Answer: hello the diagram related to your question is missing please the third image is the missing part of the question
Fx = 977.76 Ib/ft
Explanation:
<u>Estimate the force that water exerts on the pier </u>
V = 12 ft/s
D( diameter ) = 6 ft
first express the force on the first half of the cylinder as
Fx1 = -
---------------- ( 1 )
where ; Fy = 0
Ps = Po + 1/2 Pv^2 ( 1 - 4 sin^2β ) ------------- ( 2 )
Input equation (2) into equation ( 1 ) (note : assuming Po = 0 )
attached below is the remaining part of the solution
Answer:
Plans; blueprints.
Explanation:
In Engineering, it is a common and standard practice to use drawings and models in the design and development of various tools or systems that are being used for proffering solutions to specific problems in different fields such as engineering, medicine, telecommunications and industries.
Hence, a design engineer make use of drawings such as pictorial drawings, sketches, or technical drawing to communicate ideas about a design to others, to record and retain informations (ideas) so that they're not forgotten and to analyze how different components of a design work together.
Technical drawing is mainly implemented with CAD (computer-aided design) software and is typically used in plans and blueprints that show how to construct an object.
Additionally, technical drawings show in detail how the pieces of something (object) relate to each other, as well as accurately illustrating the actual (true) shape and size of an object in the design and development process.
Answer:
a) 53 MPa, 14.87 degree
b) 60.5 MPa
Average shear = -7.5 MPa
Explanation:
Given
A = 45
B = -60
C = 30
a) stress P1 = (A+B)/2 + Sqrt ({(A-B)/2}^2 + C)
Substituting the given values, we get -
P1 = (45-60)/2 + Sqrt ({(45-(-60))/2}^2 + 30)
P1 = 53 MPa
Likewise P2 = (A+B)/2 - Sqrt ({(A-B)/2}^2 + C)
Substituting the given values, we get -
P1 = (45-60)/2 - Sqrt ({(45-(-60))/2}^2 + 30)
P1 = -68 MPa
Tan 2a = C/{(A-B)/2}
Tan 2a = 30/(45+60)/2
a = 14.87 degree
Principal stress
p1 = (45+60)/2 + (45-60)/2 cos 2a + 30 sin2a = 53 MPa
b) Shear stress in plane
Sqrt ({(45-(-60))/2}^2 + 30) = 60.5 MPa
Average = (45-(-60))/2 = -7.5 MPa
Answer:
<em><u>The 'shoulder' of a road is the land to the edge of the road. On most roads without pavements, the shoulder is a strip of grass or a hedgerow. This is known as a 'soft shoulder'. On a motorway, this strip of land is hardstanding, hence the name 'hard shoulder.'</u></em>
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