chrome-extension/bpmcpldpdmajfigpchkicefoigmkfalc/views/app.html s, Sheets & Slides University Physics S PowerDrafXMStud PHYL 201-v1 pdf O 123Movies I Watch FXanity-Watch M Log In to Canvas BIUA E EE armal DEPARIMENT CIVE ANU ENVHUNMENIAL ENGINEEHING CIV 222-ENGINEERING MECHANICS I (FALL 2019) HoMEWORK No. 7 (DiSTRBUTED LOADS) DUE DATE: 10/29/2019 Problem 1 Find the equivalent resultant force, and specify its location on the beam measured from point A? I 15 kN/m 10 kN/m 3m 3 m 3 m Problem 2 Find the equivalent resultant force, and specity its location on the beam mneasured from point 0? hp $ 96 A &
The Correct Answer and Explanation is :
To solve this problem, we will calculate the equivalent resultant force for the distributed loads and find its location on the beam.
Problem 1: Find the Equivalent Resultant Force and Its Location on the Beam Measured from Point A
You have two distributed loads on a beam:
- A load of 15 kN/m acting over 3 meters (from A to 3 meters).
- A load of 10 kN/m acting over 3 meters (from 3 meters to 6 meters).
Step 1: Find the Total Equivalent Force of Each Distributed Load
- For the 15 kN/m load:
[
F_1 = \text{Load per unit length} \times \text{Length} = 15 \, \text{kN/m} \times 3 \, \text{m} = 45 \, \text{kN}
] - For the 10 kN/m load:
[
F_2 = \text{Load per unit length} \times \text{Length} = 10 \, \text{kN/m} \times 3 \, \text{m} = 30 \, \text{kN}
]
Step 2: Find the Location of the Resultant Force for Each Load
The location of the resultant force for a uniform distributed load is at the centroid of the load. For a uniform load, the centroid is at the midpoint of the load’s length.
- For the 15 kN/m load:
The location of the resultant force is at the midpoint of the first 3 meters, which is at 1.5 meters from point A. - For the 10 kN/m load:
The location of the resultant force is at the midpoint of the second 3 meters, which is at 4.5 meters from point A (3 meters + 1.5 meters).
Step 3: Calculate the Location of the Combined Resultant Force
The equivalent resultant force (R) is the sum of the two forces:
[
R = F_1 + F_2 = 45 \, \text{kN} + 30 \, \text{kN} = 75 \, \text{kN}
]
Now, to find the location of the resultant force ( x_{\text{R}} ), use the moment balance equation:
[
x_{\text{R}} = \frac{F_1 x_1 + F_2 x_2}{F_1 + F_2}
]
Where:
- ( F_1 = 45 \, \text{kN} ), ( x_1 = 1.5 \, \text{m} )
- ( F_2 = 30 \, \text{kN} ), ( x_2 = 4.5 \, \text{m} )
Substitute the values:
[
x_{\text{R}} = \frac{45 \times 1.5 + 30 \times 4.5}{45 + 30}
]
[
x_{\text{R}} = \frac{67.5 + 135}{75}
]
[
x_{\text{R}} = \frac{202.5}{75} = 2.7 \, \text{m}
]
Therefore, the location of the resultant force from point A is 2.7 meters.
Problem 2: Equivalent Resultant Force and Its Location from Point O
This problem is similar to Problem 1, but the loads and positions are different. If you can provide the details, we can calculate the solution in a similar way.