The load is placed at distance 0.4 L from the end of 
area.
<h3>What is meant by torque?</h3>
The force that can cause an object to rotate along an axis is measured as torque. Similar to how force accelerates an item in linear kinematics, torque accelerates an object in an angular direction. A vector quantity is torque.
Let the beam is of length L
Now the stress on both the end is the same now we can say that torque on the beam due to two forces must be zero
also, we know that stress at both ends are same
Now from two equations we have
solving the above equation we have
so the load is placed at distance 0.4 L from the end of area.
The complete question is:
47. the beam is supported by two rods ab and cd that have cross-sectional areas of 
and 
, respectively. determine the position d of the 6-kn load so that the average normal stress in each rod is the same.
To learn more about torque refer to:
brainly.com/question/20691242
#SPJ4
Answer:
B- extreme fit, close fit, adjustable fit
Explanation:
A human-fit design typically involves the process of manufacturing or producing products (tools) that are easy to use by the end users. Therefore, human-fit designs mainly deals with creating ideas that makes the use of a particular product comfortable and convenient for the end users.
The design for human-fit strategies include; extreme fit, close fit and adjustable fit.
Hence, when the aforementioned strategies are properly integrated into a design process, it helps to ensure the ease of use of products and guarantees comfort for the end users.
Answer: 0.95 inches
Explanation:
A direct load on a column is considered or referred to as an axial compressive load. A direct concentric load is considered axial. If the load is off center it is termed eccentric and is no longer axially applied.
The length= 64 inches
Ends are fixed Le= 64/2 = 32 inches
Factor Of Safety (FOS) = 3. 0
E= 10.6× 10^6 ps
σy= 4000ps
The square cross-section= ia^4/12
PE= π^2EI/Le^2
6500= 3.142^2 × 10^6 × a^4/12×32^2
a^4= 0.81 => a=0.81 inches => a=0.95 inches
Given σy= 4000ps
σallowable= σy/3= 40000/3= 13333. 33psi
Load acting= 6500
Area= a^2= 0.95 ×0.95= 0.9025
σactual=6500/0.9025
σ actual < σallowable
The dimension a= 0.95 inches
