Internal Loadings Practice Problems with Solutions (20 Worked Examples)

This page collects 20 original internal loadings practice problems with complete worked solutions. Internal loadings are the normal force, shear force, bending moment and torque that a member carries at a cut section, and they are the first step in every mechanics of materials problem on stress. Each solution below is typed step by step, with the free-body diagram described in words, the equilibrium equations written out and the answer checked a second way. For the next step, see combined axial loading and bending.

The method of sections for internal loadings

1. Find the support reactions for the whole body from equilibrium, if the cut will not leave a free end.
2. Pass an imaginary cut through the member at the point of interest, perpendicular to its axis.
3. Draw a free-body diagram of either piece. Show the unknown internal normal force N, shear force V, bending moment M and torque T at the cut, in their positive directions.
4. Apply the equilibrium equations (ΣF = 0 and ΣM = 0) to the piece to solve for N, V, M and T.
5. Check by repeating the calculation on the other piece. The two answers must agree.

Sign convention used in these solutions

• Normal force N is positive in tension and negative in compression.
• Shear force V is positive when it pushes the left face of the cut upward (a clockwise couple on the piece).
• Bending moment M is positive for sagging, which puts the top fibres in compression.
• Torque T is positive by the right-hand rule with the thumb pointing away from the cut surface.

Internal loadings formula cheat sheet

Quantity	Equation	Notes
Normal force	N = ΣF along the member axis	Tension positive
Shear force	V = ΣF perpendicular to the axis	Left piece: V = ΣFy up
Bending moment	M = ΣM about the cut centroid	Left piece: M = Σ F(x − xi) + couples
Torque	T = ΣM about the shaft axis	Gear torques from T = P / ω
Power and torque	T = P / ω, with ω = 2πn / 60	P in watts, n in rpm, T in N·m
Distributed load	Equivalent force = w × length	Acts at the centroid of the load
Shear–moment link	dV/dx = −w and dM/dx = V	Moment is largest where V = 0

Free-body diagrams and support reactions

Every internal loading problem starts with a correct free-body diagram. These problems practice replacing supports with reactions and using moment and force equilibrium to find them.

Problem 1: Mobile crane wheel reactions and tipping load

Problem. A mobile crane rests on a front axle F and a rear axle R that are 4 ft apart. The crane weighs 6,000 lb with its center of gravity midway between the axles. A ballast weight of 2,000 lb acts 1 ft behind the rear axle, and a load of 1,200 lb hangs from the boom 9 ft ahead of the front axle. Draw the free-body diagram, find the vertical reactions at F and R, and find the largest hanging load before the crane tips forward.

Solution.

1. Draw the crane as a horizontal bar with upward reactions F and R at the axles and downward forces at their lines of action. Measure x from the rear axle toward the front: R at x = 0, F at x = 4 ft, crane weight at x = 2 ft, ballast at x = −1 ft, hanging load at x = 13 ft.
2. Sum moments about R: F(4) − 6,000(2) − 1,200(13) + 2,000(1) = 0, so F = 6,400 lb.
3. Sum vertical forces: R + F = 6,000 + 2,000 + 1,200 = 9,200 lb, so R = 2,800 lb.
4. At the point of tipping the rear wheel just lifts off, so R = 0 and F = 6,000 + 2,000 + L. Moments about R give (8,000 + L)(4) = 6,000(2) + L(13) − 2,000(1).
5. Solving, L = 2,444 lb.

Answer. F = 6,400 lb, R = 2,800 lb (both upward). The crane tips forward when the hanging load exceeds about 2,444 lb.

Problem 2: Support reactions of a simply supported beam

Problem. A simply supported beam AB spans 8 m with a pin at A and a roller at B. It carries a 20 kN point load 2 m from A and a uniformly distributed load of 6 kN/m between x = 5 m and x = 8 m, measured from A. Find the reactions.

Solution.

1. Replace the distributed load by its resultant: 6 × 3 = 18 kN acting at the middle of the loaded length, x = 6.5 m.
2. Sum moments about A: R_B(8) − 20(2) − 18(6.5) = 0, so R_B = 19.625 kN.
3. Sum vertical forces: R_A + R_B = 20 + 18 = 38 kN, so R_A = 18.375 kN.
4. Check by taking moments about B: R_A(8) should equal 20(6) + 18(1.5) = 147.0 kN·m, and 147.0 matches.

