Given Information:

You have a beam with different types of loading as shown in figure below.

Figure 1 It is given that, L=14ft, q0=12lb/ft, P=50lb, and M0=300lb-ft.

Calculation:

To find external support reaction forces, draw free body diagram as shown in figure below.

Figure 2: Free body diagram for external reactions Take equilibrium in x−direction (horizontal direction),

∑Fx=0⇒Bx=35P=35×50 lb=30 lb

Take equilibrium of moments about point A,

∑MA=0⇒M0+ByL=12q0L2L3+45P(L+L2)⇒By=1L(−M0+12q0L2L3+45P(L+L2))⇒By=114(−300+12×12×14×2×143+45×50×(14+142))⇒By=94.57 lb

Take equilibrium in y−direction (vertical direction),

∑Fy=0⇒Ay+By=45P+12q0L⇒Ay=−By+45P+12q0L⇒Ay=−94.57+45×50+12×12×14⇒Ay=29.4 lb

To find internal forces at mid span of AB, draw the following free body diagram.

Figure 3 Free body diagram for internal forces Now take left hand side free body diagram.

Take equilibrium in x−direction (horizontal direction),

∑Fx=0⇒N=0

Take equilibrium in y−direction (vertical direction),

∑Fy=0⇒Ay=V+12q02L2⇒V=Ay−12q02L2⇒V=29.4−12×122×142⇒V=8.42 lb

Take equilibrium of moments about point A,

∑MA=0⇒M0+M=12q0L42L3×2+VL2⇒M=−M0+12q0L42L3×2+VL2⇒M=−300+12×12×144×2×143×2+8.42×142⇒M=−143.06 lb-ft

Conclusion:

Thus the external reaction forces are Bx=30lb, By=94.57lb, Ay=29.4lb and internal reaction forces are axial force N=0, shear force V=8.42lb and moment M=−143.06lb-ft.