Question

A beam of length L is loaded in the xy-plane by a uniformly distributed load, and by a concentrated tip load parallel to the z-axis, as shown in the figure. The resulting bending moment distributions about the y and the z axes are denoted by \(M_y\) and \(M_z\), respectively. Which one of the options given depicts qualitatively CORRECT variations of \(M_y\) and \(M_z\) along the length of the beam? 

• A. A
• B. B
• C. C
• D. D

Hint:

Remember the "Load-Shear-Moment" relationship graphically: - Point Load \(\rightarrow\) Constant Shear \(\rightarrow\) Linear Moment - Uniform Load \(\rightarrow\) Linear Shear \(\rightarrow\) Parabolic Moment For a cantilever, moments are always zero at the free end (unless an external moment is applied there).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The problem involves determining the bending moment diagrams for a cantilever beam subjected to loads in two different planes.
- \(M_z\) is the bending moment about the z-axis, caused by loads in the xy-plane (the uniformly distributed load \(q\)).
- \(M_y\) is the bending moment about the y-axis, caused by loads in the xz-plane (the concentrated tip load \(P\)).
We need to determine the shape of each bending moment diagram based on the loading.
Step 2: Key Formula or Approach:
The relationships between load (\(w\)), shear force (\(V\)), and bending moment (\(M\)) are key: \[ \frac{dV}{dx} = -w(x) \text{and} \frac{dM}{dx} = V(x) \] This implies:
- For a concentrated load, the shear is constant, and the moment is linear.
- For a uniformly distributed load, the shear is linear, and the moment is parabolic (quadratic).
The moment is zero at the free end (unless a moment is applied there) and maximum at the fixed support.
Step 3: Detailed Explanation:
Analysis for \(M_z\): - Loading: A uniformly distributed load \(q\) acts in the negative y-direction. This load causes bending about the z-axis.
- Shear Force (\(V_y\)): The shear force due to the UDL will vary linearly from 0 at the free end (\(x=L\)) to a maximum at the support (\(x=0\)). The slope of the shear diagram is constant and negative. \(V_y(x) = q(L-x)\).
- Bending Moment (\(M_z\)): Since the shear is linear, the moment must be parabolic (quadratic). The moment is zero at the free end (\(x=L\)). The slope of the moment diagram (\(\frac{dM_z}{dx} = V_y\)) is zero at the free end and maximum at the support. The moment diagram will be a parabola opening downwards (since the load is downwards). Specifically, \(M_z(x) = -\frac{q(L-x)^2}{2}\). This is a curve that is flat at the tip and steep at the support.
- Looking at the options for \(M_z\), (B) and (C) show a parabolic curve that is zero at the free end and non-zero at the support. (A) and (D) show linear diagrams, which is incorrect. The shape in (B) correctly shows the slope being zero at \(x=L\) and increasing towards \(x=0\).
Analysis for \(M_y\):
- Loading: A concentrated load \(P\) acts at the tip (\(x=L\)) in the positive z-direction. This load causes bending about the y-axis.
- Shear Force (\(V_z\)): The shear force due to the point load is constant along the beam, equal to \(-P\).
- Bending Moment (\(M_y\)): Since the shear is constant, the moment must vary linearly. The moment is zero at the free end (\(x=L\)) where the load is applied. It will increase linearly to a maximum at the support (\(x=0\)). Specifically, \(M_y(x) = P(L-x)\). This is a straight line.
- Looking at the options for \(M_y\), (A) and (B) show a linear (triangular) diagram, which is correct. (C) and (D) are parabolic, which is incorrect.
Conclusion:
- \(M_y\) must be linear (triangular).
- \(M_z\) must be parabolic.
Combining these, only option (B) has a linear \(M_y\) diagram and a parabolic \(M_z\) diagram with the correct shapes (zero at the free end, non-zero at the support, correct curvature for \(M_z\)).
Step 4: Final Answer:
Option (B) correctly depicts the variations of \(M_y\) and \(M_z\).
Step 5: Why This is Correct:
The shapes of the moment diagrams are directly determined by integrating the load distributions. A point load leads to a triangular moment diagram, and a uniformly distributed load leads to a parabolic moment diagram for a cantilever beam. Option (B) is the only one that correctly represents both of these facts.