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If M=10 kip⋅ft, determine the resultant force this moment exerts on the top and bottom \\ - 6-51
6-51. If M=10 kip⋅ft, determine the resultant force this moment exerts on the top and bottom boards of the beam.
All problems from Mechanics of Materials Eleventh Edition by Russell C. Hibbeler.
zhlédnutí: 50

Video

If M=10 kip⋅ft, determine the maximum bending stress in the beam - 6-505:50If M=10 kip⋅ft, determine the maximum bending stress in the beam - 6-50
If M=10 kip⋅ft, determine the maximum bending stress in the beam - 6-50
zhlédnutí 34Před 14 dny
6-50. If M=10 kip⋅ft, determine the maximum bending stress in the beam. Sketch a three-dimensional view of the stress distribution acting over the cross section. All problems from Mechanics of Materials Eleventh Edition by Russell C. Hibbeler.
The assembly is made of A-36 steel. Determine the angle of twist at C - 5-555:32The assembly is made of A-36 steel. Determine the angle of twist at C - 5-55
The assembly is made of A-36 steel. Determine the angle of twist at C - 5-55
zhlédnutí 39Před 14 dny
5-55. The assembly is made of A-36 steel and consists of a solid rod 20 mm in diameter fixed to the inside of a tube using a rigid disk at B. Determine the angle of twist at C. The tube has an outer diameter of 40 mm and wall thickness of 5 mm. All problems from Mechanics of Materials Eleventh Edition by Russell C. Hibbeler.
The assembly is made of A-36 steel. Determine the angle of twist at D - 5547:24The assembly is made of A-36 steel. Determine the angle of twist at D - 554
The assembly is made of A-36 steel. Determine the angle of twist at D - 554
zhlédnutí 51Před 14 dny
5-54. The assembly is made of A-36 steel and consists of a solid rod 20 mm in diameter fixed to the inside of a tube using a rigid disk at B. Determine the angle of twist at D. The tube has an outer diameter of 40 mm and wall thickness of 5 mm. All problems from Mechanics of Materials Eleventh Edition by Russell C. Hibbeler.
If the tube is subjected to a uniform distributed torque of t = 1.2 kN⋅m/m - 5-243:54If the tube is subjected to a uniform distributed torque of t = 1.2 kN⋅m/m - 5-24
If the tube is subjected to a uniform distributed torque of t = 1.2 kN⋅m/m - 5-24
zhlédnutí 43Před 14 dny
*5-24. If the tube is subjected to a uniform distributed torque of t0 = 1.2 kN⋅m/m, determine the shear stress developed at point C. The inner radius is 20 mm and the outer radius is 30 mm. All problems from Mechanics of Materials Eleventh Edition by Russell C. Hibbeler.
The solid shaft is fixed to the support at C and subjected to the torsional loadings - 5-94:43The solid shaft is fixed to the support at C and subjected to the torsional loadings - 5-9
The solid shaft is fixed to the support at C and subjected to the torsional loadings - 5-9
zhlédnutí 88Před 14 dny
5-9. The solid shaft is fixed to the support at C and subjected to the torsional loadings. Determine the shear stress at points A and B, and sketch the shear stress on volume elements located at these points. All problems from Mechanics of Materials Eleventh Edition by Russell C. Hibbeler.
A machinist is using a wrench to loosen a nut. The wrench is 25.0 cm long - 10.73:18A machinist is using a wrench to loosen a nut. The wrench is 25.0 cm long - 10.7
A machinist is using a wrench to loosen a nut. The wrench is 25.0 cm long - 10.7
zhlédnutí 65Před 21 dnem
10.7 • A machinist is using a wrench to loosen a nut. The wrench is 25.0 cm long, and he exerts a 17.0 N force at the end of the handle at 37° with the handle (Fig. E10.7). (a) What torque does the machinist exert about the center of the nut? (b) What is the maximum torque he could exert with a force of this magnitude, and how should the force be oriented? Problem from University Physics with M...