Answer. R_A = 18.38 kN up, R_B = 19.62 kN up.

Problem 3: Reactions of an overhanging beam

Problem. A beam ABC is pinned at A (x = 0) and supported by a roller at B (x = 6 m). It extends to a free end C at x = 8 m. It carries a uniform load of 10 kN/m over its full length and a 15 kN downward point load at C. Find the reactions at A and B.

Solution.

1. Resultant of the distributed load: 10 × 8 = 80 kN at x = 4 m.
2. Moments about A: R_B(6) − 80(4) − 15(8) = 0, so R_B = 73.33 kN.
3. Vertical equilibrium: R_A = 80 + 15 − 73.33 = 21.67 kN.

Answer. R_A = 21.67 kN up, R_B = 73.33 kN up. Because the overhang loads the beam beyond B, the reaction at B is much larger than at A.

Internal axial force in bars and columns

Axial force is found by cutting the member and balancing all the load on one side. Weight that is spread along the member makes the force change with position.

Problem 4: Column with two weights per length

Problem. A vertical column supports a 40 kN load at its top. The upper 3 m of the column weighs 1.2 kN/m and the lower 4 m weighs 2.0 kN/m. Find the internal normal force at a section 1 m below the top, at a section 2 m below the upper/lower junction, and at the base.

Solution.

1. Cut the column horizontally and keep the piece above the cut. The internal normal force must balance the load on top plus the weight of the piece, so N is compressive.
2. Section 1 m below the top: N = 40 + 1.2(1) = 41.2 kN.
3. Weight of the whole upper segment: 1.2(3) = 3.6 kN, so just below the junction N = 40 + 3.6 = 43.6 kN.
4. Section 2 m below the junction: N = 43.6 + 2.0(2) = 47.6 kN.
5. At the base: N = 43.6 + 2.0(4) = 51.6 kN.

Answer. N = 41.2 kN, 47.6 kN and 51.6 kN, all compressive.

Problem 5: Steel bar hanging from a ceiling

Problem. A steel bar of constant cross-section A = 400 mm² and length 6 m hangs from a ceiling and carries a 2,000 N load at its lower end. Steel density is 7,850 kg/m³. Find the internal normal force at the ceiling, at mid-length, and 1.5 m above the lower end.

Solution.

1. Weight per unit length: w = ρAg = 7,850(400×10⁻⁶)(9.81) = 30.80 N/m.
2. Cut the bar and keep the lower piece. Equilibrium requires the internal force to support the end load plus the weight of the piece below the cut, so the bar is in tension.
3. At the ceiling: N = 2,000 + 30.80(6) = 2,184.8 N.
4. At mid-length (3 m of bar below the cut): N = 2,000 + 30.80(3) = 2,092.4 N.
5. 1.5 m above the lower end: N = 2,000 + 30.80(1.5) = 2,046.2 N.

Answer. N = 2,185 N (ceiling), 2,092 N (mid-length) and 2,046 N (1.5 m above the end), all tension. The self-weight adds only about 9.2% to the end load.

Problem 6: Bar with three axial loads

Problem. A straight bar ABCD is fixed to a wall at A. Axial loads act along the bar at B, C and D: 30 kN at B pulling away from the wall, 20 kN at C pushing toward the wall, and 10 kN at D pulling away from the wall. Find the internal normal force in segments AB, BC and CD.

Solution.

1. Take the direction away from the wall as positive. Cutting a segment and keeping the free (right-hand) piece avoids having to find the wall reaction first.
2. Segment CD: only the 10 kN load at D lies to the right of the cut, so N = +10 kN (tension).
3. Segment BC: the loads to the right are −20 kN at C and +10 kN at D, so N = −20 + 10 = −10 kN, which is 10 kN of compression.
4. Segment AB: all three loads lie to the right, so N = 30 − 20 + 10 = +20 kN (tension).
5. Check: the wall reaction must equal −20 kN (toward the wall), which matches N in AB.

