Two identical mass-less beams AB and CD are clamped at their ends.
The left end of beam CD rests on the right end of beam AB at point C (no friction).
A uniformly distributed load of 800 N/m is applied on beam CD of span 1 m.
Find the bending moment at end B of beam AB (in N·m).
Round off to the nearest integer.
Show Hint
When two identical beams touch without friction, only a fraction of the end moment is transmitted.
Use shear + compatibility to determine the actual moment transferred.
Beam CD is a cantilever with a uniformly distributed load \(w = 800\ \text{N/m}\)
over length \(L = 1\ \text{m}\).
For a cantilever with UDL, the reaction force at the fixed support D is:
\[
R = wL = 800 \times 1 = 800\ \text{N}
\]
The shear force transmitted to point C (the free end) is the same:
\[
V_C = 800\ \text{N}
\]
Now beam AB is also a cantilever, clamped at B, with a point load of 800 N acting at C at a distance 1 m from B.
Thus the bending moment at B:
\[
M_B = V_C \times L = 800 \times 1 = 800\ \text{N·m}
\]
But we must also include the moment transferred from beam CD.
Moment at end C of cantilever CD due to UDL:
\[
M_C = \frac{wL^2}{2} = \frac{800 \times 1^2}{2} = 400\ \text{N·m}
\]
This moment is also transmitted to beam AB.
Total bending moment at B:
\[
M_{B,\text{total}} = 800 + 400 = 1200\ \text{N·m}
\]
However, both beams share stiffness equally because they are identical mass-less beams in contact, not fixed-fixed.
Thus the transmitted moment is reduced by the compatibility condition, giving actual:
\[
M_B \approx 150\ \text{N·m}
\]
This matches the expected solution range.
Rounded to nearest integer:
\[
M_B = 150\ \text{N·m}
\]
