For the joint shown in the figure, if F = 28 kN and do = 12.8 mm, calculate: 7.5 mm F F 12 mm 12 mm 7.5 mm F 18 mm 125 mm 18 mm 1. The largest bearing stress in MPa between the pins and the members; a. 146.23 MPa b. 182.29 MPa c. 174.56 MPa d. 199.37 MPa 2. The average shear stress in MPa in the pin and; a. 108.80 MPa b. 116.14 MPa C. 136.23 MPa d. 98.01 MPa 3. The largest average normal stress in MPa in the members. a. 181.23 MPa b. 175.76 MPa c. 191.26 MPa d. 150.32 MPa

For the joint shown in the figure, if F = 28 kN and do = 12.8 mm, calculate: 7.5 mm F F 12 mm 12 mm 7.5 mm F 18 mm 125 mm 18 mm 1. The largest bearing stress in MPa between the pins and the members; a. 146.23 MPa b. 182.29 MPa c. 174.56 MPa d. 199.37 MPa 2. The average shear stress in MPa in the pin and; a. 108.80 MPa b. 116.14 MPa C. 136.23 MPa d. 98.01 MPa 3. The largest average normal stress in MPa in the members. a. 181.23 MPa b. 175.76 MPa c. 191.26 MPa d. 150.32 MPa

Mechanics of Materials (MindTap Course List)

9th Edition

ISBN:9781337093347

Author:Barry J. Goodno, James M. Gere

Publisher:Barry J. Goodno, James M. Gere

Chapter1: Tension, Compression, And Shear

Section: Chapter Questions

Problem 1.8.10P: A steel punch consists of two shafts: upper shaft and lower shaft. Assume that the upper shaft has a...

See similar textbooks

Related questions

Q: Q1) Determine the size of the rivets to be used for the joint of the bracket shown in Figure (1).…

A: Draw a diagram for the given cross-section. Fp and Fs indicate primary shear force and secondary…

Q: P1-22: Consider the bolted joint shown in Figure. The force P is 30 kN and the diameter of the bolt…

A: Given, P = 30 kN d = 10 mm

Q: 1-69.M Figure P1-69 shows a bolt subjected to a tensile load. One failure mode would be if the…

A: Given:Tensile load, P = 22.3 kN or 22300 NDiameter of circular shank, d = 12 mmHeight of the head, h…

Q: In Fig. 1-12, assume that a 20-mm- diameter rivet joins the plates that are each 110 mm wide. The…

A: Write the given data. Diameter of the rivet, d=20 mmWidth of the rivet, w=110 mmAllowable bearing…

Q: Two 2 by 4 cm 1062 steel bars are butt-spliced with two 3 by 4 cm 1016 splice plates using six M14…

A: We have to find the correct formula for finding shear stress

Q: B 20 mm- 50 mm 20 mm 100 mm 50 mm 100 mm M = 8 kN-m

Q: For the punching operation shown in the figure, compute the shear stress (in GPa) in the material if…

A: GivenP=5500kNt=1.5mm=0.0015md=15mm=0.015mL=20mm=0.020m

Q: a) The maximum normal stress occurring in the plate, b) Find the shear stress on the M10 bolt and…

Q: A wood structure is assembled to support the force applied. All pieces of wood have a width of 5 in.…

A: Find outDetermine the normal force in member AC, Determine the normal force in member AB, Determine…

Q: Assume that the axial load P = 40KN applied to the lap joint is distributed equally among the three…

Q: Each of the four links in Figure 3 has an 8x36mm uniform rectangular cross section and each of the…

Q: A one-inch thick steel plate is attached to a one-inch thick vertical column using three SAE Grade…

A: The given bolt system as shown below, Calculate the centroid of the bolt system as follows,…

Q: Two circular rods of 50 mm diameter are connected by a knuckle joint, as shown in Figure 1, by a pin…

Q: In the following armor find the forces in the members CD, KD and JK of section 1-1 indicated in the…

A: A pin support will exert both horizontal and vertical components of reaction as the reaction forces.…

Question

For the joint shown in the figure, if F = 28 kN and dn = 12.8 mm, calculate:
7.5 mm
F
F
12 mm
12 mm
7.5 mm
- de -
F
18 mm -
F
18 mm
125 mm
1. The largest bearing stress in MPa between the pins and the members;
a. 146.23 MPa
b. 182.29 MPa
c. 174.56 MPa
d. 199.37 MPa
2. The average shear stress in MPa in the pin and;
a. 108.80 MPa
b. 116.14 MPa
c. 136.23 MPa
d. 98.01 MPa
3. The largest average normal stress in MPa in the members.
a. 181.23 MPa
b. 175.76 MPa
c. 191.26 MPa
d. 150.32 MPa