Answer. N_CD = 10 kN (T), N_BC = 10 kN (C), N_AB = 20 kN (T).

Internal torque in shafts

Torque is the internal loading that twists a shaft. Cut between the gears or pulleys and balance the moments about the shaft axis.

Problem 7: Shaft with four gears

Problem. A shaft ABCD turns freely in bearings. Gear A is driven by a motor with a torque of 800 N·m. Gears B, C and D take off torques of 300 N·m, 350 N·m and 150 N·m. Find the internal torque in segments AB, BC and CD.

Solution.

1. Check equilibrium of the whole shaft: 800 − 300 − 350 − 150 = 0, so the shaft turns at constant speed.
2. Cut the segment and sum torques about the shaft axis on the piece to the left, treating the input torque as positive.
3. Segment AB: T = 800 N·m.
4. Segment BC: T = 800 − 300 = 500 N·m.
5. Segment CD: T = 800 − 300 − 350 = 150 N·m.
6. Check from the right: only the 150 N·m at D lies to the right of a cut in CD, so T = 150 N·m. This matches.

Answer. T_AB = 800 N·m, T_BC = 500 N·m, T_CD = 150 N·m. The largest torque is in AB, so that segment governs the shaft design.

Problem 8: Shaft with a distributed torque

Problem. A shaft of length 3 m is fixed at wall A. A uniformly distributed torque of 400 N·m per metre acts along its full length, and a concentrated torque of 600 N·m acts at the free end B. Find the internal torque at a section 1 m from the free end and the reaction torque at the wall.

Solution.

1. Let x be the distance from the free end. The piece between the free end and the section carries the 600 N·m end torque plus 400x of distributed torque.
2. Internal torque: T(x) = 600 + 400x.
3. At x = 1 m: T = 600 + 400(1) = 1,000 N·m.
4. At the wall, x = 3 m: T = 600 + 400(3) = 1,800 N·m, which must equal the wall reaction torque.

Answer. T(1 m) = 1,000 N·m. The wall reaction torque is 1,800 N·m.

Problem 9: Torque from motor power

Problem. A motor at A delivers 30 kW to a shaft turning at 1,200 rpm. Gear B takes off 12 kW and gear C takes off the remaining 18 kW. Find the internal torque in segments AB and BC.

Solution.

1. Angular speed: ω = 2πn/60 = 2π(1,200)/60 = 125.66 rad/s.
2. Torque from power, T = P/ω. Motor torque: T_A = 30,000 / 125.66 = 238.7 N·m.
3. Gear B takes T_B = 12,000 / 125.66 = 95.5 N·m, and gear C takes T_C = 18,000 / 125.66 = 143.2 N·m.
4. Segment AB carries the full input torque: T = 238.7 N·m.
5. Segment BC carries what remains after gear B: T = 238.7 − 95.5 = 143.2 N·m, which equals T_C.

Answer. T_AB = 238.7 N·m and T_BC = 143.2 N·m.

Internal shear and bending moment in beams

Cut the beam at the position you want, keep either piece, and balance forces and moments. Always verify with the opposite piece.

Problem 10: Shear and moment in a simply supported beam

Problem. A simply supported beam spans 10 m and carries a 24 kN point load 4 m from the left support A. Find the internal shear force and bending moment at a section 2 m from A and at a section 6 m from A.

Solution.

1. Reactions: R_B = 24(4)/10 = 9.6 kN and R_A = 24 − 9.6 = 14.4 kN.
2. Section at x = 2 m (left of the load). Keep the left piece: V = R_A = 14.4 kN and M = R_A(2) = 28.8 kN·m.
3. Section at x = 6 m (right of the load). Keep the left piece: V = 14.4 − 24 = −9.6 kN and M = 14.4(6) − 24(2) = 38.4 kN·m.
4. Check with the right piece: V = −R_B = −9.6 kN and M = R_B(4) = 38.4 kN·m.

Answer. At 2 m: V = 14.4 kN, M = 28.8 kN·m. At 6 m: V = −9.6 kN, M = 38.4 kN·m. Sign convention: positive shear pushes the left piece up at the cut, and positive moment causes compression in the top fibres.