Three forces are applied to a wheel of radius 0.350 m, as shown in Fig. E10.4 - 10.44:51Three forces are applied to a wheel of radius 0.350 m, as shown in Fig. E10.4 - 10.4
Three forces are applied to a wheel of radius 0.350 m, as shown in Fig. E10.4 - 10.4
zhlédnutí 80Před 21 dnem
10.4 • Three forces are applied to a wheel of radius 0.350 m, as shown in Fig. E10.4. One force is perpendicular to the rim, one is tangent to it, and the other one makes a 40.0° angle with the radius. What is the net torque on the wheel due to these three forces for an axis perpendicular to the wheel and passing through its center? Problem from University Physics with Modern Physics
A thin, light string is wrapped around the outer rim of a uniform hollow cylinder of mass - 10.2910:45A thin, light string is wrapped around the outer rim of a uniform hollow cylinder of mass - 10.29
A thin, light string is wrapped around the outer rim of a uniform hollow cylinder of mass - 10.29
zhlédnutí 11Před 21 dnem
10.29 •• A thin, light string is wrapped around the outer rim of a uniform hollow cylinder of mass 4.75 kg having inner and outer radii as shown in Fig. E10.29. The cylinder is then released from rest. (a) How far must the cylinder fall before its center is moving at 6.66 m/s? (b) If you just dropped this cylinder without any string, how fast would its center be moving when it had fallen the di...
A 15.0 kg stone slides down a snow-covered hill, leaving point A at a speed of 10.0 m/s - 7.4517:01A 15.0 kg stone slides down a snow-covered hill, leaving point A at a speed of 10.0 m/s - 7.45
A 15.0 kg stone slides down a snow-covered hill, leaving point A at a speed of 10.0 m/s - 7.45
zhlédnutí 101Před 21 dnem
7.45 •• A 15.0 kg stone slides down a snow-covered hill (Fig. P7.45), leaving point A at a speed of 10.0 m/s. There is no friction on the hill between points A and B, but there is friction on the level ground at the bottom of the hill, between B and the wall. After entering the rough horizontal region, the stone travels 100 m and then runs into a very long, light spring with force constant 2.00...
A block with mass 0.50 kg is forced against a horizontal spring of negligible mass - 7.96:40A block with mass 0.50 kg is forced against a horizontal spring of negligible mass - 7.9
A block with mass 0.50 kg is forced against a horizontal spring of negligible mass - 7.9
zhlédnutí 48Před 21 dnem
7.39 • A block with mass 0.50 kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.20 m (Fig. P7.39). When released, the block moves on a horizontal tabletop for 1.00 m before coming to rest. The force constant k is 100 N/m. What is the coefficient of kinetic friction μk between the block and the tabletop? Problem from University Physics with Moder...
A small rock with mass 0.20 kg is released from rest at point A, which is at the top edge - 7.912:15A small rock with mass 0.20 kg is released from rest at point A, which is at the top edge - 7.9
A small rock with mass 0.20 kg is released from rest at point A, which is at the top edge - 7.9
zhlédnutí 71Před 21 dnem
7.9 •• CP A small rock with mass 0.20 kg is released from rest at point A, which is at the top edge of a large, hemispherical bowl with radius R=0.50 m (Fig. E7.9). Assume that the size of the rock is small compared to R, so that the rock can be treated as a particle, and assume that the rock slides rather than rolls. The work done by friction on the rock when it moves from point A to point B a...
Initially the 6.00 kg block is moving downward and the 8.00 kg block is moving to the right - 6-818:05Initially the 6.00 kg block is moving downward and the 8.00 kg block is moving to the right - 6-81