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step 3

VIEW

Step by step

Solved in 3 steps with 3 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

Similar questions

A steel punch consists of two shafts: upper shaft and lower shaft. Assume that the upper shaft has a diameter d1= 24 mm and the bottom shaft has a diameter d2= 16 mm. The punch is used to insert a hole in a 4 mm plate, as shown in the figure. If a force P - 70 kN is required to create the hole, what is the average shear stress in the plate and the average compressive stress in the upper and lower shaft of the punch?
Solve the preceding problem if the internal pressure is 3,85 MPa, the diameter is 20 m, the yield stress is 590 MPa, and the factor of safety is 3.0. (a) Determine the required thickness to the nearest millimeter. (b) If the tank wall thickness is 85 mm, what is the maximum permissible internal pressure?
The hollow drill pipe for an oil well (sec figure) is 6,2 in. in outer diameter and 0.75 in. in thickness. Just above the bit, the compressive force in the pipe (due to the weight of the pipe) is 62 kips and the torque (due to drilling) is 185 kip-in. Determine the maximum tensile, compressive, and shear stresses in the drill pipe.
Solve the preceding problem if F =90 mm, F = 42 kN, and t = 40°MPa
I need handwritten otherwise skip all parts with option or I'll dislike
We will use both shearing and bearing analysis to determine the quantities of rivets needed to fasten BC and BE. We can determine two values then choose what will be more safe. ■ For Member BC: ■ Based on Shearing: V T == ■ Av Based on Bearing: P Ab 70MPa = 96000 N .n. (19) ²x n 1/4. π n = 4.84 5 rivets 96000 N (19mm) (6mm) xn 140MPa n = 46.015≈ 7 rivets For member BC, we must use 7 rivets since we must need a minimum of 7 rivets to resist bearing stress produced by the connection. If we only use 5 rivets, then the bearing stress will become 168.42 MPa. It will result to failure since the allowable bearing stress is only 140 Mpa. ■ For member BE: (Practice at Home) Pac EXAMPLE: 5. The figure shows a roof truss and the detail of the riveted connection at joint B. Using allowable stresses of t = 70 MPa and ob= 140 MPa, how many 19-mm diameter rivets are required to fasten member BC to the gusset plate? Member BE? What is the largest average tensile or compressive stress in BC and BE? Say…
Please answer this NEATLY, COMPLETELY, and CORRECTLY for an UPVOTE. Draw appropriate FBDs showing how the internal forces were determined. A square bar (side length = 50 mm) is welded to a fixed supporting slab. A 30-mm diameter pin is inserted in the bar. When the loading shown is applied, determine the following: a. maximum normal stress in the bar (MPa) and bearing stress between the pin and the bar (MPa)b. stress in the weld between the bar and the supporting slab (MPa)c. shear stress in the pin (MPa) and shear stress in the bar (MPa)
C- ODD: Assume that the axial load P = 40KN applied to the lap joint is distributed equally among the three rivets. What is the smallest diameter of the 130 mm rivet if the allowable stresses are 90 MPa for 25 mm bearing between a plate and a rivet, and 40 25 mm MPa for shear in rivets?
The figure below shows two plates connected by two rivets. The plates are made from aluminum 6061-T6 and the rivets are made from aluminum 2014-T4. Evaluate the maximum allowable force on the connection to meet the following design criteria: (a) The shear stress in a rivet cannot exceed 14 of the ultimate strength in shear. (b) The tensile stress in the plates cannot exceed 13 of the yield strength. (c) The bearing stress on either the plates or the rivets cannot exceed the design bearing stress 9 mm 75 mm 38 mm 38 mm 12 mm dia. 9 mm
The lap joint carries an P axial load of P - 50 kN. It is connected by 3-18 mm 150 mm 20 mm 20 mm dia. rivets. 38. What is the bearing stress between a plate and a rivet? 39. What is the shear stress in rivet? 40. What is the maximum average tensile stress in each plate?
For the joint shown in the figure below, calculate (a) the largest bearing stress between the pin and the members; (b) the average shear stress in the pin; and (c) the largest average normal stress in the members. 7.5 mm 25 kN 25 kN 12 mm 12 mm 7.5 mm 12.5 mm 25 kN 25 kN 18 mm !25 mm 18 mm
2. Using the same figure from item number one, find the largest safe value of P if the allowable stresses are 125 MPa for steel, 70 MPa for Aluminum, and 110 MPa for bronze. All applied loads are still axial forces and the materials are still rigidly connected. Bronze Aluminum Steel A=400 mm2 A=600 mm2 A= 300 mm2 2P ЗР 4P -3m -5m- -4 m