Finding support reactions for shear and moment diagrams

[mwjrain]

Draw the shear and moment diagrams for the beam in the figure. Clearly show all values, including the maximum shear and maximum moment and their positions on the diagrams. Assume the supports at A and C are rollers and D is a pin support. note there is a hinge at point B.

I am having trouble finding the support reactions.
My attempt is:
(Assuming positive Y direction is up)
Moment about b = 10(60)+10A-2(10)5=0
A=-50 kN

Forces in y direction = C+D-50-2(10)-5-18=0
C+D=93

Moment about D= C(12)+60(32)-(50)32-(20)(27)-5(16)-18(6)=0
C=34

D=93-34=59

 

[PhanthomJay]

Draw the shear and moment diagrams for the beam in the figure. Clearly show all values, including the maximum shear and maximum moment and their positions on the diagrams. Assume the supports at A and C are rollers and D is a pin support. note there is a hinge at point B.

I am having trouble finding the support reactions.
My attempt is:
(Assuming positive Y direction is up)
Moment about b = 10(60)+10A-2(10)5=0
A=-50 kN

Forces in y direction = C+D-50-2(10)-5-18=0
C+D=93

Moment about D= C(12)+60(32)-(50)32-(20)(27)-5(16)-18(6)=0
C=34

D=93-34=59

Your work is generally very good, however, your error is in the calculation of the moment due to the applied couple at A. You are incorrectly multiplying that couple by a moment arm. The moment of a couple about any point is the couple itself.

 

[mwjrain]

So it would be
Moment about b = 60 +10A-2(10)5=0
A=4 kN

Moment about D = C(12)+60-50(32)-20(27)-5(16)-18(6)=0
C=189kN

Forces in y direction=C+D+4-(2)10-5-18=0
D=-150kN

 

[PhanthomJay]

So it would be
Moment about b = 60 +10A-2(10)5=0
A=4 kN

yes

Moment about D = C(12)+60-50(32) -20(27)-5(16)-18(6)=0
C=189kN

You still used A = -50 instead of A = +4

Forces in y direction=C+D+4-(2)10-5-18=0
D=-150kN

make necessary corrections. It is a good idea to look at the FBD of the beam between B and D to check results.

Welcome to PF!

Note: My lightweight monitor made it easy to turn it upside down.

 

[mwjrain]

Sorry about the image being upside down! I didn't realize that. And thank you for the warm welcome.

So
Moment about D= C12-18(6)-5(16)-2(10)27+4(32)+60=0
C=45

Forces in y =D+4+45-20-18-5=0
D=-6

Note (the new pictures going to be sideways. I can't figure out how to fix it on my iPad.) sorry!

 

[PhanthomJay]

Sorry about the image being upside down! I didn't realize that. And thank you for the warm welcome.

So
Moment about D= C12-18(6)-5(16)-2(10)27+4(32)+60=0
C=45

Forces in y =D+4+45-20-18-5=0
D=-6

Note (the new pictures going to be sideways. I can't figure out how to fix it on my iPad.) sorry!

You should show the negative 6 support reaction at D as a downward force of magnitude 6. Otherwise your reactions, shear, and moment diagrams are literally perfect, correctly curved, etc. , this is excellent work.

 

[mwjrain]

Thank you for all the help! Preparing for finals.