Problem 11: Beam with a partial distributed load

Problem. A simply supported beam spans 8 m. A uniform load of 12 kN/m acts on the left 4 m. Find V and M at x = 3 m and x = 6 m from the left support A.

Solution.

1. Resultant of the load: 12 × 4 = 48 kN at x = 2 m.
2. Moments about A: R_B(8) = 48(2), so R_B = 12.0 kN and R_A = 48 − 12.0 = 36.0 kN.
3. x = 3 m: V = 36.0 − 12(3) = 0.0 kN and M = 36.0(3) − 12(3)²/2 = 54.0 kN·m. Zero shear means this is the maximum moment.
4. x = 6 m: the whole load is to the left, so V = 36.0 − 48 = −12.0 kN and M = 36.0(6) − 48(6 − 2) = 24.0 kN·m.
5. Check with the right piece: M = R_B(2) = 24.0 kN·m.

Answer. x = 3 m: V = 0.0 kN, M = 54.0 kN·m. x = 6 m: V = −12.0 kN, M = 24.0 kN·m.

Problem 12: Cantilever beam with distributed and point loads

Problem. A cantilever beam is fixed at A and extends 5 m to a free end B. It carries a uniform load of 8 kN/m over its full length and a 15 kN downward point load at B. Find the shear force and bending moment at a section 2 m from the fixed end, and the reactions at A.

Solution.

1. Keep the free piece to the right of the section, which has length 3 m, so the wall reactions are not needed.
2. Shear: V = 15 + 8(3) = 39.0 kN.
3. Moment: M = −[15(3) + 8(3)²/2] = −[45 + 36] = −81.0 kN·m. The negative sign means the top fibres are in tension.
4. Wall reactions: V_A = 8(5) + 15 = 55.0 kN up, and M_A = 15(5) + 8(5)²/2 = 175.0 kN·m (hogging).
5. Check from the wall: V = 55.0 − 8(2) = 39.0 kN, and M = −175.0 + 55.0(2) − 8(2)²/2 = −81.0 kN·m.

Answer. V = 39.0 kN and M = −81.0 kN·m at 2 m from the wall. At the wall, V_A = 55.0 kN and M_A = 175.0 kN·m (hogging).

Problem 13: Beam with an applied couple

Problem. A simply supported beam of span 6 m carries a clockwise couple of 18 kN·m applied at midspan. Find the shear force and bending moment at x = 1.5 m and x = 4.5 m from the left support.

Solution.

1. A pure couple has no vertical force. Moments about A: R_B(6) − 18 = 0, so R_B = 3.0 kN up, and then R_A = −3.0 kN, which is 3.0 kN downward.
2. x = 1.5 m: V = −3.0 kN and M = R_A(1.5) = −4.5 kN·m.
3. x = 4.5 m: the couple is now on the left piece and adds +18 kN·m to the moment, so M = −3.0(4.5) + 18 = 4.5 kN·m, while V = −3.0 kN.
4. Check with the right piece at x = 4.5 m: M = R_B(1.5) = 4.5 kN·m. This matches.

Answer. x = 1.5 m: V = −3.0 kN, M = −4.5 kN·m. x = 4.5 m: V = −3.0 kN, M = 4.5 kN·m. The moment jumps by 18 kN·m across the couple, while the shear does not change.

Problem 14: Beam with an inclined load

Problem. A beam is pinned at A and supported by a roller at B, 6 m from A. A 20 kN force acts at a point 2 m from A, directed 60° below the horizontal and pointing toward B. Find N, V and M at x = 1 m and x = 4 m.

Solution.

1. Components of the load: F_x = 20cos60° = 10.00 kN toward B and F_y = 20sin60° = 17.32 kN downward.
2. The roller has no horizontal reaction, so the pin takes it all: A_x = 10.00 kN directed away from B.
3. Moments about A: R_B(6) = 17.32(2), so R_B = 5.774 kN and R_A = 17.32 − 5.774 = 11.547 kN.
4. x = 1 m (left of the load): N = 10.00 kN tension (the pin pulls the left piece away from the cut), V = 11.547 kN, M = 11.547 kN·m.
5. x = 4 m (right of the load): the horizontal components of A_x and F_x cancel, so N = 0. V = 11.547 − 17.32 = −5.774 kN and M = 11.547 kN·m.

