**PROBLEM 5.1 **

*For the beam and loading shown, (a) draw the shear and bending-moment *
*diagrams, (b) determine the equations of the shear and bending-moment *
curves.
**SOLUTION **
Reactions:
0: 0
*C*
*Pb*
*M* *LA* *bP* *A*
*L*
Σ = − = =
0: 0
*A*
*Pa*
*M* *LC* *aP* *C*
*L*
Σ = − = =
*From A to B: * 0< < *x* *a*
0: 0
*y*
*Pb*
*F* *V*
*L*
Σ = − =
*Pb*
*V*
*L*
=
0: 0
*J*
*Pb*
*M* *M* *x*
*L*
Σ = − _{= }*Pbx*
*M*
*L*
= _{ }*From B to C: * *a* < < *x* *L*
0: 0
*y*
*Pa*
*F* *V*
*L*
Σ = + =
*Pa*
*V*
*L*
= −
0: ( ) 0
*K*
*Pa*
*M* *M* *L* *x*
*L*
Σ = − + − ** _{= }**
( )

*Pa L*

*x*

*M*

*L*− =

_{ }*At section B:*2

*Pab*

*M*

*L*=

**PROBLEM 5.2 **

*For the beam and loading shown, (a) draw the shear and bending-moment *
*diagrams, (b) determine the equations of the shear and bending-moment *
curves.
**SOLUTION **
Reactions:
** ** *B* 0: _{2} 0 _{2}
*L* *wL*
*M* *AL* *wL* *A*
Σ = − + ⋅ = =
** ** *A* 0: _{2} 0 _{2}
*L* *wL*
*M* *BL* *wL* *B*
Σ = − ⋅ = =

Free body diagram for determining reactions.
Over whole beam, 0 < < *x* *L*

*Place section at x. *

Replace distributed load by equivalent concentrated load.

0: 0
2
*y*
*wL*
*F* *wx* *V*
Σ = − − =
2
*L*
*V* = *w*_{} −*x*_{}
0: 0
2 2
*J*
*wL* *x*
*M* *x* *wx* *M*
Σ = − + + ** _{= }**
2
( )
2

*w*

*M*=

*Lx*−

*x*( ) 2

*w*

*M*=

*x L*

**− **

*x*Maximum bending moment occurs at .

2
*L*
*x* =
2
max
8
*wL*
*M* =

**PROBLEM 5.15 **

For the beam and loading shown, determine the maximum
*normal stress due to bending on a transverse section at C. *

**SOLUTION **
*Reaction at A: *

0:*M _{B}* = −4.5

*A*+(3.0)(3)+(1.5)(3) +(1.8)(4.5)(2.25) = 0

*=7.05 kN ↑*

**A***Use AC as free body.*

3
0: (7.05)(1.5) (1.8)(1.5)(0.75) 0
8.55 kN m 8.55 10 N m
*C* *C*
*C*
*M* *M*
*M*
Σ = − + =
= ⋅ = × ⋅
3 3 6 4
6 4
1 1
(80)(300) 180 10 mm
12 12
180 10 m
1
(300) 150 mm 0.150 m
2
*I* *bh*
*c*
−
= = = ×
= ×
= = =
3
6
6
(8.55 10 )(0.150)
7.125 10 Pa
180 10
*Mc*
*I*
σ = = × _{−} = ×
× 7.13 σ = MPa

F
n
**SOLUTION **
U
U

**PROBLEM 5.20 **

For the beam and loading shown, determine the max
*normal stress due to bending on a transverse section at C*

Use entire beam as free body.
0 :
*B*
*M*
_{=}
4.8*A* (3.6)(216) (1.6)(150) (0.8)(150) 0
− + + + =
237 kN
=
**A**

*Use portion AC as free body. *
0 :
*C*
*M*
_{=}
(2.4)(237) (1.2)(216) 0
309.6 kN m
*M*
*M*
− + =
= ⋅
For W460×113, *S* = 2390×10 mm6 3
Normal stress:
3
6 3
6
309.6 10 N m
2390 10 m
129.5 10 Pa
*M*
*S*
σ = = × _{−} ⋅
×
= ×
129.5
σ = M
ximum
*C. *
MPa