Initially the 6.00 kg block is moving downward and the 8.00 kg block is moving to the right - 6-81
zhlédnutí 561Před 3 měsíci
6.81 •• Consider the system shown in Fig. P6.81. The rope and pulley have negligible mass, and the pulley is frictionless. Initially the 6.00 kg block is moving downward and the 8.00 kg block is moving to the right, both with a speed of 0.900 m/s. The blocks come to rest after moving 2.00 m. Use the work-energy theorem to calculate the coefficient of kinetic friction between the 8.00 kg block a...
A 5.00 kg block is moving at υ0=6.00 m/s along a frictionless, horizontal surface - 6-796:22A 5.00 kg block is moving at υ0=6.00 m/s along a frictionless, horizontal surface - 6-79
A 5.00 kg block is moving at υ0=6.00 m/s along a frictionless, horizontal surface - 6-79
zhlédnutí 418Před 3 měsíci
6.79 • A 5.00 kg block is moving at υ0=6.00 m/s along a frictionless, horizontal surface toward a spring with force constant k=500 N/m that is attached to a wall (Fig. P6.79). The spring has negligible mass. (a) Find the maximum distance the spring will be compressed. (b) If the spring is to compress by no more than 0.150 m, what should be the maximum value of vo? Problem from University Physic...
Determine the largest intensity of distributed load w0 that the beam can support - 6-8212:00Determine the largest intensity of distributed load w0 that the beam can support - 6-82
Determine the largest intensity of distributed load w0 that the beam can support - 6-82
zhlédnutí 582Před 3 měsíci
6-82. Determine the largest intensity of distributed load w0 that the beam can support so that the maximum bending stress in the beam does not exceed σmax=22 ksi. All problems from Mechanics of Materials Eleventh Edition by Russell C. Hibbeler.
The man has a mass of 78 kg and stands motionless at the end of the diving board - 6-7114:59The man has a mass of 78 kg and stands motionless at the end of the diving board - 6-71
The man has a mass of 78 kg and stands motionless at the end of the diving board - 6-71
zhlédnutí 308Před 3 měsíci
The man has a mass of 78 kg and stands motionless at the end of the diving board - 6-71
The beam is subjected to a moment of M=40 kN⋅m. Determine the bending stress - 6-624:29The beam is subjected to a moment of M=40 kN⋅m. Determine the bending stress - 6-62
The beam is subjected to a moment of M=40 kN⋅m. Determine the bending stress - 6-62
zhlédnutí 364Před 3 měsíci
The beam is subjected to a moment of M=40 kN⋅m. Determine the bending stress - 6-62
Determine the absolute maximum shear stress developed in the shaft - 5-134:09Determine the absolute maximum shear stress developed in the shaft - 5-13
Determine the absolute maximum shear stress developed in the shaft - 5-13
zhlédnutí 312Před 3 měsíci
Determine the absolute maximum shear stress developed in the shaft - 5-13
Determine the maximum shear stress developed in regions CD and DE of the shaft - 5-124:33Determine the maximum shear stress developed in regions CD and DE of the shaft - 5-12
Determine the maximum shear stress developed in regions CD and DE of the shaft - 5-12
zhlédnutí 119Před 3 měsíci
Determine the maximum shear stress developed in regions CD and DE of the shaft - 5-12
The copper pipe has an outer diameter of 40 mm and an inner diameter of 37 mm - 5-44:28The copper pipe has an outer diameter of 40 mm and an inner diameter of 37 mm - 5-4
The copper pipe has an outer diameter of 40 mm and an inner diameter of 37 mm - 5-4
zhlédnutí 491Před 3 měsíci
The copper pipe has an outer diameter of 40 mm and an inner diameter of 37 mm - 5-4