Answer. x = 1 m: N = 10.00 kN (T), V = 11.55 kN, M = 11.55 kN·m. x = 4 m: N = 0, V = −5.77 kN, M = 11.55 kN·m.

Internal loadings in frames and machine members

Here the cut member is inclined or bent, so all three loadings N, V and M appear together. Resolve every force along and across the section.

Problem 15: L-shaped bracket with an inclined force

Problem. An L-shaped bracket has a vertical leg and a horizontal arm. The arm extends 0.30 m to the right from the top of the leg, D, to the loaded end B. A 150 N force acts at B, 30° from the vertical, pointing down and to the right. Section A passes horizontally through the vertical leg 0.20 m below D. Find the resultant internal loading at A.

Solution.

1. Components of the force at B: F_x = 150sin30° = 75.0 N to the right and F_y = 150cos30° = 129.9 N downward.
2. Cut the leg at A and keep the upper piece (leg above A, the arm and the load). On the cut face draw an axial force N, a shear force V and a moment M.
3. Vertical equilibrium: N = 129.9 N, and the leg is compressed because it holds up the load.
4. Horizontal equilibrium: V = 75.0 N, acting to the left on the upper piece.
5. Moments about A: the vertical component acts 0.30 m to the right of A and the horizontal component acts 0.20 m above A. M = 129.9(0.30) + 75.0(0.20) = 53.97 N·m.

Answer. N = 129.9 N (compression), V = 75.0 N, M = 54.0 N·m.

Problem 16: Locking pliers gripping a bolt

Problem. A pair of pliers pivots about pin A. A hand force of 50 N acts perpendicular to the handle 140 mm from A, and the jaws grip a bolt 35 mm from A on the opposite side of the pin. Find the gripping force, and the internal loadings at section C in the handle 70 mm from A and at section D in the jaw 20 mm from A.

Solution.

1. Moments about the pin: F_jaw(35) = 50(140), so F_jaw = 200 N.
2. Section C: keep the handle piece beyond C, which is 70 mm long and carries the hand force. N = 0, V = 50 N, and M = 50(70) = 3,500 N·mm = 3.50 N·m.
3. Section D: keep the jaw piece beyond D, which is 15 mm long and carries the jaw force. N = 0, V = 200 N, and M = 200(15) = 3,000 N·mm = 3.00 N·m.
4. Pin force check: 50 + 200 = 250 N, so the pin carries about five times the hand force.

Answer. Gripping force = 200 N. Section C: V = 50 N, M = 3.50 N·m. Section D: V = 200 N, M = 3.00 N·m.

Problem 17: Beam held by a cable

Problem. A horizontal beam AB is 4 m long and pinned to a wall at A. A cable runs from B to a point D on the wall 3 m directly above A. A load P acts downward at the midpoint C of the beam. The cable can carry at most 6 kN. Find the largest P, and then N, V and M at C.

Solution.

1. The cable is 5 m long, so its horizontal and vertical components are T(0.8) and T(0.6).
2. Moments about A: T(0.6)(4) = P(2). With T = 6 kN, P = 7.2 kN.
3. Wall reactions: A_x = T cosθ = 4.8 kN (toward the beam) and A_y = P − T sinθ = 7.2 − 3.6 = 3.6 kN.
4. Keep the piece AC. The pin pushes it with 4.8 kN horizontally, so N = 4.8 kN in compression.
5. Vertical equilibrium gives V = A_y = 3.6 kN, and moments about C give M = A_y(2) = 7.2 kN·m.

Answer. P_max = 7.2 kN. At C: N = 4.8 kN (compression), V = 3.6 kN, M = 7.2 kN·m.

Problem 18: Jib crane internal loadings

Problem. A jib crane has a horizontal boom 10 ft long welded to the top of a vertical column 12 ft tall. The boom weighs 50 lb/ft and the column weighs 60 lb/ft. A hoist and load of 800 lb hang from the free end of the boom. Find the internal loadings at section A in the boom 4 ft from the free end, and at section C in the column 5 ft above the ground.

