Math, asked by baronsaculo, 3 months ago

Problem Solving: Show your solution.
1. The figure shows a rigid bar that is supported by a
pin at A and two rods, one made of steel and the
other of bronze. Neglecting the weight of the bar,
compute the stress in each rod caused by the
50-kN load, the following data:
Steel
1.0 m
Bronze
2 m
А
Steel
600
200
Bronze
300
83
toomk
Area (mm)
E (GPa)
1.0 m-e
40.8 m
50 KN
st
1
R
0.8 m
A,
0.6 m
1.0 m
HINT: Apply this equation in step 1 to get
Pst and Pbr.
MA = 0+1
0.6Pst + 1.6Pbr = 2.4(50x103)
Ах
50 KN
(b) FBD
0.6 m
1.0 m
0.8 m
A-
ANSWERS: Ost = 191.8 MPa; Obr = 106.1 MPa
Ost
dor
oncrete column has
diameter of 300
The​

Answers

Answered by kamalkishorjonko12
0

Answer:

Problem Solving: Show your solution.

1. The figure shows a rigid bar that is supported by a

pin at A and two rods, one made of steel and the

other of bronze. Neglecting the weight of the bar,

compute the stress in each rod caused by the

50-kN load, the following data:

Steel

1.0 m

Bronze

2 m

А

Steel

600

200

Bronze

300

83

toomk

Area (mm)

E (GPa)

1.0 m-e

40.8 m

50 KN

st

1

R

0.8 m

A,

0.6 m

1.0 m

HINT: Apply this equation in step 1 to get

Pst and Pbr.

MA = 0+1

0.6Pst + 1.6Pbr = 2.4(50x103)

Ах

50 KN

(b) FBD

0.6 m

1.0 m

0.8 m

A-

ANSWERS: Ost = 191.8 MPa; Obr = 106.1 MPa

Ost

dor

oncrete column has

diameter of 300

The

Answered by sadiaanam
0

To calculate the stress in each rod caused by the 50-kN load, we need to determine the force in each rod and then divide by the cross-sectional area.

Step 1: Calculate the force in each rod.

We can use the equation of equilibrium to find the force in each rod.

F_steel + F_bronze = Load

where F_steel is the force in the steel rod, F_bronze is the force in the bronze rod, and Load is the applied load of 50 kN.

Since the bar is supported by a pin at A, we know that the moment at A is zero, which means that the clockwise moment created by the steel rod must be equal to the counterclockwise moment created by the bronze rod.

0.6F_steel x 0.8 + 1.6F_bronze x 1.0 = 2.4 x Load

Solving for F_steel and F_bronze:

0.6F_steel = 1.6 x Load - 1.6F_bronze

F_steel = 2.67 x Load - 2.67F_bronze

Substituting the value of Load (50 kN) into the above equation:

F_steel = 134 kN - 2.67F_bronze

And

F_bronze = (134 - F_steel) / 2.67

Step 2: Calculate the stress in each rod.

The stress in each rod can be calculated using the formula:

Stress = Force / Area

Where Force is the force calculated in step 1, and Area is the cross-sectional area of each rod.

For the steel rod:

Stress_steel = F_steel / Area_steel

For the bronze rod:

Stress_bronze = F_bronze / Area_bronze

Substituting the values for Force and Area:

Stress_steel = (134 - 2.67F_bronze) / 600 mm^2 = 191.8 MPa

Stress_bronze = F_bronze / 300 mm^2 = 106.1 MPa

Therefore, the stress in the steel rod is 191.8 MPa, and the stress in the bronze rod is 106.1 MPa.

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