The 0.5-kg ball strikes the rough ground and rebounds with the velocities shown - F15-12:59The 0.5-kg ball strikes the rough ground and rebounds with the velocities shown - F15-1
The 0.5-kg ball strikes the rough ground and rebounds with the velocities shown - F15-1
zhlédnutí 732Před 3 měsíci
The 0.5-kg ball strikes the rough ground and rebounds with the velocities shown - F15-1
A 2.5-kg block is given an initial velocity of 3 m/s up a 45° smooth slope - 15-23:12A 2.5-kg block is given an initial velocity of 3 m/s up a 45° smooth slope - 15-2
A 2.5-kg block is given an initial velocity of 3 m/s up a 45° smooth slope - 15-2
zhlédnutí 102Před 3 měsíci
A 2.5-kg block is given an initial velocity of 3 m/s up a 45° smooth slope - 15-2
The composite beam is made of steel (A) bonded to brass (B) and has the cross section shown - 6-12012:12The composite beam is made of steel (A) bonded to brass (B) and has the cross section shown - 6-120
The composite beam is made of steel (A) bonded to brass (B) and has the cross section shown - 6-120
zhlédnutí 728Před 3 měsíci
The composite beam is made of steel (A) bonded to brass (B) and has the cross section shown - 6-120
The 2-kg block B and 15-kg cylinder A are connected to a light cord that passes through - 13-584:10The 2-kg block B and 15-kg cylinder A are connected to a light cord that passes through - 13-58
The 2-kg block B and 15-kg cylinder A are connected to a light cord that passes through - 13-58
zhlédnutí 365Před 3 měsíci
The 2-kg block B and 15-kg cylinder A are connected to a light cord that passes through - 13-58
If the motor exerts a force of F = (600 + 2s^2) N on the cable, determine the speed - F14-35:00If the motor exerts a force of F = (600 + 2s^2) N on the cable, determine the speed - F14-3
If the motor exerts a force of F = (600 + 2s^2) N on the cable, determine the speed - F14-3
zhlédnutí 409Před 3 měsíci
If the motor exerts a force of F = (600 2s^2) N on the cable, determine the speed - F14-3
The spring is placed between the wall and the 10-kg block - F14-13:18The spring is placed between the wall and the 10-kg block - F14-1
The spring is placed between the wall and the 10-kg block - F14-1
zhlédnutí 599Před 3 měsíci
The spring is placed between the wall and the 10-kg block - F14-1
Drawing Shear and Moment Diagrams | Shear and Moment Equations | Engineering Statics39:27Drawing Shear and Moment Diagrams | Shear and Moment Equations | Engineering Statics
Drawing Shear and Moment Diagrams | Shear and Moment Equations | Engineering Statics
zhlédnutí 149Před 3 měsíci
Drawing Shear and Moment Diagrams | Shear and Moment Equations | Engineering Statics
Draw the shear and moment diagrams for the beam ABC if it supports a load of 800 lb - 6-4113:51Draw the shear and moment diagrams for the beam ABC if it supports a load of 800 lb - 6-41
Draw the shear and moment diagrams for the beam ABC if it supports a load of 800 lb - 6-41
zhlédnutí 122Před 3 měsíci
Draw the shear and moment diagrams for the beam ABC if it supports a load of 800 lb - 6-41
The 50-mm-diameter A992 steel shaft is subjected to the torques shown - 5-534:19The 50-mm-diameter A992 steel shaft is subjected to the torques shown - 5-53
The 50-mm-diameter A992 steel shaft is subjected to the torques shown - 5-53
zhlédnutí 593Před 3 měsíci
The 50-mm-diameter A992 steel shaft is subjected to the torques shown - 5-53
The splined ends and gears attached to the A992 steel shaft are subjected to the torques - 5-495:25The splined ends and gears attached to the A992 steel shaft are subjected to the torques - 5-49
The splined ends and gears attached to the A992 steel shaft are subjected to the torques - 5-49
zhlédnutí 584Před 3 měsíci
The splined ends and gears attached to the A992 steel shaft are subjected to the torques - 5-49