Solution.

1. Section A: keep the 4 ft tip piece of the boom. Its weight is 50(4) = 200 lb, acting 2 ft from the cut.
2. N_A = 0, V_A = 800 + 200 = 1,000 lb, and M_A = 800(4) + 200(2) = 3,600 lb·ft.
3. Section C: keep everything above C. The boom weighs 50(10) = 500 lb acting 5 ft from the column, and the column above C is 7 ft long and weighs 60(7) = 420 lb.
4. Axial force: N_C = 800 + 500 + 420 = 1,720 lb, compressive.
5. Shear: V_C = 0 because every load is vertical and no horizontal force acts on the column. Moment about the centroid of section C: M_C = 800(10) + 500(5) = 10,500 lb·ft.

Answer. Section A: N = 0, V = 1,000 lb, M = 3,600 lb·ft. Section C: N = 1,720 lb (compression), V = 0, M = 10,500 lb·ft.

Problem 19: Gear tooth at its root

Problem. A gear tooth is loaded by a 900 N force that makes an angle of 20° with the direction perpendicular to the tooth centerline, as shown by the pressure angle. The force acts at the tooth tip, 14 mm from the root section a–a, and its line passes through the centerline. Find the resultant internal loading at the centroid of section a–a.

Solution.

1. Resolve the force along and across the tooth centerline: axial component F_a = 900sin20° = 307.8 N and transverse component F_t = 900cos20° = 845.7 N.
2. Cut the tooth at the root and keep the tip piece. The axial component pushes along the tooth, so N = 307.8 N in compression.
3. The transverse component must be balanced by shear: V = 845.7 N.
4. Moment about the centroid of the root section: M = F_th = 845.7(14 mm) = 11,840 N·mm = 11.84 N·m.

Answer. N = 307.8 N (compression), V = 845.7 N, M = 11.84 N·m. The moment is the largest contributor to root stress, which is why tooth root bending governs gear design.

Problem 20: C-clamp frame

Problem. A C-clamp squeezes a workpiece with a force of 1,200 N. The line of the clamping force is 80 mm from the centroid of the frame cross-section at section a–a, which is perpendicular to the frame’s back. Find the internal loading at section a–a.

Solution.

1. Cut the frame at a–a and keep the upper jaw. The clamping force of 1,200 N acts parallel to the back of the frame and 80 mm from the centroid of the section.
2. Axial force: N = 1,200 N in tension, because the frame is being pulled open.
3. There is no force perpendicular to the section, so V = 0.
4. Moment about the centroid: M = 1,200(80/1000) = 96 N·m.
5. Both N and M put the inside of the C in tension, so the stress there is the sum of the axial and bending terms. See combined loading for that step.

Answer. N = 1,200 N (tension), V = 0, M = 96 N·m.

Internal loadings FAQ

What are internal loadings in mechanics of materials?

Internal loadings are the resultant normal force, shear force, bending moment and torque that act on a cut cross-section of a member. They balance the external loads on the piece that was removed.

How do you find internal loadings with the method of sections?

Cut the member at the section of interest, draw a free-body diagram of one piece with the unknown N, V, M and T shown, and apply the six equations of equilibrium to that piece. Check the answer using the other piece.

What is the sign convention for shear force and bending moment?

A common convention takes shear as positive when it pushes the left face of the cut upward, and bending moment as positive when it causes sagging, with compression in the top fibres. Always state your convention before solving.

Where is the bending moment maximum?

The bending moment is largest where the shear force crosses zero, because dM/dx = V. It can also peak at a point where a concentrated couple is applied.

How do you find torque in a shaft with several gears?

Cut the shaft between each pair of gears and balance moments about the shaft axis. For a motor, the torque follows from T = P / ω, where ω = 2πn / 60 is the angular speed in rad/s.

What do you do with the internal loadings after you find them?

They feed the stress formulas: σ = N/A for axial force, τ = VQ/It for shear, σ = My/I for bending and τ = Tρ/J for torsion. When several act together, superpose the stresses.

Keep learning

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• Understanding moment of inertia
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• Uniaxial tension test
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• Fatigue analysis and S-N curves