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Question 5
For the beam shown below;
a) Calculate the support reactions.
Ans: Ay=127.5 kN, By $=212.5 \mathrm{kN}$
b) Draw a free body diagram (FBD) of the left section of the beam, with the cut made 4.5 m to the right of support A. Using your FBD and applying the equilibrium equations $\Sigma \mathrm{F}_{\mathrm{y}}=0$ and $\Sigma \mathrm{M}_{\mathrm{z}}=0$, calculate the value of the shear and the moment at this location.
(For additional practice, in your own time, you can consider an FBD to the right of the cut including support B.)
c) Draw a FBD of the left section of the beam, now cutting through the beam between points $A$ and $B$. In your FBD define the distance between point $A$ and the cut using the parameter ' $x$ ', where $0 \mathrm{~m}

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help
Question 5
For the beam shown below;
a) Calculate the support reactions.
Ans: Ay=127.5 kN, By $=212.5 \mathrm{kN}$
b) Draw a free body diagram (FBD) of the left section of the beam, with the cut made 4.5 m to the right of support A. Using your FBD and applying the equilibrium equations $\Sigma \mathrm{F}_{\mathrm{y}}=0$ and $\Sigma \mathrm{M}_{\mathrm{z}}=0$, calculate the value of the shear and the moment at this location.
(For additional practice, in your own time, you can consider an FBD to the right of the cut including support B.)
c) Draw a FBD of the left section of the beam, now cutting through the beam between points $A$ and $B$. In your FBD define the distance between point $A$ and the cut using the parameter ' $x$ ', where $0 \mathrm{~m}<x<8.0 \mathrm{~m}$ and $x=0$ lies at $A$. Using your FBD and applying equilibrium equations, derive the general mathematical expressions for the shear force and bending moment in the beam in terms of the parameter ' $x$ '. Check your answer in part b) by substituting $x=4.5 \mathrm{~m}$ into your general moment equation.
(For additional practice, in your own time, derive equations where the origin lies at the cantilever tip C. You can check the equivalency of your equations by substituting in numerical values.)
d) Draw the Shear force and Bending moment diagrams and sketch the corresponding deflected shape. It is recommended you draw your SFD and BMD aligned with the beam based on the lines shown below.
(You will not be asked to draw a SFD or BMD in the Content test)

Accepted Answer

∻Solution∻ ‖ a ⋮ To calculate the support reactions at points A and B, we first apply the equilibrium equations. The sum of vertical forces must equal zero: $$\Sigma F_y = 0$$. The vertical forces include the reactions at supports A and B, the distributed load, and the concentrated load. The distributed load over the 8 m section is $$40 \, ext{kN/m} imes 8 \, ext{m} = 320 \, ext{kN}$$ acting downward, and the concentrated load at C is 20 kN also acting downward. Thus, we have: $$A_y + B_y - 320 - 20 = 0$$. This simplifies to $$A_y + B_y = 340 \, ext{kN}$$. ‖ ‖ b ⋮ Next, we apply the moment equilibrium about point A to find the reaction at B. The clockwise moment at A is 50 kNm, and the counterclockwise moment due to the distributed load and concentrated load can be calculated as follows: The moment due to the distributed load (acting at its centroid, 4 m from A) is $$320 \, ext{kN} imes 4 \, ext{m} = 1280 \, ext{kNm}$$, and the moment due to the concentrated load at C (3 m from B, or 8 m from A) is $$20 \, ext{kN} imes 8 \, ext{m} = 160 \, ext{kNm}$$. The moment equilibrium equation is: $$-50 + B_y imes 8 - 1280 - 160 = 0$$. Solving for $$B_y$$ gives us $$B_y = 212.5 \, ext{kN}$$. ‖ ‖ c ⋮ Now, substituting $$B_y$$ back into the vertical force equilibrium equation gives us: $$A_y + 212.5 = 340$$, leading to $$A_y = 127.5 \, ext{kN}$$. Thus, the support reactions are confirmed as $$A_y = 127.5 \, ext{kN}$$ and $$B_y = 212.5 \, ext{kN}$$. ‖ ‖ d ⋮ For the free body diagram (FBD) of the left section of the beam, we cut the beam 4.5 m to the right of support A. The shear force $$V$$ at this section can be calculated as: $$V = A_y - ext{(distributed load over 4.5 m)}$$. The distributed load over 4.5 m is $$40 \, ext{kN/m} imes 4.5 \, ext{m} = 180 \, ext{kN}$$. Therefore, $$V = 127.5 - 180 = -52.5 \, ext{kN}$$. The bending moment $$M$$ at this section is given by: $$M = A_y imes 4.5 - ext{(moment due to distributed load)}$$. The moment due to the distributed load is $$180 \, ext{kN} imes 2.25 \, ext{m} = 405 \, ext{kNm}$$. Thus, $$M = 127.5 imes 4.5 - 405 = 573.75 - 405 = 168.75 \, ext{kNm}$$. ‖ ∻Answer∻ ⚹ A_y = 127.5 kN, B_y = 212.5 kN ⚹ ∻Key Concept∻ ⚹ Static Equilibrium: The sum of forces and moments acting on a body at rest must be zero. Equations: $$\sum F_y = 0$$ and $$\sum M = 0$$. ⚹ ∻Explanation∻ ⚹ By applying the equilibrium equations, we can determine the support reactions and analyze the shear and moment at specific points along the beam. This is essential for understanding the internal forces and moments in structural elements. ⚹

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