text{mm}}{2} = 75 \, \text{mm} \] So, \[ Q_A = 600 \, \text{mm}^2 \times 75 \, \text{mm} = 45000 \, \text{mm}^3\] Now, calculate the shear stress at point A: \[ \tau_A = \frac{11000 \, \text{N} \times 45000 \, \text{mm}^3}{5.626 \times 10^6 \, \text{mm}^4 \times 4 \, \text{mm}} = \frac{495000000 \, \text{N} \cdot \text{mm}}{22504 \, \text{mm}^4} = 21998.2 \, \text{N/mm}^2 = 22 \, \text{MPa} \] #### Shear Stress at Point BTo find \(Q\) at point B, we need to consider the area to the left of point B: \[ Q_B = \text{Area} \times \text{distance from centroid of area to neutral axis} \] The area to the left of point B is a rectangle with height 150 mm and width 4 mm: \[ \text{Area} = 150 \, \text{mm} \times 4 \, \text{mm} = 600 \, \text{mm}^2\] The distance from the centroid of this area to the neutral axis is: \[ \text{Distance} = \frac{150 \, \text{mm}}{2} = 75 \, \text{mm} \] So, \[ Q_B = 600 \, \text{mm}^2 \times 75 \, \text{mm} = 45000 \, \text{mm}^3\] Now, calculate the shear stress at point B: \[ \tau_B = \frac{11000 \, \text{N} \times 45000 \, \text{mm}^3}{5.626 \times 10^6 \, \text{mm}^4 \times 4 \, \text{mm}} = \frac{495000000 \, \text{N} \cdot \text{mm}}{22504 \, \text{mm}^4} = 21998.2 \, \text{N/mm}^2 = 22 \, \text{MPa} \] #### Shear Stress at the Neutral AxisAt the neutral axis, the shear stress is maximum. The first moment of area \(Q\) at the neutral axis is the sum of the first moments of the areas above and below the neutral axis. \[ Q_{\text{neutral}} = \frac{1}{2} \times \text{Area} \times \text{distance from centroid