The uniform beam is supported by two rods AB and CD that have crosssectional areas of 10 mm2 and 15 mm2 , respectively. Determine the intensity w of the distributed load so that the average normal stress in each rod does not exceed 300 kPa.

w=2.25

Explanation:

It is necessary to determine the maximum w so that the normal stress in the AB and CD rods does not exceed the permitted normal stress.  

The surface of the cross-section of the stapes was determined:  

A_ab= 10 mm^2

A-cd=  15 mm^2

The maximum load is determined from the condition that the normal stresses is not higher than the permitted normal stress σ_allow.

σ_ab = F_ab/A_abσ_allow

σ_cd =  F_cd/A_cdσ_allow

In the next step we will determine the static size: Picture b).  

We apply the conditions of equilibrium:  

∑F_x=0

∑F_y=0

  ∑M=0

∑M_a=0 ==> -w*6*0.5*6*0.75*F_cd*6 =0

              ==> F_cd = 2*w*k*N

∑F_y=0 ==> F_cd+F_ab - 6*w*0.5 ==>2*w+F_ab -6*w*0.5 =0

              ==> F_ab = w*k*N

Now we determine the load w  

Sector AB:  

σ_ab = F_ab/A_ab σ_allow=300 KPa

         = w/10*10^-6 σ_allow=300 KPa

w_ab = 3*10^-3 kN/m

Sector CD:  

σ_cd = F_cd/A_cd σ_allow=300 KPa

         = 2*w/15*10^-6 σ_allow=300 KPa

w_cd = 2.25*10^-3 kN/m

w=min{w_ab;w_cd} ==> w=min{3*10^-3;2.25*10^-3}

                                ==> w=2.25 * 10^-3 kN/m

The solution is:  

                                w=2.25 N/m

note:

find the attached graph

answer:

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