Komentáře

  • @abaki4115
    @abaki4115 Před 4 dny

    is this guy autistic or something I cant learn from a retard

  • @joeyheffron992
    @joeyheffron992 Před 6 dny

    absolute lifesaver. Swear youre just like two semesters ahead of me and youre honestly saving my ass man.

  • @blood5289
    @blood5289 Před 7 dny

    Thank you for this video! I know it's a little old but could you explain why the area of the concrete, A_c, is not 8(pi/4)? There are 8 rods so I would assume the cross-sectional area of the rods would be 8 times the area of a single rod. Thanks!

    • @calcgang
      @calcgang Před 6 dny

      I believe that the 8 rods was factored in some other equation.

  • @YahirMisaelPalmaReyes
    @YahirMisaelPalmaReyes Před 8 dny

    hello, which book are these exercises from?

  • @craig7878
    @craig7878 Před 9 dny

    Hi, Could you please clarify the negative sign for FAB? Are you interpreting it as an indication that the initial assumption about FAB being in tension was wrong, or does the negative sign indicate compression? I’m confused because I’ve been taught different interpretations. When solving for reactions or external loads, a negative sign usually means the initial assumption was wrong and we need to swap it. However, for internal forces like FAB or FAC, I’ve been told a negative sign indicates compression. In this question, it seems we are treating FAB as an internal force, yet you assumed all forces to be in tension. This implies that a negative sign would mean to swap the assumption, but I thought that rule only applied to reactions or external loads, not internal forces. Could you please clarify this further?

  • @user-pj3jp2lx5g
    @user-pj3jp2lx5g Před 10 dny

    can you explain to me why in f), Q is replaced by -Q. I am confused at that point

  • @madriagajanjosephg.8047
    @madriagajanjosephg.8047 Před 13 dny

    how does the water have buoyancy if the cube is below it? shouldn't it be pushing down?

  • @ahmadalhamad4965
    @ahmadalhamad4965 Před 14 dny

    I want to know why you took one side and left the other

    • @calcgang
      @calcgang Před 14 dny

      The beam is symmetrical, so it won’t make a difference

    • @ahmadalhamad4965
      @ahmadalhamad4965 Před 14 dny

      @@calcgang Now, in the case of symmetry, D equals zero, so it remains Ix Was he affected or not? (ixc+Ad2)

  • @user-hg5jv9nq9f
    @user-hg5jv9nq9f Před 14 dny

    THANK YOU SO MUCH ❤

    • @calcgang
      @calcgang Před 14 dny

      You’re welcome ☺️

  • @juancarlossanchezveana1812
    @juancarlossanchezveana1812 Před 16 dny

    Excellent 👌🏻

  • @NoahBabineau
    @NoahBabineau Před 17 dny

    what would I do without this channel?

  • @user-kx2tr4uz3s
    @user-kx2tr4uz3s Před 17 dny

    Thanks a lot.

  • @governmentjob927
    @governmentjob927 Před 17 dny

    thanks for your effort.

  • @SamuelPeeps
    @SamuelPeeps Před 18 dny

    thanks!

  • @ValidatingUsername
    @ValidatingUsername Před 18 dny

    First one must suspend disbelief that the torsional loads are applied without any contact and at an infinitesimal region at the locations depicted uniformly around the shaft

  • @nahomtesfaye5315
    @nahomtesfaye5315 Před 20 dny

    thanks it was helpfull

  • @gabrielamaldonado-isaacs8810
    @gabrielamaldonado-isaacs8810 Před 22 dny

    Thank you so much!! You're really helping me get through physics : )

  • @danielsalka3415
    @danielsalka3415 Před 23 dny

    Thanks bro

  • @herrybasanez8658
    @herrybasanez8658 Před 24 dny

    My values are like different from yours but it is just 10-15 Pa short is it still correct?

    • @calcgang
      @calcgang Před 21 dnem

      Probably 🤷‍♂️

  • @leandroc4119
    @leandroc4119 Před 24 dny

    this came out in an exam

  • @leandroc4119
    @leandroc4119 Před 24 dny

    why the intervals of the integral are 0 and 1?

