Answer:
I'm going to make a list of everything you need to consider for the supervision and design of the bridge.
1. the materials with which you are going to build it.
2. the length of the bridge.
3. The dynamic and static load to which the bridge will be subjected.
4. How corrosive is the environment where it will be built.
5.wind forces
6. The force due to possible earthquakes.
7. If it is going to be built in an environment where snow falls.
8. The bridge is unique,so the shape has a geometry that resists loads?.
9. bridge costs.
10. Personal and necessary machines.
11. how much the river grows
Answer:
All Brake lights are dimmer than normal because high resistance in the brake switch could be the cause according to Technician B.
Explanation:
According to Technician A
When the bulb is faulty then no current will flow through bulb and it will be open circuit.So no light will produce in bulb .
According to Technician B
When a high resistance inserted in series circuit the voltage across each resistance is reduced and this cause the light glow dimly.
Formula of resistance in series circuit
Rt=r1+r2+r3......
Less, if it’s too big: hard to control and maneuverability for shooting wouldn’t be that good. a smaller wheelchair allows for faster movement and control, along with easier shooting and upper body movement
Answer: The net force in every bolt is 44.9 kip
Explanation:
Given that;
External load applied = 245 kip
number of bolts n = 10
External Load shared by each bolt (P_E) = 245/10 = 24.5 kip
spring constant of the bolt Kb = 0.4 Mlb/in
spring constant of members Kc = 1.6 Mlb/in
combined stiffness factor C = Kb / (kb+kc) = 0.4 / ( 0.4 + 1.6) = 0.4 / 2 = 0.2 Mlb/in
Initial pre load Pi = 40 kip
now for Bolts; both pre load Pi and external load P_E are tensile in nature, therefore we add both of them
External Load on each bolt P_Eb = C × PE = 0.2 × 24.5 = 4.9 kip
So Total net Force on each bolt Fb = P_Eb + Pi
Fb = 4.9 kip + 40 kip
Fb = 44.9 kip
Therefore the net force in every bolt is 44.9 kip
Answer:
c = 18.0569 mm
Explanation:
Strategy
We will find required diameter based on angle of twist and based on shearing stress. The larger value will govern.
Given Data
Applied Torque
T = 750 N.m
Length of shaft
L = 1.2 m
Modulus of Rigidity
G = 77.2 GPa
Allowable Stress
г = 90 MPa
Maximum Angle of twist
∅=4°
∅=4*
/180
∅=69.813 *10^-3 rad
Required Diameter based on angle of twist
∅=TL/GJ
∅=TL/G*/2*c^4
∅=2TL/G**c^4
c=![\sqrt[4]{2TL/\pi G }](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B2TL%2F%5Cpi%20G%20%7D)
∅
c=18.0869 *10^-3 rad
Required Diameter based on shearing stress
г = T/J*c
г = [T/(J*/2*c^4)]*c
г =[2T/(J**c^4)]*c
c=17.441*10^-3 rad
Minimum Radius Required
We will use larger of the two values
c= 18.0569 x 10^-3 m
c = 18.0569 mm
