Question

Determine the internal normal force, shear force, and bending moment at point C in the beam.

Answer 1

Draw the free-body diagram of the beam.

Apply the condition of equilibrium.

Consider the algebraic sum of the moments about \(A\) is to be zero. Assume counter clockwise

moments are positive.

\(\sum M_{A}=0\)

\(20(2)\left(\frac{1}{2} \times 2\right)+B_{y}(4)=0\)

Here, \(B_{y}\) is the vertical reaction force at point \(B\).

Rewrite the above equation to find the \(B_{y}\)

\(40+B_{y}(4)=0\)

\(B_{y}=-10 \mathrm{kN}\)

Consider the equilibrium equations in \(x\) -direction.

\(\sum F_{x}=0\)

\(B_{x}=0\)

Here, \(B_{x}\) is the horizontal reaction force at point \(B\).

Draw the free body diagram of the segment to the right of \(C\).

Calculate the internal force at \(C\) as follows:

Apply the condition of equilibrium of forces in \(x\) direction.

\(\sum F_{x}=0\)

\(N_{C}-B_{x}=0\)

\(N_{C}=B_{x}\)

Here, \(N_{C}\) is internal force.

Substitute 0 for \(B_{x}\)

$$ \begin{aligned} N_{C} &=B_{x} \\ &=0 \end{aligned} $$

Therefore, the internal reaction force at \(C\) is 0.

Calculate the shear force at \(C\) as follows:

Apply the condition of equilibrium of forces in \(y\) direction.

\(\sum F_{y}=0\)

\(V_{C}+B_{y}=0\)

Here, \(V_{C}\) is shear force.

Substitute \(-10 \mathrm{kN}\) for \(B_{y}\)

\(V_{C}+(-10)=0\)

\(V_{C}-10=0\)

\(V_{C}=10 \mathrm{kN}\)

Therefore, the shear force at \(C\) is \(10 \mathrm{kN}\).

Calculate moment about \(C\) as follows:

Consider the algebraic sum of the moments about \(C\) is to be zero.

\(\sum M_{c}=0\)

\(M_{C}+B_{y}(2)=0\)

Substitute \(-10 \mathrm{kN}\) for \(B_{y}\)

\(M_{c}+(-10)(2)=0\)

\(M_{C}=20 \mathrm{kN} \cdot \mathrm{m}\)

Therefore, the bending moment at point \(C\) is \({ 2 0 \mathrm { kN } \cdot \mathrm { m } ( \mathrm { CCW } ) }\)

Answer 2

shear force at c is zero(20 N on AB part of beam is balnced by 10N at ecah end A and B.) free body diagram of Ac load 10 n donward,shear force at A 10nipward..this gives shear force at c zer0)

bending monent at c is +10

internal normal foce is zero beacuse at b normal foece is zero as at the end of hanging part there is no normal froce.this leads to normal force at Azero and at C zero