# Absolute max shear and moment

1. Jan 23, 2017

### fonseh

1. The problem statement, all variables and given/known data
For the maximum moment occur under the 1.5k load , why when we consider the left part of the beam, we just need to consider the forces 2k and 1.5k only ?

2. Relevant equations

3. The attempt at a solution
How about FR = 4.5k and also the 1k ?
Cant i just consider all of them ? That's means i make an iamginary cut after the 1k load ...

#### Attached Files:

File size:
41.4 KB
Views:
30
• ###### 548.jpg
File size:
24.1 KB
Views:
28
2. Jan 23, 2017

### PhanthomJay

if you look carefully, those are not the forces used when making an imaginary cut
of the left section at a plane just left of the 1.5k load, which is the assumed location of the absolute maximum moment.
the 4.5k load is the resultant sum of the applied loads, so you must not double count these loads.
but a cut after the 1k load is not the point of absolute maximum moment.

The best way in my opinion to determine the magnitude and location of the absolute maximum moment and shear is to draw a shear and moment diagram.

3. Jan 23, 2017

### fonseh

do you mean we just need to make an imaginary cut up to 1.5 k because the author assume the max moment will occur under the 1.5k load ?

4. Jan 23, 2017

### PhanthomJay

yes

5. Jan 23, 2017

### fonseh

Actually how did the author know that the absolute max moment occur under the 1.5k load ? why not the 1k load or 2k load ? Why did the author didnt show the calculation for the 1k load ?

6. Jan 24, 2017

### PhanthomJay

Author did not know. The point of max moment occurs under one of the concentrated loads near the resultant location. The author had to check both the 1.5 and 2k load locations to find out which controlled for max moment. The moment at the 1k load point did not have to be checked because it was not near the resultant.