Question 10.2.3: A rider is pushing on each arm of the bicycle handlebars in ...

Goal Find the internal forces and bending moments in the x, y, and z directions at the intersection of the handlebars and the stem.
Given Information about the dimensions and loading of the handlebars.
Assume The weight of the handlebars is negligible (no other information is given to estimate weight).
Draw Each arm of the handlebars is a cantilever beam supported by the stem, and loaded by force components in the y and z directions. We cut the handlebars at the stem, isolate the left arm, and draw the internal loads (Figure 2). To clarify the loading for the analysis we break the 50-lb force into components in the y and z directions (F_{y} = −50 sin 30° = −25.0 lb and F_{z} = −50 cos 30° = − 43.3 lb). because the system and its loading are symmetric, we know that both arms of the handlebars have the same internal loads.
Formulate Equations and Solve We apply the six nonplanar equilibrium equations to the free-body diagram in Figure 2 to solve for the internal loads. Analyzing force equilibrium gives the following three
results based on (5.3A), (5.3b), and (5.3C).

\sum{M_{z }}=0                            (5.5C)
\sum{F_{x}}=0      \Rightarrow N_{x}=0  
\sum{F_{y}}= 0 = −25.0  lb – V_{y} \Rightarrow  V_{y} =-25.0  lb
\sum{F_{z}}= – 43.3  lb + V_{z}= 0 \Rightarrow  V_{z}= 43.3 lb

Because the handlebars are bent out of the xy plane the force F_{y} is offset from that plane by a distance d (Figure 3), creating a moment about the x axis. Similarly F_{z} is offset from the xz plane by 1.5 in., creating a moment about the x axis. To determine the distance d, we use similar triangles. From Figure 1b we know that the end of the arm is offset 2 in. from the x axis. In addition, F is applied 2 in. from the end of the arm; therefore analysis of similar triangles gives

\frac{d}{6  in.} =\frac{2  in.}{8  in.} \Rightarrow d=\frac{(6  in.)2  in.}{8  in.} =15  in.

We now determine internal moments about the x, y, and z axes by analyzing moment equilibrium with the moment center at B in Figure 2. When writing our equations, we define positive moments as those that create a positive moment about an axis (the right hand rule is very helpful here for keeping signs straight).

\sum{M_{x @ B} } =0=F_{y}d – F_{z} (1.5  in.) +M_{bx}
(25.0  lb)(1.5  in.) – (43.3  lb)(1.5  in.)  + M_{bx} =0  \Rightarrow M_{bx} = 27.5  lb.in.
\sum{M_{y @ B} } =0= – F_{z}(10  in.) – M_{by}
-(43.3  lb)(10  in.) – M_{by}=0 \Rightarrow M_{by}= – 433  lb.in.
\sum{M_{z @ B} } =0=F_{y}(10  in.) + M_{bz}
(25.0  lb)(10  in.) +M_{bz} =0 \Rightarrow M_{bz}= – 250  lb.in.

M_{bx} (often called a torque) causes twisting about the x axis. The moment M_{by} causes the handlebars to bend toward the front of the bicycle and M_{bz} causes them to bend downward. The negative signs in the results indicate that the internal loads are in the opposite direction from those shown in Figure 2.
Check To check our result, we can apply the calculated internal loads to the free-body diagram in Figure 2 and perform an equilibrium analysis.
Question: Can you think of how using the cross product might have made the solution easier?