**P**

D
an
d
**SOLUTION**

**P**

**ROBLEM 5.22 **

Draw the shear and bending-moment diagrams for the
nd loading shown and determine the maximum normal
ue to bending.
Reactions:
0 : 4 64 (24)(2)(1) 0 28
*D*
*M* *A*
_{=} _{−} _{−} _{=} **A**_{=}
0 : 28 (24)(2) 0 76 k
*y*
*F* *D*
= − + − = * D* =

*A to C:*0 < <

*x*2m 0 : 28 0 28 kN

*y*

*F*

*V*

*V*

_{=}

_{− −}

_{=}= − 0 : 28 0 ( 28 ) kN m

*J*

*M*

*M*

*x*

*M*

*x* = + = = − ⋅

*C to D:*2m < <

*x*4m 0 : 28 0 2

*y*

*F*

*V*

*V*

_{=}

_{− −}

_{=}= − 0 : 28 64 ( 28 64) kN m

*J*

*M*

*M*

*x*

*M*

*x*

_{=}

_{+}

_{−}= − + ⋅

*D to B:*4m < <

*x*6m 0 : 24(6 ) 0 ( 24 144) kN

*y*

*F*

*V*

*x*

*V*

*x* = − − = = − + 2 0 : 6 24(6 ) 2 12(6 ) kN m

*J*

*M*

*x*

*M*

*x*

*M*

*x* = − − −

_{− } = − − ⋅ 3 max

*M*=56 kN m ⋅ =56×10 N m⋅ For S250×52 section,

*S*= 482×10 mm3 3 Normal Stress: 3 6 3 56 10 N m 116.2 10 482 10 m

*M*

*S*σ = = ×

_{−}⋅ = × × 116.2 MPσ = beam stress 8 kN kN N 28 kN 4 0 m = 0 m = 6 0 Pa Pa

**PROBLEM 5.42 **

Using the method of Sec. 5.3, Solve Prob. 5.9

**PROBLEM 5.9 **Draw the shear and bending-moment diagrams for the
beam and loading shown, and determine the maximum absolute value
*(a) of the shear, (b) of the bending moment. *

**SOLUTION **
Reactions:
0 : 2 (12)(2)(1) (40)(1) 0
8 kN
*C*
*M* *A*
= + − =
= ↓
* A*
0 : 2 (12)(2)(1) (40)(3) 0
72 kN

*A*

*M*

*C*

_{=}

_{−}

_{−}

_{=}= ↑

*Shear diagram:*

**C***V*= −8 kN

_{A}*A to C: 0*< <

*x*2 m

*w*=12kN/m 2 2 0 012 24 kN 24 8 32 kN

*C*

*A*

*C*

*V*

*V*

*wdx*

*dx*

*V*− = − = − = − = − − = −

##

##

*C to B: 32V*= − +72 = 40 kN Areas of shear diagram:

_{B}*A to C: * 1( 8 32)(2) 40 kN m
2
*Vdx*
_{=} _{− −} _{= −} _{⋅ }
*C to B: * *Vdx*=(1)(40)=40 kN m⋅
Bending moments:
0
0 40 40 kN m
40 40 0
*A*
*C* *A*
*B* *C*
*M*
*M* *M* *Vdx*
*M* *M* *Vdx*
=
= + = − = − ⋅
= + = − + =
*(a) Maximum V* = 40.0 kN
*(b) Maximum * *M* = 40.0 kN m⋅

**PROBLEM 5.93 **

*Beams AB, BC, and CD have the cross section shown *
*and are pin-connected at B and C. Knowing that the *
allowable normal stress is +110 MPa in tension and
*−150 MPa in compression, determine (a) the largest *
**permissible value of P if beam BC is not to be ***overstressed, (b) the corresponding maximum *
*distance a for which the cantilever beams AB and CD *
are not overstressed.

