The beam shown in the Figure below is a stepped, flat bar having a constant thickness of 1.20 in. It carries a single concentrated load of C of 1500 lb. Compare the stresses in the following locations:

The beam shown in the Figure below is a stepped, flat bar having a constant thickness of 1.20 in.  It carries a single concentrated load of C of 1500 lb.  Compare the stresses in the following locations:
In the vicinity of the load
At the section through the smaller hole to the right of the section C
At the section through the larger hole to the right of section C
Near section B where the bar changes height.

Useful Formulas (You may find the following useful):

Members under axial and shear load
Axial stress    σ=F/A
Shear stress    〖τ=F_s/A〗_s
Hook’s law    E=σ/ϵ;       δ=FL/AE;    δ=∑_i▒(F_i L_i)/(A_i E_i )
Factor of safety
F.S.=σ_fail/σ_allow =τ_fail/τ_allow ;   N=s_u/σ_max
Thermal stresses
Thermal displacement    δ_T=∝ ∆T L
Bending of Beams
Bending stress    σ_x=-(M_z y)/I_Z   ,The maximum Stress:  σ=Mc/I

Deflection of Beams and Shafts    EI (d^2 y)/(dx^2 )=M(x)
Members under Torsion
For circular sections    J_solid=(πD^4)/32=π/2 c^4  ;   J_hollow=(π(〖D_o〗^4-〖D_i〗^4))/32=π/2(〖c_o〗^4-〖c_i〗^4)
Angle of twist    ϕ=TL/GJ      ;      ϕ=∑_i▒(T_i L_i)/(G_i J_i )
Torsional shear stress    τ=Tc/J
Power/Torque/Speed    P=T×ω ;    ω=2πf= 2πn/60
Transverse Shear
Shear Stress
τ= VQ/It      ;         Q=∑▒〖(y ̅^’ 〗 A^’)

Combined Loadings
Equivalent Torque (Torsion & Bending Moment)    T_e=√(M^2+T^2 )
Geometric properties
I_solid=(πd^4)/64

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