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inn [45]
2 years ago
15

Universal Containers has a junction object called "Job Production Facility". With 2 master-detail relationships to the Job and P

roduction Facility custom objects. Both master records have a private sharing model. What statement is true if a user's profile allows access (Create/Read) to the Job, but no access to the Production Facility Object?
Engineering
1 answer:
alexdok [17]2 years ago
3 0

Answer:

C. The user will not be able to see the junction object records or the field values.

Explanation:

For the profile of the user to give access permission such as create and read to the job without granting access permission to the production facility object, the value of the field or records of the junction object will not be seen by the user. This is one of the necessary criteria or principle for the universal container with a junction object.

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A flexural member is fabricated from two flange plates 7-1/2 x ½ and a web plate 17 x 3/8. The yield stress of steel is 50 ksi.
MissTica

Answer:

(a)

Plastic section modulus Z =92.72 in^3

Plastic moment Mp = 4636 kip-in

(b)

Elastic section modulus S = 80.88 in^3

Yield moment My = 4044 kip-in

Explanation:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached files.i hope my explanation will help you in understanding this particular question.

7 0
2 years ago
2. A ¼ in. diameter rod must be machined on a lathe to a smaller diameter for use as a specimen in a tension test. The rod mater
gladu [14]

Answer:

we use

sigma=force /area\\A=\pi r^2\\D=2*r

5 0
2 years ago
Which statement concerning symbols used on plans is true?
Zielflug [23.3K]

Answer:

The symbols are to be noted on the title sheet or other introductory sheet of the plans.

Explanation:

in able to be understood for example a map key is always on a map so the reader can use it efficently without confusion

8 0
2 years ago
A steady tensile load of 5.00kN is applied to a square bar, 12mm on a side and having a length of 1.65m. compute the stress in t
Shtirlitz [24]

Answer:

The stress in the bar is 34.72 MPa.

The design factor (DF) for each case is:

A) DF=0.17

B) DF=0.09

C) DF=0.125

D) DF=0.12

E) DF=0.039

F) DF=1.26

G) DF=5.5

Explanation:

The design factor is the relation between design stress and failure stress. In the case of ductile materials like metals, the failure stress considered is the yield stress. In the case of plastics or ceramics, the failure stress considered is the breaking stress (ultimate stress). If the design factor is less than 1, the structure or bar will endure the applied stress. By the opposite side, when the DF is higher than 1, the structure will collapse or the bar will break.

we will calculate the design stress in this case:

\displaystyle \sigma_{dis}=\frac{T_l}{Sup}=\frac{5.00KN}{(12\cdot10^{-3}m)^2}=34.72MPa

The design factor for metals is:

DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{y}}

The design factor for plastic and ceramics is:

DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{u}}

We now need to know the yield stress or the ultimate stress for each material. We use the AISI and ASTM charts for steels, materials charts for non-ferrous materials and plastics safety charts for the plastic materials.

For these cases:

A) The yield stress of AISI 120 hot-rolled steel (actually is AISI 1020) is 205 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{205MPa}=0.17

B) The yield stress of AISI 8650 OQT 1000 steel is 385 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{385MPa}=0.09

C) The yield stress of ductile iron A536-84 (60-40-18) is 40Kpsi, this is 275.8 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{275.8MPa}=0.125

D) The yield stress of aluminum allot 6061-T6 is 290 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{290MPa}=0.12

E) The yield stress of titanium alloy Ti-6Al-4V annealed (certified by manufacturers) is 880 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{880MPa}=0.039

F) The ultimate stress of rigid PVC plastic (certified by PVC Pipe Association) is 4Kpsi or 27.58 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{27.58 MPa}=1.26

In this case, the bar will break.

F) You have to consider that phenolic plastics are used as matrix in composite materials and seldom are used alone with no reinforcement. In this question is not explained if this material is reinforced or not, therefore I will use the ultimate stress of most pure phenolic plastics, in this case, 6.31 MPa:

DF=\displaystyle\frac{34.72MPa}{6.31 MPa}=5.5

This material will break.

3 0
2 years ago
Determine the design angle ϕ (0∘≤ϕ≤90 ∘) between struts AB and AC so that the 400 lb horizontal force has a component of 600 lb
Svetllana [295]

Answer:

design angle ∅ = 4.9968 ≈ 5⁰

Explanation:

First calculate the force Fac :

Fac = \sqrt{400^2 + 650^2 - 2(400)(650)cos30}

      = \sqrt{160000 + 422500 - 80210}

      = 708.72 Ib

using the sine law to determine the design angle

\frac{sin}{400}  = \frac{sin 30}{Fac}

hence ∅ = sin^{-1} (\frac{sin 30 *400}{708.72} )

              = sin^{-1} 0.0871 =  4.9968 ≈ 5⁰

4 0
2 years ago
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