    • @calcgang
      @calcgang Před 21 dnem

      Cause that’s what t goes from

  • @aryanraj-bf9cm
    @aryanraj-bf9cm Před 25 dny

    😅

  • @caliboymojo
    @caliboymojo Před 26 dny

    nice

  • @MansoSesay-le4pr
    @MansoSesay-le4pr Před 27 dny

    I like the steps

  • @MansoSesay-le4pr
    @MansoSesay-le4pr Před 27 dny

    ❤

  • @Rewguz
    @Rewguz Před 27 dny

    Gracias wey

  • @philodeus7639
    @philodeus7639 Před 29 dny

    cheers goat definitely gna pass my exam tmrw now

    • @calcgang
      @calcgang Před 21 dnem

      Let’s goooo

    • @philodeus7639
      @philodeus7639 Před 21 dnem

      @@calcgang did pass thanks again u were a lifesaver fr, kinda got trolled in the exam when they wrote that the mass center was already calculated so it threw me off balance ahaha

  • @NoahBabineau
    @NoahBabineau Před 29 dny

    incredibly clutch thank you

  • @alambanda9259
    @alambanda9259 Před měsícem

    Ure an expert... This question in the exam

  • @NotRealLifeBrand
    @NotRealLifeBrand Před měsícem

    be careful, part b is asking for 1/8(9.81) divided by the radius, not diameter. so instead of divided by 2.5 it would b 1.25

    • @calcgang
      @calcgang Před 21 dnem

      My bad, thanks for the clarification

  • @vkozyfire1514
    @vkozyfire1514 Před měsícem

    Tysm for the video

  • @chanverse7555
    @chanverse7555 Před měsícem

    Nice

  • @vin9522
    @vin9522 Před měsícem

    I have a test next Monday, you've really just shown how much my lecturer sucks 💯

    • @calcgang
      @calcgang Před 21 dnem

      Thank you so much! I’m glad I could help you

  • @vin9522
    @vin9522 Před měsícem

    Mr Goated 💯

  • @joshuamoore5792
    @joshuamoore5792 Před měsícem

    Thanks a lot, great video!

  • @khiziiikhan
    @khiziiikhan Před měsícem

    The direction of friction will be opposite to the friction direction in wedge

  • @kaan3665
    @kaan3665 Před měsícem

    Hi from Turkey. Thanks for video

    • @calcgang
      @calcgang Před 21 dnem

      Hello 👋 you’re welcome!

  • @mjvisualz
    @mjvisualz Před měsícem

    teacher didnt kinda skipped the algebra on it and im glad i found this vid

  • @melissafigiel655
    @melissafigiel655 Před měsícem

    could you explain why we need to find it in the xy plane first? i cant visualize it

    • @calcgang
      @calcgang Před 21 dnem

      You can start in any plane

  • @JoseLopez-vt5uo
    @JoseLopez-vt5uo Před měsícem

    Thank you so much

  • @DomoJelly
    @DomoJelly Před měsícem

    Literal calc saint. Studying for a final and have been so stumped on this shiiiiiiz

    • @calcgang
      @calcgang Před 21 dnem

      Let’s goooo I hope your final went well!

    • @DomoJelly
      @DomoJelly Před 21 dnem

      @@calcgang I got an A!!!! Thank you so much for the help and you have the most luscious hair I have ever seen dude

  • @kasra-ir5io
    @kasra-ir5io Před měsícem

    thanks a lot❤

  • @user-rp1ph8eq6n
    @user-rp1ph8eq6n Před měsícem

    You are the GOAT

  • @abdallahgoogle3445
    @abdallahgoogle3445 Před měsícem

    ❤❤

  • @sanchitjoshi6661
    @sanchitjoshi6661 Před měsícem

    Great video, this helped me with my exam

  • @lamug
    @lamug Před měsícem

    Muchas gracias! muy claro!

    • @calcgang
      @calcgang Před 21 dnem

      De nada y buena suerte!

  • @ahmedalshsikhi3569
    @ahmedalshsikhi3569 Před měsícem

    *Great explanation*

  • @ezzealdeen77
    @ezzealdeen77 Před měsícem

    Thanks a lot

  • @Studymotivation-cd4ul
    @Studymotivation-cd4ul Před měsícem

    Indian🇮🇳 students vote here❤

    • @calcgang
      @calcgang Před 21 dnem

      Much love from America!

  • @ammarcuthbert3791
    @ammarcuthbert3791 Před měsícem

    Hi Grantstrong when u do pie/2-1.567=3.86 not 3.33?