Concept explainers
The beam AB supports a uniformly distributed load of 480 lb/ft and two concentrated loads P and Q. The normal stress due to bending on the bottom edge of the lower flange is +14.85 ksi at D and +10.65 ksi at E. (a) Draw the shear and bending-moment diagrams for the beam. (b) Determine the maximum normal stress due to bending that occurs in the beam.
Fig. P5.63
(a)
Draw the shear and bending-moment diagrams for the beam.
Explanation of Solution
Given information:
The normal stress due to bending at the point D is
The normal stress due to bending at the point E is
Refer to Appendix C “Properties of Rolled-Steel Sections” in the textbook.
The section modulus (S) for
Determine the bending moment at point D
Here, the normal stress at point D is
Substitute 14.85 ksi for
Determine the bending moment at point E
Here, the normal stress at point E is
Substitute 10.65 ksi for
Show the free-body diagram of the region DE as in Figure 1.
Determine the vertical reaction at point D by taking moment about point E.
Show the free body diagram of the region ACD as in Figure 2.
Determine the magnitude of the load P by taking moment about the point A.
Determine the vertical reaction at point A by resolving the vertical component of forces.
Show the free body diagram of the entire beam as in Figure 3.
Determine the magnitude of the load P by taking moment about the point B.
Determine the vertical reaction at point A by resolving the vertical component of forces.
Shear force:
Show the calculation of shear force as follows;
Show the calculated shear force values as in Table 1.
Location (x) ft | Shear force (V) kips |
A | 24.54 |
C (Left) | 23.82 |
C (Right) | –2.01 |
F (Left) | –4.41 |
F (Right) | –13.12 |
B | –13.84 |
Plot the shear force diagram as in Figure 4.
Bending moment:
Show the calculation of the bending moment as follows;
Show the calculated bending moment values as in Table 2.
Location (x) ft | Bending moment (M) kips-ft |
A | 0 |
C | 36.27 |
F | 19.77 |
B | 0 |
Plot the bending moment diagram as in Figure 5.
Refer to Figure 5;
The maximum absolute bending moment is
(b)
The maximum normal stress due to bending.
Answer to Problem 63P
The maximum normal stress due to bending is
Explanation of Solution
Given information:
Refer to Appendix C “Properties of Rolled-Steel Sections” in the textbook.
The section modulus (S) for
The maximum absolute bending moment is
Determine the maximum normal stress
Substitute
Therefore, the maximum normal stress due to bending is
Want to see more full solutions like this?
Chapter 5 Solutions
EBK MECHANICS OF MATERIALS
- Q2: The plate material of a pressure vessel is AISI 1050 QT 205 °C. The plate is rolled to a diameter of 1.2 m. The two sides of the plate are connected via a riveted joint as shown below. If the rivet material is G10500 with HB=197 and all rivet sizes M31. Find the required rivet size when the pressure vessel is subjected to an internal pressure of 500 MPa. Take safety factor = 2. 1.2m A B' A Chope olm 10.5 0.23 hopearrow_forwardContinuity equation A y x dx D T معادلة الاستمرارية Ly X Q/Prove that ди хе + ♥+ ㅇ? he me ze ོ༞“༠ ?arrow_forwardQ Derive (continuity equation)? I want to derive clear mathematics.arrow_forward
- motor supplies 200 kW at 6 Hz to flange A of the shaft shown in Figure. Gear B transfers 125 W of power to operating machinery in the factory, and the remaining power in the shaft is mansferred by gear D. Shafts (1) and (2) are solid aluminum (G = 28 GPa) shafts that have the same diameter and an allowable shear stress of t= 40 MPa. Shaft (3) is a solid steel (G = 80 GPa) shaft with an allowable shear stress of t = 55 MPa. Determine: a) the minimum permissible diameter for aluminum shafts (1) and (2) b) the minimum permissible diameter for steel shaft (3). c) the rotation angle of gear D with respect to flange A if the shafts have the minimum permissible diameters as determined in (a) and (b).arrow_forwardFirst monthly exam Gas dynamics Third stage Q1/Water at 15° C flow through a 300 mm diameter riveted steel pipe, E-3 mm with a head loss of 6 m in 300 m length. Determine the flow rate in pipe. Use moody chart. Q2/ Assume a car's exhaust system can be approximated as 14 ft long and 0.125 ft-diameter cast-iron pipe ( = 0.00085 ft) with the equivalent of (6) regular 90° flanged elbows (KL = 0.3) and a muffler. The muffler acts as a resistor with a loss coefficient of KL= 8.5. Determine the pressure at the beginning of the exhaust system (pl) if the flowrate is 0.10 cfs, and the exhaust has the same properties as air.(p = 1.74 × 10-3 slug/ft³, u= 4.7 x 10-7 lb.s/ft²) Use moody chart (1) MIDAS Kel=0.3 Q3/Liquid ammonia at -20°C is flowing through a 30 m long section of a 5 mm diameter copper tube(e = 1.5 × 10-6 m) at a rate of 0.15 kg/s. Determine the pressure drop and the head losses. .