When the shear force diagram passes through the zero axis, the corresponding bending moment at the same location will be...?Question 5Select one:a.equal to zerob.below the zero axis line, i.e., positivec.above the zero axis line, i.e., negatived.a local maximum or minimum

Which one of the following statements is incorrect?a.If an external moment is applied at a point in FBD, then there will be a discontinuity in BMD at that point with equal in magnitude and opposite in sense.b.The slope of the shear force diagram at a point will give the loading intensity value at that point.c.The area in the loading intensity (from SFD) till a point will give the negative shear force value at that point.d.The negative slope of the BMD at a point will give the value of Shear force at that point

Determine the normal force, shear force, and bending moment acting at point B of thetwo-member frame

If the moment of force at left side is clockwise and right side is anticlockwise, Bending moment is considered:*NegativeBoth negative and positivePositiveNone of above

Define the terms Shear force and Bending Moment?

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