**SOLUTION **
0
*B* *C*
*B* *C*
*M* *M*
*V* *V* *P*
= =
= − =

*Area B to E of shear diagram: 2.4 P *

0 2.4 P 2.4 P

*E* *F*

*M* = + = =*M*

Centroid and moment of inertia:

Part *A*(mm )2 *y*(mm) *Ay*(mm )3 *d*(mm) *Ad*2(mm )4 *I* (mm )4
2500 156.25 390625 34.82 6
3.031 10× 0.0326 10× 6
1875 75 140625 46.43 4.042 10× 6 3.516 10× 6
Σ 4375 531250 7.073 10× 6 3.548 10× 6
2 6 4
531250
121.43 mm
4375
10.621 10 mm
*Y*
*I* *Ad* *I*
= =
= Σ + Σ = ×
Location *y*(mm) *I y*/ (10 mm )3 3 ← also (10−6m )3
Top 41.07 258.6
Bottom −121.43 −87.47

**PROBLEM 5.93 (Continued) **

Bending moment limits: *M* = −σ*I y*/

Tension at *E and F: * −(110 10 ) ( 87.47 10 )× 6 − × −6 =9.622 10 N m× 3 ⋅
Compression at *E and F: * − −( 150 10 )(258.6 10 )× 6 × −6 =38.8 10 N m× 3 ⋅
Tension at *A and D: * −(110 10 ) (258.6 10 )× 6 × −6 = −28.45 10 N m× 3 ⋅

Compression at *A and D: * − −( 150 10 )( 87.47 10 )× 6 − × −6 = −13.121 10 N m× 3 ⋅

*(a) Allowable *load *P: * 2.4 P=9.622 10× 3 *P*=4.01 10 N× 3 *P*=4.01 kN
Shear at *A: * *V _{A}*=

*P*

Area *A to B of shear diagram: * *aV _{A}*=

*aP*

*Bending moment at A: * *MA* = −*aP* = −4.01 10× 3*a*

*(b) Distance a: * −4.01 10× 3*a*= −13.121 10× 3 3.27 *a*= m

**PROBLEM 5.120 **

Using a computer and step functions, calculate the shear and bending
*moment for the beam and loading shown. Use the specified increment ΔL, *
*starting at point A and ending at the right-hand support. *

**SOLUTION **
1
0: 6 (4)(120) (1) (3)(36) 0
2
*D* *A*
*M* *R*
Σ = − + + _{ } =
89 kN
*A*
*R* =
1 1
36
3 12 3
3
*w*= − = − *x* *x*
0 2
89 120 2 6 3 kN
*V* = − − − − *x* *x*
1 3
89 120 2 2 3 kN m
*M* = *x*− − − − *x* *x* ⋅
*x V * *M *
m kN kN ⋅ m
0.0 89.0 0.0
0.3 89.0 22.3
0.5 89.0 44.5
0.8 89.0 66.8
1.0 89.0 89.0
1.3 89.0 111.3
1.5 89.0 133.5
1.8 89.0 155.8
2.0 −31.0 178.0
2.3 −31.0 170.3
2.5 −31.0 162.5
2.8 −31.0 154.8
3.0 −31.0 147.0
3.3 −31.4 139.2
3.5 −32.5 131.3
3.8 −34.4 122.9
4.0 −37.0 114.0
4.3 −40.4 104.3
4.5 −44.5 93.8
4.8 −49.4 82.0
*x V * *M *
m kN kN ⋅ m
5.0 −55.0 69.0
5.3 −61.4 54.5
5.5 −68.5 38.3
5.8 −76.4 20.2
6.0 −85.0 −0.0

**PROBLEM 5.121 **

Using a computer and step functions, calculate the shear and bending
moment for the beam and loading shown. Use the specified increment
*ΔL, starting at point A and ending at the right-hand support. *

**SOLUTION **
0: (5.2)(12) 4 (2)(4)(16) 0
*C*
*M* *B*
Σ = − + =
47.6 kN
= ↑
* B*
0: (1.2)(12) (2)(4)(16) 4 0

*B*

*M*

*C*Σ = − + = 28.4 kN = ↑

*0 16 1.2*

**C***dV*

*w*

*x*

*dx*= − = − 1 0 16 1.2 12 47.6 1.2

*V*= − − − +

*x* −

*x* 2 1 8 1.2 12 47.6 1.2

*M*= − − −

*x*

*x*+ −

*x*

*x V*

*M*m kN kN ⋅ m 0.0 −12.0 0.00 0.4

_{−12.0 }

_{−4.80 }0.8 −12.0 −9.60 1.2 35.6 −14.40 1.6 29.2 −1.44 2.0 22.8 8.96 2.4 16.4 16.80 2.8 10.0 22.08 3.2 3.6 24.80 3.6 −2.8 24.96 4.0 −9.2 22.56 4.4

_{−15.6 }17.60 4.8 −22.0 10.08 5.2 −28.4 −0.00