μ= 2.36 × 10-4 kg/m.s)p = 665.1 kg/m³arrow_forward2/Y Y+1 2Cp Q1/ Show that Cda Az x P1 mactual Cdf Af R/T₁ 2pf(P1-P2-zxgxpf) Q2/ A simple jet carburetor has to supply 5 Kg of air per minute. The air is at a pressure of 1.013 bar and a temperature of 27 °C. Calculate the throat diameter of the choke for air flow velocity of 90 m/sec. Take velocity coefficient to be 0.8. Assume isentropic flow and the flow to be compressible. Quiz/ Determine the air-fuel ratio supplied at 5000 m altitude by a carburetor which is adjusted to give an air-fuel ratio of 14:1 at sea level where air temperature is 27 °C and pressure is 1.013 bar. The temperature of air decreases with altitude as given by the expression The air pressure decreases with altitude as per relation h = 19200 log10 (1.013), where P is in bar. State any assumptions made. t = ts P 0.0065harrow_forward
- 36 2) Use the method of MEMBERS to determine the true magnitude and direction of the forces in members1 and 2 of the frame shown below in Fig 3.2. 300lbs/ft member-1 member-2 30° Fig 3.2. https://brightspace.cuny.edu/d21/le/content/433117/viewContent/29873977/Viewarrow_forwardCan you solve this for me?arrow_forward5670 mm The apartment in the ground floor of three floors building in Fig. in Baghdad city. The details of walls, roof, windows and door are shown. The window is a double glazing and air space thickness is 1.3cm Poorly Fitted-with Storm Sash with wood strip and storm window of 0.6 cm glass thickness. The thickness of door is 2.5 cm. The door is Poor Installation. There are two peoples in each room. The height of room is 280 cm. assume the indoor design conditions are 25°C DBT and 50 RH, and moisture content of 8 gw/kga. The moisture content of outdoor is 10.5 gw/kga. Calculate heat gain for living room : الشقة في الطابق الأرضي من مبنى ثلاثة طوابق في مدينة بغداد يظهر في مخطط الشقة تفاصيل الجدران والسقف والنوافذ والباب. النافذة عبارة عن زجاج مزدوج وسمك الفراغ الهوائي 1.3 سم ضعيف الاحكام مع ساتر حماية مع إطار خشبي والنافذة بسماكة زجاج 0.6 سم سماكة الباب 2.5 سم. الباب هو تركيب ضعيف هناك شخصان في كل غرفة. ارتفاع الغرفة 280 سم. افترض أن ظروف التصميم الداخلي هي DBT25 و R50 ، ومحتوى الرطوبة 8…arrow_forward
- How do i solve this problem?arrow_forwardQ4/ A compressor is driven motor by mean of a flat belt of thickness 10 mm and a width of 250 mm. The motor pulley is 300 mm diameter and run at 900 rpm and the compressor pulley is 1500 mm diameter. The shaft center distance is 1.5 m. The angle of contact of the smaller pulley is 220° and on the larger pulley is 270°. The coefficient of friction between the belt and the small pulley is 0.3, and between the belt and the large pulley is 0.25. The maximum allowable belt stress is 2 MPa and the belt density is 970 kg/m³. (a) What is the power capacity of the drive and (b) If the small pulley replaced by V-grooved pulley of diameter 300 mm, grooved angle of 34° and the coefficient of friction between belt and grooved pulley is 0.35. What will be the power capacity in this case, assuming that the diameter of the large pulley remain the same of 1500 mm.arrow_forwardYou are tasked with designing a power drive system to transmit power between a motor and a conveyor belt in a manufacturing facility as illustrated in figure. The design must ensure efficient power transmission, reliability, and safety. Given the following specifications and constraints, design drive system for this application: Specifications: Motor Power: The electric motor provides 10 kW of power at 1,500 RPM. Output Speed: The output shaft should rotate at 150 rpm. Design Decisions: Transmission ratio: Determine the necessary drive ratio for the system. Shaft Diameter: Design the shafts for both the motor and the conveyor end. Material Selection: Choose appropriate materials for the gears, shafts. Bearings: Select suitable rolling element bearings. Constraints: Space Limitation: The available space for the gear drive system is limited to a 1-meter-long section. Attribute 4 of CEP Depth of knowledge required Fundamentals-based, first principles analytical approach…arrow_forward
- Elements Of ElectromagneticsMechanical EngineeringISBN:9780190698614Author:Sadiku, Matthew N. O.Publisher:Oxford University PressMechanics of Materials (10th Edition)Mechanical EngineeringISBN:9780134319650Author:Russell C. HibbelerPublisher:PEARSONThermodynamics: An Engineering ApproachMechanical EngineeringISBN:9781259822674Author:Yunus A. Cengel Dr., Michael A. BolesPublisher:McGraw-Hill Education
- Control Systems EngineeringMechanical EngineeringISBN:9781118170519Author:Norman S. NisePublisher:WILEYMechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage LearningEngineering Mechanics: StaticsMechanical EngineeringISBN:9781118807330Author:James L. Meriam, L. G. Kraige, J. N. BoltonPublisher:WILEY